Display device

Abstract

A display device includes a display panel having a display region overlapping a plurality of pixels arranged in a matrix shape in a plan view, and having a first side and a second side opposite to the first side; a light source device disposed on the first side, which emits light incident on the display panel from the first side; and a drive circuit which calculates an output gradation value based on an input gradation value possessed by an image signal, and outputs a pixel drive signal corresponding to the output gradation value to a corresponding one of the plurality of pixels, the plurality of pixels including a first pixel and a second pixel arranged in a first direction from the first side toward the second side, the second pixel being located between the first pixel and the second side, and the drive circuit making the output gradation value corresponding to the second pixel greater than the output gradation value corresponding to the first pixel when the input gradation value corresponding to the first pixel and the input gradation value corresponding to the second pixel are equal.

Description

Technical Field

[0001] This application relates to a display device. Background Technology

[0002] Patent Document 1 discloses a display device configured to visually confirm the background on one side of a display panel from one side of the panel. The display device in Patent Document 1 is a so-called see-through display, comprising a display panel having a liquid crystal layer including a polymer-dispersed liquid crystal, and a light source disposed opposite to the side of the display panel.

[0003] Existing technical documents

[0004] Patent documents

[0005] Patent Document 1: Japanese Patent Application Publication No. 2020-160254 Summary of the Invention

[0006] The technical problem that the invention aims to solve

[0007] In the display device of Patent Document 1, light from a light source (light source device) enters from a first side of the display panel and propagates within the display panel towards a second side opposite to the first side. Switching elements and electrodes are disposed within the display panel, and a portion of the light from the light source device propagating within the display panel is consumed by these elements. Therefore, as the light from the light source device propagates within the display panel, the light intensity decreases. Consequently, the light intensity of the light source device on the second side of the display panel is less than that on the first side. Consequently, in this case, when multiple grayscale values ​​corresponding to multiple pixels are equal, the brightness of the pixels on the second side becomes lower than the brightness of the pixels on the first side. Therefore, the desired pixel brightness may not be obtained on the second side of the display panel.

[0008] The purpose of this application is to achieve uniform brightness of pixels when multiple grayscale values ​​corresponding to multiple pixels are equal in a display device with a display panel having a light source device that is incident on the side.

[0009] Solutions for solving technical problems

[0010] The display device of this application comprises: a display panel having a display area overlapping a plurality of pixels arranged in a matrix when viewed from above, and having a first side and a second side opposite to the first side; a light source device disposed on the first side and emitting light incident from the first side to the second side; and a driving circuit that calculates an output gray value (output gray level) based on an input gray value (input gray level) of an image signal, and outputs a pixel driving signal corresponding to the output gray value to a corresponding pixel among the plurality of pixels, the plurality of pixels including a first pixel and a second pixel arranged along a first direction from the first side toward the second side, the second pixel being located between the first pixel and the second side, wherein when the input gray value corresponding to the first pixel and the input gray value corresponding to the second pixel are equal, the driving circuit causes the output gray value corresponding to the second pixel to be greater than the output gray value corresponding to the first pixel.

[0011] Furthermore, the display device of this application comprises: a display panel having a first side surface; a plurality of light emitters facing the first side surface; a first pixel disposed on the display panel; a second pixel disposed on the display panel, located on the side opposite to the first side surface, separated from the first pixel; and a driving circuit for receiving image signals having a first input grayscale value corresponding to the first pixel and a second input grayscale value corresponding to the second pixel. The driving circuit outputs a first pixel driving signal having a first output grayscale value corresponding to the first input grayscale value to the first pixel, and outputs a second pixel driving signal having a second output grayscale value corresponding to the second input grayscale value to the second pixel. When the first input grayscale value and the second input grayscale value are equal, the second output grayscale value is greater than the first output grayscale value.

[0012] Furthermore, the display device of this application includes: a display panel having a liquid crystal layer including a polymer-dispersed liquid crystal, a first substrate and a second substrate separated by the liquid crystal layer, and a third substrate, the second substrate being located between the first substrate and the third substrate; a plurality of light emitters facing a side surface of the third substrate; a first pixel disposed on the display panel; a second pixel disposed on the display panel at a position arranged with the first pixel in a direction intersecting the side surface; and a driving circuit for inputting an image signal having a first input grayscale value corresponding to the first pixel and a second input grayscale value corresponding to the second pixel, wherein light emitted from the plurality of light emitters is incident on the third substrate from the side surface and exits from the main surface of the third substrate, the driving circuit outputs a first pixel driving signal having a first output grayscale value corresponding to the first input grayscale value to the first pixel, and outputs a second pixel driving signal having a second output grayscale value corresponding to the second input grayscale value to the second pixel, wherein when the first input grayscale value and the second input grayscale value are equal, the second output grayscale value is greater than the first output grayscale value. Attached Figure Description

[0013] Figure 1 This is a diagram illustrating the structure of a display device according to an embodiment of this application.

[0014] Figure 2 This is a top view of the display device.

[0015] Figure 3 This is a diagram showing the circuit structure of the display device.

[0016] Figure 4 This is a cross-sectional view of the display panel.

[0017] Figure 5 This is a magnified cross-sectional view of the display panel.

[0018] Figure 6 This diagram illustrates the operation of the first and second drive circuits when an image is displayed on the display panel.

[0019] Figure 7 It is a top view schematically showing the propagation of light from a luminous body.

[0020] Figure 8 This is a schematic diagram of the display area.

[0021] Figure 9 This is a schematic diagram showing the display area of ​​light incident on a light source of a pixel and light emitted from a light source of a pixel when viewed from above.

[0022] Figure 10This is a flowchart executed when the first driving circuit calculates the output grayscale value.

[0023] Figure 11 It is a graph showing the total amount of incident light in each of multiple pixels.

[0024] Figure 12 It is a graph showing the output grayscale values ​​of multiple pixels.

[0025] Figure 13 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is 0° (n=0).

[0026] Figure 14 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is +10° (n=+1).

[0027] Figure 15 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is -10° (n = -1).

[0028] Figure 16 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is +20° (n=+2).

[0029] Figure 17 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is -20° (n = -2).

[0030] Figure 18 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is +30° (n=+3).

[0031] Figure 19 This is a diagram showing the amount of incident light and outgoing light in each pixel of the first row, and the amount of incident light in each pixel of the second row, when the tilt angle is -30° (n = -3).

[0032] Figure 20 This is a diagram showing the pixels incident on by outgoing light that is tilted in a direction that is positive (+) relative to the Y direction.

[0033] Figure 21 This is a diagram showing the incident pixel P of outgoing light that is tilted in a direction that is negative (-) relative to the Y direction.

[0034] Figure 22 This is a diagram showing the amount of incident light and outgoing light in each of the pixels in the second row, and the amount of incident light in each of the pixels in the third row, when the tilt angle is 0° (n=0).

[0035] Figure 23 This is a diagram showing the amount of incident light and outgoing light in each pixel of the second row, and the amount of incident light in each pixel of the third row, when the tilt angle is +10° (n=+1).

[0036] Figure 24 This is a diagram showing the amount of incident light and outgoing light in each of the pixels in the second row, and the amount of incident light in each of the pixels in the third row, when the tilt angle is -10° (n = -1).

[0037] Figure 25 This is a diagram showing the amount of incident light and outgoing light in each pixel of the second row, and the amount of incident light in each pixel of the third row, when the tilt angle is +20° (n=+2).

[0038] Figure 26 This is a diagram showing the amount of incident light and outgoing light in each pixel of the second row, and the amount of incident light in each pixel of the third row, when the tilt angle is -20° (n = -2).

[0039] Figure 27 This is a diagram showing the amount of incident light and outgoing light in each pixel of the second row, and the amount of incident light in each pixel of the third row, when the tilt angle is +30° (n=+3).

[0040] Figure 28 This is a diagram showing the amount of incident light and outgoing light in each of the pixels in the second row, and the amount of incident light in each of the pixels in the third row, when the tilt angle is -30° (n = -3).

[0041] Figure 29 This illustrates a case where, in a display device according to an embodiment of this application, the output grayscale value is calculated by a first driving circuit using a correction coefficient. Figure 8 , Figure 11 as well as Figure 12 The graph shows the brightness and grayscale values ​​of pixel P in columns i+3 and i+4. Detailed Implementation

[0042] Hereinafter, various embodiments of this application will be described with reference to the accompanying drawings. This application is not limited to the content described in the following embodiments. In addition, the constituent elements described below include elements that can be easily conceived by those skilled in the art, and substantially the same elements. Furthermore, the constituent elements described below can be appropriately combined.

[0043] Furthermore, this application is merely an example, and appropriate modifications that can be readily conceived by those skilled in the art while maintaining the spirit of this application are of course included within the scope of this application. Additionally, to make the description clearer, the drawings sometimes schematically represent the width, thickness, shape, etc., of various parts compared to the actual form; however, these are merely examples and do not limit the interpretation of this application. Furthermore, in this specification and the various figures, the same reference numerals are used for the same elements as described above for figures that have already appeared, and detailed descriptions are sometimes appropriately omitted.

[0044] The X and Y directions shown in the attached figures correspond to directions parallel to the surface of the substrate included in the display device 1. The +X and -X sides of the X direction, and the +Y and -Y sides of the Y direction correspond to the sides of the display device 1. The Z direction corresponds to the thickness direction of the display device 1, the +Z side of the Z direction corresponds to the front side of the display device 1 where the image is displayed, and the -Z side of the Z direction corresponds to the back side of the display device 1. Furthermore, in this specification, "top view" refers to viewing the display device 1 along the Z direction from the +Z side towards the -Z side. Additionally, the X direction corresponds to a third direction, and the Y direction corresponds to a first direction. The X, Y, and Z directions are merely examples, and this application is not limited to these directions.

[0045] Figure 1 This is a diagram showing the structure of the display device 1 according to an embodiment of this application. Figure 2 This is a top view of the display device 1. The display device 1 displays images based on image signals output from an external device (not shown) electrically connected via a first flexible wiring substrate 2a. The display device 1 includes a display panel 10, a light source device 20, a first driving circuit 30, and a second driving circuit 40.

[0046] Display panel 10 is a so-called see-through display (transparent display). In display panel 10, the background on the other side (e.g., back side 10b) can be visually viewed from one side of the display panel 10 (e.g., front side 10a). Display panel 10 has a display area DA on the panel (front side 10a) for displaying images. Display panel 10 includes a first substrate 11, a second substrate 12, a liquid crystal layer 13, a first substrate 14, and a second substrate 15 (equivalent to a "third substrate").

[0047] The first substrate 11 and the second substrate 12 are rectangular in shape when viewed from above and are transparent. The first substrate 11 and the second substrate 12 are made of resin such as polyethylene terephthalate or glass. The first substrate 11 has an exposed portion E that protrudes from the second substrate 12 when viewed from above. The first substrate 11 is disposed on the back side 12b of the second substrate 12. The front side 11a of the first substrate 11 and the back side 12b of the second substrate 12 face each other. A liquid crystal layer 13 is disposed between the first substrate 11 and the second substrate 12.

[0048] like Figure 2 As shown, the display area DA overlaps with multiple pixels P arranged in a matrix along the X and Y directions when viewed from above. Pixel P appears square when viewed from above. Details of the liquid crystal layer 13 and pixels P will be described later.

[0049] Figure 1 as well as Figure 2 The first substrate 14 and the second substrate 15 shown protect the first substrate 11, the second substrate 12, and the liquid crystal layer 13. The first substrate 14 and the second substrate 15 are rectangular in shape when viewed from above and are translucent. The first substrate 14 and the second substrate 15 are made of, for example, glass or resin. The first substrate 14 is attached to the back side 11b of the first substrate 11 via a first adhesive portion 16. The second substrate 15 is attached to the front side 12a of the second substrate 12 via a second adhesive portion 17. The first adhesive portion 16 and the second adhesive portion 17 are translucent and are formed by curing the adhesive.

[0050] The front side 11a and back side 11b of the first substrate 11, the front side 12a and back side 12b of the second substrate 12, the front side 14a and back side 14b of the first substrate 14, and the front side 15a (equivalent to the "main surface of the third substrate") and back side 15b of the second substrate 15 are all planar and parallel to each other. Furthermore, the front side 15a of the second substrate 15 corresponds to the front side 10a of the display panel 10, and the back side 14b of the first substrate 14 corresponds to the back side 10b of the display panel 10.

[0051] Furthermore, the -X side surfaces of the first substrate 11, second substrate 12, first substrate 14, and second substrate 15, namely the first XL side surface 11c, the second XL side surface 12c, the third XL side surface 14c, and the fourth XL side surface 15c, are all planar and parallel to each other. Furthermore, the +X side surfaces of the first substrate 11, second substrate 12, first substrate 14, and second substrate 15, namely the first XR side surface 11d, the second XR side surface 12d, the third XR side surface 14d, and the fourth XR side surface 15d, are all planar and parallel to each other.

[0052] Furthermore, the -Y side surfaces of the first substrate 11, the second substrate 12, the first substrate 14, and the second substrate 15, namely the first YB side surface 11e, the second YB side surface 12e, the third YB side surface 14e, and the fourth YB side surface 15e, are all planar and parallel to each other. The first YB side surface 11e, the second YB side surface 12e, the third YB side surface 14e, and the fourth YB side surface 15e correspond to the first side surface 10c of the display panel 10.

[0053] Furthermore, the +Y side surfaces of the first substrate 11, the second substrate 12, the first substrate 14, and the second substrate 15, namely the first YF side surface 11f, the second YF side surface 12f, the third YF side surface 14f, and the fourth YF side surface 15f, are all planes and parallel to each other. The first YF side surface 11f, the second YF side surface 12f, the third YF side surface 14f, and the fourth YF side surface 15f correspond to the second side surface 10d of the display panel 10. The second side surface 10d is located on the side opposite to the first side surface 10c.

[0054] The light source device 20 is disposed on the side of the display panel 10. Specifically, the light source device 20 is located on the first side 10c side of the display panel 10, opposite the fourth YB side 15e of the second substrate 15. The light source device 20 emits light incident from the first side 10c side to the second side 10d side (details will be described later). The light source device 20 is fixed to the second substrate 15 via a support 18. The light source device 20 includes a light-emitting part 21 and a light-guiding part 22.

[0055] There are multiple light-emitting units 21 arranged along the X direction. Each light-emitting unit 21 includes a first light-emitting element 21a of a first color, a second light-emitting element 21b of a second color, and a third light-emitting element 21c of a third color. The first, second, and third colors are different from each other; the first color is red, the second color is green, and the third color is blue. That is, the light emitted by the first light-emitting element 21a is red, the light emitted by the second light-emitting element 21b is green, and the light emitted by the third light-emitting element 21c is blue. Hereinafter, without distinguishing between the first light-emitting element 21a, the second light-emitting element 21b, and the third light-emitting element 21c, they will be simply referred to as "light-emitting element SL".

[0056] Furthermore, the number of light-emitting elements SL in the light-emitting unit 21 and the color of the light are not limited to the numbers and colors described above. For example, the number of light-emitting elements SL could be one, and the color of the light-emitting element SL could be white. In this case, a grayscale image or a black and white image would be displayed in the display area DA.

[0057] The light-emitting element SL emits light toward the light-guiding section 22. The light-emitting element SL is, for example, an LED (Light Emitting Diode). The light from the light-emitting element SL is equivalent to the light from the light source device 20.

[0058] The light guide portion 22 is rectangular in shape, having a facing surface 22a opposite to the light emitter SL, and a facing surface 22b opposite to the facing surface 22a and opposite to the fourth YB side surface 15e of the second substrate 15. Furthermore, the light guide portion 22 has a continuous shape from the fourth XL side surface 15c to the fourth XR side surface 15d when viewed from above. The light guide portion 22 is translucent. Light from the light emitter SL enters the light guide portion 22 from the facing surface 22a, diffuses within the light guide portion 22, and then exits from the facing surface 22b toward the fourth YB side surface 15e of the second substrate 15 in a state of uniform light intensity.

[0059] Light from the light-emitting element SL incident from the fourth YB side 15e of the second substrate 15 propagates within the display panel 10 from the first side 10c toward the second side 10d, which is opposite to the first side 10c. Specifically, the light from the light-emitting element SL is reflected within the display panel 10 by the front surfaces 11a, 12a, 14a, 15a and the back surfaces 11b, 12b, 14b, 15b of the first substrate 11, the second substrate 12, the first substrate 14 and the second substrate 15, respectively, and propagates to the second side 10d.

[0060] Figure 3 This is a diagram showing the circuit structure of display device 1. (For example...) Figure 1 as well as Figure 3 As shown, a first driving circuit 30 is disposed on a first substrate 11. The first driving circuit 30 calculates an output grayscale value (described later) based on an image signal transmitted from an external device. The first driving circuit 30 applies voltages to a plurality of pixels P in accordance with the output grayscale value (details are described later). The first driving circuit 30 includes a signal processing circuit 31, a signal output circuit 32, and a scanning circuit 33.

[0061] The signal processing circuit 31 generates multiple pixel driving signals based on the image signal (details are described later) and outputs the multiple pixel driving signals to the signal output circuit 32. In addition, the signal processing circuit 31 outputs a clock signal to the signal output circuit 32 and the scanning circuit 33 to synchronize the operation of the signal output circuit 32 with the operation of the scanning circuit 33.

[0062] The signal output circuit 32 outputs multiple pixel driving signals to their respective pixels P. For example... Figure 3 As shown, the signal output circuit 32 and multiple pixels P are electrically connected via multiple signal lines Lb extending along the Y direction.

[0063] The scanning circuit 33 scans multiple pixels P synchronously with the output of the pixel driving signal based on the signal output circuit 32. The scanning circuit 33 and the multiple pixels P are electrically connected via multiple scan lines Lc extending along the X direction.

[0064] Multiple pixels P have a switching element SW, a pixel electrode PE, a common electrode CE, a liquid crystal capacitor LC, and a holding capacitor KC.

[0065] The switching element SW is, for example, composed of a thin-film transistor (TFT). In the switching element SW, the source electrode is electrically connected to the signal line Lb, and the gate electrode is electrically connected to the scan line Lc.

[0066] The pixel electrode PE is connected to the drain electrode of the switching element SW. Multiple common electrodes CE are configured corresponding to multiple scan lines Lc. Both the pixel electrode PE and the common electrode CE are transparent.

[0067] The liquid crystal capacitor LC is the capacitive component of the liquid crystal material in the liquid crystal layer 13 (described later) located between the pixel electrode PE and the common electrode CE. The holding capacitor KC is positioned between an electrode at the same potential as the common electrode CE and an electrode at the same potential as the pixel electrode PE.

[0068] Figure 4 This is a cross-sectional view of the display panel 10. Furthermore, in Figure 4 The first substrate 14 and the second substrate 15 are omitted. Signal lines Lb (not shown), pixel electrodes PE, and scan lines Lc are disposed on the front side 11a of the first substrate 11 in an electrically insulated state. In the display panel 10, the area divided by two adjacent signal lines Lb and two adjacent scan lines Lc when viewed from above corresponds to a pixel P.

[0069] Furthermore, a first alignment film AL1 is disposed on the front side 11a of the first substrate 11. The alignment direction of the first alignment film AL1 is along the Y direction. A signal line Lb, a pixel electrode PE, and a scan line Lc are disposed between the first substrate 11 and the first alignment film AL1.

[0070] A common electrode CE and a second alignment film AL2 are disposed on the back surface 12b of the second substrate 12. The common electrode CE is disposed between the second substrate 12 and the second alignment film AL2. The alignment direction of the second alignment film AL2 is along the Y direction. That is, the alignment direction of the first alignment film AL1 is parallel to the alignment direction of the second alignment film AL2. Alternatively, the alignment directions of the first alignment film AL1 and the second alignment film AL2 may be orthogonal to each other.

[0071] Figure 5 This is a partially enlarged cross-sectional view of the display panel 10. The liquid crystal layer 13 includes a polymer-dispersed liquid crystal. Specifically, the liquid crystal layer 13 has a three-dimensional mesh-like polymer network 51 and liquid crystal molecules 52.

[0072] The polymer network 51 is formed by polymerizing monomers oriented by the first alignment film AL1 and the second alignment film AL2 using ultraviolet light and heat. Liquid crystal molecules 52 are located in the gaps between the polymer network 51.

[0073] like Figure 1 As shown, the second driving circuit 40 is disposed on the support 18. The second driving circuit 40 drives the light source device 20 based on a light control signal output from an external device (not shown) electrically connected via the second flexible wiring substrate 2b. The light control signal includes information on the amount of light emitted by the light emitter SL (the amount of light emitted by the light emitter) determined based on the image signal.

[0074] Next, the basic operation of the display device 1 when displaying an image on the display panel 10 will be explained. The first driving circuit 30 and the second driving circuit 40 drive the display panel 10 and the light source device 20 in a field-sequence manner.

[0075] First, the case where the image signal and light control signal are not sent to the display device 1 and the display panel 10 does not display an image will be explained. In this case, the first driving circuit 30 does not output a pixel driving signal and does not apply voltage to the pixel electrode PE. In addition, the second driving circuit 40 does not drive the light source device 20 and does not emit light from the light emitter SL.

[0076] Without applying voltage to the pixel electrode PE, such as Figure 5 As shown, the optical axis AX1 of the polymer network 51 and the optical axis AX2 of the liquid crystal molecule 52 are constrained by the orientation directions of the first alignment film AL1 and the second alignment film AL2. In this embodiment, when no voltage is applied to the pixel electrode PE, the optical axis of the polymer network 51 and the optical axis of the liquid crystal molecule 52 are parallel to each other and along the Y direction.

[0077] The refractive index of the polymer network 51 and the refractive index of the liquid crystal molecules 52 are equal. Therefore, when no voltage is applied to the pixel electrode PE, the difference between the refractive index of the polymer network 51 and the refractive index of the liquid crystal molecules 52 is zero in all directions. Consequently, light propagating within the display panel 10 is not scattered. That is, in this case, the liquid crystal layer 13 becomes a transmissive state that prevents light propagating within the display panel 10 from being scattered.

[0078] When the liquid crystal layer 13 is in a transparent state, the background of the other side (e.g., the back side 10b) of the display panel 10 can be visually confirmed from one side (e.g., the front side 10a). Furthermore, when the liquid crystal layer 13 is in a transparent state, the light emitted by the light-emitting element SL propagating within the display panel 10 is almost not scattered. Therefore, even when the light from the light-emitting element SL is emitted based on a light control signal, the background of the other side of the display panel 10 can be visually confirmed from one side of the display panel 10 when the liquid crystal layer 13 is in a transparent state.

[0079] Next, the situation where the image signal and the light control signal are sent to the display device 1 and the display panel 10 displays the image will be explained. First, the state where the first drive circuit 30 outputs a pixel drive signal and applies a voltage to the pixel electrode PE will be explained.

[0080] When a voltage is applied to the pixel electrode PE, the optical axis AX2 of the liquid crystal molecule 52 is tilted relative to the Y direction, corresponding to the magnitude of the voltage. On the other hand, even when a voltage is applied to the pixel electrode PE, the optical axis AX1 of the polymer network 51 is not tilted, but remains along the Y direction. That is, the optical axis AX2 of the liquid crystal molecule 52 is tilted relative to the optical axis AX1 of the polymer network 51.

[0081] This creates a difference between the refractive index of the polymer network 51 and the refractive index of the liquid crystal molecules 52. At this time, if the second driving circuit 40 emits light from the light emitter SL based on the light control signal, the light propagating within the display panel 10 is scattered. That is, in this case, the liquid crystal layer 13 becomes a scattering state that causes the light propagating within the display panel 10 to scatter. The light scattered by the liquid crystal layer 13 is emitted from the front side 10a and the back side 10b of the display panel 10 to the outside of the display panel 10, which can be visually confirmed from both sides of the display panel 10.

[0082] Furthermore, the amount of light scattered in the liquid crystal layer 13 varies depending on the degree of scattering by the liquid crystal layer 13. The degree of scattering by the liquid crystal layer 13 is determined by the slope of the liquid crystal molecules 52, i.e., the magnitude of the voltage applied to the pixel electrode PE. The magnitude of the voltage is determined based on the output grayscale value included in the pixel driving signal. The output grayscale value is determined by the first driving circuit 30 for each group of multiple pixels P based on the input grayscale value of the image signal. The input grayscale value and the output grayscale value are values ​​(grayscale values) indicating the grayscale of pixel P. The input grayscale value and the output grayscale value are determined for the color of each pixel. There are three colors for the pixels, and the color of the light emitter SL corresponds to the color of the pixel.

[0083] The first driving circuit 30 generates a pixel driving signal for each group of pixels P, including an output grayscale value determined based on the input grayscale value, and sends the pixel driving signal to the multiple pixels P. As a result, in each of the multiple pixels P, a voltage corresponding to the output grayscale value is applied to the pixel electrode PE. The liquid crystal molecule 52 corresponding to the pixel P tilts accordingly to the magnitude of the output grayscale value, changing the degree of scattering by the liquid crystal layer 13, thereby changing the amount of light emitted to the outside. The larger the output grayscale value, the larger the voltage applied to the pixel electrode PE, the more light is emitted to the outside, and the higher the brightness of the pixel P visually confirmed from both sides (front 10a and back 10b) of the display panel 10.

[0084] Figure 6 This diagram illustrates the operation of the first drive circuit 30 and the second drive circuit 40 when an image is displayed on the display panel 10. Figure 6 The operation of the first driving circuit 30 and the second driving circuit 40 in each frame F is shown. A frame F has a first subframe SF1, a second subframe SF2 and a third subframe SF3 in sequence.

[0085] In the first subframe SF1, light of the first color (red) included in the image is emitted from pixel P. Specifically, the first driving circuit 30 scans multiple pixels P during the first scan period TS1, selects the pixel P that emits light of the first color, and sends a pixel driving signal corresponding to the first color to the selected pixel P. As a result, the liquid crystal layer 13 corresponding to the selected pixel P becomes a scattering state corresponding to the output grayscale value corresponding to the first color. The voltage applied to the pixel electrode PE is maintained during the first light emission period TL1 and is reset at the end of the first subframe SF1.

[0086] Furthermore, during the first emission period, the second driving circuit 40 causes the first light emitter 21a to emit light via TL1. The light of the first color from the first light emitter 21a propagates within the display panel 10. Consequently, in the liquid crystal layer 13 corresponding to the pixel P selected by the first driving circuit 30, the light of the first color is scattered according to the degree of scattering by the liquid crystal layer 13 and emitted to the outside. That is, light of the first color with a grayscale corresponding to the output grayscale value is emitted from the pixel P selected by the first driving circuit 30.

[0087] In the second subframe SF2, the second color (green) light included in the image is emitted from pixel P. Specifically, the first driving circuit 30 scans multiple pixels P during the second scan period TS2, selects the pixel P that emits the second color light, and sends a pixel driving signal corresponding to the second color to the selected pixel P. As a result, the liquid crystal layer 13 corresponding to the selected pixel P becomes a scattering state corresponding to the output grayscale value corresponding to the second color. The voltage applied to the pixel electrode PE is maintained during the second light emission period TL2 and is reset at the end of the second subframe SF2.

[0088] Furthermore, during the second emission period, the second driving circuit 40 causes the second light emitter 21b to emit light via TL2. The second-color light from the second light emitter 21b propagates within the display panel 10. Consequently, in the liquid crystal layer 13 corresponding to the pixel P selected by the first driving circuit 30, the second-color light is scattered according to the degree of scattering by the liquid crystal layer 13 and emitted to the outside. That is, the pixel P selected by the first driving circuit 30 emits second-color light with a grayscale value corresponding to the output grayscale value.

[0089] In the third subframe SF3, the third color (blue) light included in the image is emitted from pixel P. Specifically, the first driving circuit 30 scans multiple pixels P during the third scan period TS3, selects the pixel P that emits the third color light, and sends a pixel driving signal corresponding to the third color to the selected pixel P. As a result, the liquid crystal layer 13 corresponding to the selected pixel P becomes a scattering state corresponding to the output grayscale value corresponding to the third color. The voltage applied to the pixel electrode PE is maintained during the third emission period TL3 and is reset at the end of the third subframe SF3.

[0090] Furthermore, during the third emission period, the second driving circuit 40 causes the third light emitter 21c to emit light via TL3. The third-color light from the third light emitter 21c propagates within the display panel 10. Consequently, in the liquid crystal layer 13 corresponding to the pixel P selected by the first driving circuit 30, the third-color light is scattered according to the degree of scattering by the liquid crystal layer 13 and emitted to the outside. That is, the pixel P selected by the first driving circuit 30 emits third-color light with a gray level corresponding to the output gray level value.

[0091] The duration of one frame F is defined as the time it takes for the human eye to perceive the light obtained by combining the first, second, and third colors emitted in one frame F. That is, the human eye perceives the color and grayscale of the light obtained by combining the first, second, and third colors. Thus, by emitting the first, second, and third colors from multiple pixels P as described above, the image can be visually confirmed. Furthermore, in this case, when viewing the display area DA from one side of the display panel 10, the background on the other side of the display panel 10 is visually confirmed in a state of overlap with the image.

[0092] In such a display device 1, there is a technical problem as described below. As described above, the light from the light-emitting element SL propagates within the display panel 10 from a first side 10c toward a second side 10d. The display panel 10, as described above, has components such as signal lines Lb, pixel electrodes PE, and switching elements SW, and the light from the light-emitting element SL is scattered due to its illumination by these components. In other words, the light from the light-emitting element SL is consumed by these components.

[0093] Furthermore, when the liquid crystal layer 13 corresponding to any pixel P is in a scattering state, since a portion of the light emitted by the light source SL is emitted outward from pixel P, the amount of light from the light source SL propagating from pixel P to the second side surface 10d is reduced compared to when the liquid crystal layer 13 corresponding to pixel P is in a transmitting state. In other words, the light from the light source SL is consumed by the liquid crystal layer 13 in a scattering state. Thus, if the light from the light source SL is consumed, the amount of light from the light source SL reaching the second side surface 10d is reduced.

[0094] Figure 7 This is a top view schematically illustrating the propagation of light from the light source SL in the display area DA. If the light from the light source SL is consumed as described above, then... Figure 7 As shown, the light intensity of the light-emitting element SL decreases as it moves along the Y direction from the -Y side to the +Y side, that is, from the first side 10c to the second side 10d. In this case, when the display device 1 operates in a manner that makes the entire display area DA display white, that is, in each of the plurality of pixels P, when the first driving circuit 30 sets the output grayscale value corresponding to the first color, the second color, and the third color to be equal to the maximum value, the brightness of the pixel P, that is, the brightness of the display area DA, decreases as it moves from the first side 10c to the second side 10d.

[0095] For example, if the brightness of the pixel P located on the -Y side among the plurality of pixels P is set to 1, the brightness of the pixel P located on the +Y side among the plurality of pixels P is less than 1, for example, 0.6. That is, although the display device 1 operates in a manner that makes the entire display area DA display white, the desired brightness of the pixel P cannot be obtained, and the grayscale on the second side 10d side of the display area DA is reduced, displaying gray.

[0096] Furthermore, sometimes, although the display device 1 operates in a manner that displays a portion of the display area DA in red, the display area DA on the second side 10d side of that portion of the display area DA displays the complementary color of red (cyan (blue-green)). Thus, sometimes, because the color represented by the input grayscale value is different from the color displayed in the display area DA, an image different from the expected image is displayed in the display area DA.

[0097] In order to solve the above-mentioned technical problems, the first driving circuit 30 of this embodiment of the present application calculates the output grayscale value as described below.

[0098] Figure 8This is a schematic diagram of the display area DA. As described above, in the display area DA, multiple pixels P are arranged in a matrix. The number of columns of pixels P arranged along the Y direction and the number of rows of pixels P arranged along the X direction are determined by the size of the display panel 10 and the total number of pixels P. The number of rows of pixels P is a predetermined value (e.g., 1080). The number of columns of pixels P is determined corresponding to the predetermined value; for example, it is 1920 when the number of rows of pixels P is 1080.

[0099] For the sake of simplicity, the variable representing the row number of pixel P will be set to "k", and the row number will increase by 1 successively from the -Y side towards the +Y side. Furthermore, the maximum value of the row number is a predetermined value (e.g., 1080). Additionally, the variable representing the column number of pixel P will be set to "i", and the column number will increase by 1 successively from the -X side towards the +X side.

[0100] exist Figure 8 The image shows a portion of multiple pixels P. Specifically, in Figure 8 The image shows multiple pixels P with 4 rows and 6 columns. Figure 8 The four rows (k=1, 2, 3, 4) of pixels P shown correspond to the pixels P in the first to fourth rows starting from the -Y side (i.e., the first side 10c side). That is, there are no pixels P on the -Y side closer to the pixel P in the first row (k=1), and multiple pixels P are arranged on the +Y side closer to the pixel P in the fourth row (k=4).

[0101] in addition, Figure 8 The pixel P in the 6 columns (i+1, i+2, i+3, i+4, i+5, i+6) shown is located approximately at the center of the X direction among the multiple pixels P arranged in the X direction. That is, multiple pixels P are arranged on the -X side of the pixel P in the i+1 column, and multiple pixels P are also arranged on the +X side of the pixel P in the i+6 column.

[0102] Hereafter, the pixel P in the k-th row and i-th column will be recorded as pixel P(k, i). Additionally, in Figure 8 In the diagram, the value shown within pixel P represents the input grayscale value. In this specification, the minimum value for both the input and output grayscale values ​​is 0 (zero), and the maximum value is 1. Furthermore, Figure 8 The input grayscale values ​​shown correspond to the first color. The calculation of the output grayscale values ​​corresponding to the first color will be explained below. Furthermore, the output grayscale values ​​for the second and third colors are calculated in the same manner as described below. Figure 8 The input grayscale values ​​shown are those after gamma correction.

[0103] For simplicity, in this embodiment, from the first row up to the predetermined row number, the input grayscale values ​​of pixels P in the (i+3)th and (i+4)th columns are all set to 0.8. That is, the grayscale values ​​of pixels P in the (i+3)th and (i+4)th columns of the image signal are equal to each other. Therefore, it is desirable that the brightness of pixels P in the (i+3)th and (i+4)th columns is uniform.

[0104] Additionally, from the first row up to the predetermined row number, the input grayscale values ​​of pixels P in all columns except for the (i+3)th and (i+4)th columns are set to zero. Figure 8 The image shows that, from the first row to the fourth row (k = 1, 2, 3, 4), the input grayscale value of pixel P in column i+3 and column i+4 is 0.8, and the input grayscale value of pixel P in column i+1, column i+2, column i+5, and column i+6 is zero. The following explains... Figure 8 In the 6 columns of pixels P shown, based on Figure 8 The process of calculating the output grayscale value from the input grayscale value is shown.

[0105] Figure 9 This is a schematic diagram showing the display area DA of light incident on a light emitter SL of a pixel P and light emanating from a pixel P, as viewed from above. As described above, the light from the light emitter SL propagates within the display panel 10 from the first side 10c toward the second side 10d. That is, the light from the light emitter SL passes through multiple pixels P from the -Y side toward the +Y side.

[0106] Furthermore, the light from the light-emitting element SL propagates within the display panel 10 along the Y direction (corresponding to the first direction) and in a direction inclined relative to the Y direction (corresponding to the second direction) when viewed from above. That is, the light from the light-emitting element SL is incident on pixel P from the -Y side along the Y direction and in a direction inclined relative to the Y direction when viewed from above. When viewed from above, the light from the light-emitting element SL incident on a pixel P consists of 7 incident light beams J. n (n is one of the seven integers from -3 to +3). Hereinafter, with the Y direction set as the reference orientation and the clockwise direction in the view from above set as the positive (+) direction, the angle between the Y direction and the direction of light travel is called the tilt angle.

[0107] The incident light J0 is tilted at an angle of 0°. That is, the incident light J0 travels along the Y direction. +1 The tilt angle is +10°. Incident light J +2 The tilt angle is +20°. Incident light J +3 The tilt angle is +30°.

[0108] In addition, incident light J -1 The tilt angle is -10°. Incident light J -2The tilt angle is -20°. Incident light J -3 The tilt angle is -30°. The incident light J is tilted relative to the Y direction. -1 J -2 J -3 J +1 J +2 J +3 The direction of travel is equivalent to the second direction.

[0109] Furthermore, when viewed from above, the light emitted from the light source SL, which originates from one pixel P, consists of seven emitted light sources I. n (n is one of the seven integers from -3 to +3) represents this.

[0110] The outgoing light I0 is parallel to the incident light J0. The tilt angle of the outgoing light I0 is 0°. +1 With incident light J +1 Parallel, outgoing light I +1 The tilt angle is +10°. Outgoing beam I +2 With incident light J +2 Parallel, outgoing light I +2 The tilt angle is +20°. Outgoing beam I +3 With incident light J +3 Parallel, outgoing light I +3 The tilt angle is +30°.

[0111] Additionally, emitted light I -1 With incident light J -1 Parallel, outgoing light I -1 The tilt angle is -10°. Outgoing beam I -2 With incident light J -2 Parallel, outgoing light I -2 The tilt angle is -20°. Outgoing beam I -3 With incident light J -3 Parallel, outgoing light I -3 The tilt angle is -30°.

[0112] Incident light J n and emitted light I n The subscript 'n' in the attached figures corresponds to the tilt angle. That is, the tilt angle is -30° when n = -3, -20° when n = -2, -10° when n = -1, and 0° when n = 0. Furthermore, the tilt angle is +10° when n = +1, +20° when n = +2, and +30° when n = +3.

[0113] The first driving circuit 30 uses the aforementioned input grayscale value, as well as the amount of light emitted by the light source SL incident on the pixel P along the Y direction (first direction) and in a direction inclined relative to the Y direction (second direction) (i.e., the amount of light from the light source device 20), to calculate the output grayscale value.

[0114] Figure 10 This is a flowchart executed when the first driving circuit 30 calculates the output grayscale value. If the first driving circuit 30 acquires the input grayscale value (refer to...), then... Figure 8 Then begin Figure 10 The process of the program. In step S1, the first driving circuit 30 sets k = 1. That is, the first driving circuit 30 sets the row number (k) to 1.

[0115] Next, the first driving circuit 30 selects pixel P in the k-th row in step S2. When k=1, the first driving circuit 30 selects pixel P located in the first row.

[0116] Furthermore, in step S3, the first driving circuit 30 calculates the incident light J for each tilt angle for each pixel P. n The amount (light intensity). Towards respectively Figure 8 The incident light J is incident on pixel P in the first row (k=1). n The amount of light (the amount of light per pixel P in the first row) is determined by the amount of light emitted by the light emitter SL and the configuration of the multiple light emitters SL, specifically stored in the first table shown in Table 1. The first table is stored in the storage section (not shown) of the first driving circuit 30.

[0117] [Table 1]

[0118] (Table 1)

[0119]

[0120] In each pixel P of the first row, the incident light J n The quantities are equal to each other. Furthermore, the incident light J... n The amount can also be different among pixels P in the first row. Additionally, in Table 1, among pixels P in the first row, the incident light J in a pixel P... n The total amount of light emitted is equivalent to the first light quantity (equivalent to "first incident light quantity"). The first light quantity is the amount of light that enables the desired brightness to be obtained in pixel P. In this embodiment, the first light quantity is 1. Furthermore, the amount of light emitted by the light source device 20 corresponding to the first light quantity is equivalent to the "first emitted light quantity".

[0121] In this embodiment, as described above, the maximum value of the grayscale value (input grayscale value and output grayscale value) is 1, which is equal to the value of the first light intensity. Therefore, the fact that the brightness of pixel P is equal to the input grayscale value means that the desired brightness is obtained in pixel P. For example, if the amount of light incident on pixel P is the first light intensity (1 as described above in this embodiment) and the input grayscale value is 0.8, if the brightness of pixel P is 0.8, then the brightness of pixel P is equal to the input grayscale value, and the desired brightness can be obtained in pixel P.

[0122] In step S3, the first driving circuit 30 obtains the values ​​stored in the first table, and directs the incident light J to each pixel P in the first row. n The value is set as the result of multiplying the acquired value by a correction factor. Details of the correction factor will be described later. The correction factor is greater than 1, and in this embodiment it is 1.6.

[0123] That is, in each pixel P of the first row, the incident light J -3 The amount is 0.0912 (=0.057×1.6), and the incident light J -2 The amount is 0.2064 (=0.129×1.6), and the incident light J -1 The amount of incident light J0 is 0.3192 (=0.1995×1.6), and the amount of incident light J0 is 0.3664 (=0.229×1.6). +1 The amount is 0.3192 (=0.1995×1.6), and the incident light J +2 The amount is 0.2064 (=0.129×1.6), and the incident light J +3 The amount is 0.0912 (=0.057×1.6).

[0124] Next, in step S4, the first driving circuit 30 calculates the incident light J for each selected pixel P. n The total amount. Specifically, the first driving circuit 30 provides a total of incident light J for each pixel P in the first row. -3 J -2 J -1 J0, J +1 J +2 J +3 The amount of incident light J. n The total is 1.6 (=0.0912+0.2064+0.3192+0.3664+0.3192+0.2064+0.0912).

[0125] Figure 11 It shows the incident light J in each of multiple pixels P. n A graph showing the total amount. Figure 11 In, with Figure 8 Similarly, a portion of multiple pixels P is shown. In Figure 11 In the image, the value shown within pixel P indicates the incident light J. n The total amount. In each pixel P of the first row (k=1), the incident light J n The total amount is 1.6.

[0126] Thus, when the first driving circuit 30 calculates the output grayscale value using the correction coefficient, in the first row of pixels P, the incident light J in one pixel P... n The total amount of light becomes a second light quantity (1.6 (=first light quantity (1) × correction coefficient (1.6)): equivalent to “second incident light quantity”) which is greater than the first light quantity (1). This means that the light quantity of the light source SL when the first driving circuit 30 calculates the output gray value using the correction coefficient is greater than the light quantity of the light source SL when the first driving circuit 30 does not calculate the output gray value using the correction coefficient. That is, the first light quantity does not necessarily correspond to the maximum brightness of the light source SL. The light source SL is driven at a brightness below the maximum brightness (e.g., about 50% to 85% of the maximum brightness) under normal use conditions, and the brightness can be increased to the maximum brightness. For example, when the first light quantity corresponds to 50% of the maximum brightness of the light source SL, the second light quantity corresponds to approximately 80% of the maximum brightness of the light source (=50% × 1.6). This correction coefficient can be appropriately changed. In this embodiment, the light quantity of the light source device 20 is adjusted to a light quantity corresponding to the second light quantity (equivalent to “second emitted light quantity”).

[0127] Furthermore, in step S5, the first driving circuit 30 calculates and outputs grayscale values ​​for each selected pixel P. Specifically, the first driving circuit 30 uses... Figure 8 The input grayscale values ​​shown are Figure 11 The incident light J shown n The total amount is calculated by dividing the input grayscale value by the incident light J for each pixel P in the first row. n The total amount of light is used to calculate the output grayscale value (= input grayscale value / incident light J). n (The total amount).

[0128] Figure 12 This is a graph showing the output grayscale values ​​of multiple pixels P. Figure 12 In, with Figure 8 as well as Figure 11 Similarly, a portion of multiple pixels P is shown. In Figure 12In the diagram, the value shown within pixel P represents the final output grayscale value (or the corrected output grayscale value). For these pixels P in columns i+3 and i+4, the output grayscale value differs for each row, but as described later, the brightness of pixel P is uniformized to appear the same to the user. In other words, each output grayscale value is corrected in a way that makes the brightness of pixel P uniform and appears the same to the user.

[0129] Specifically, in the first row of pixels P, the output grayscale value of pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is zero (=0 / 1.6). Additionally, the output grayscale value of pixels P(1, i+3) and (1, i+4) is 0.5000 (=0.8 / 1.6).

[0130] Next, in step S6, the first driving circuit 30 determines whether k is greater than or equal to a predetermined value. As described above, the predetermined value is the maximum number of rows of pixel P (e.g., 1080). If k is greater than or equal to the predetermined value ("yes" in step S6), the grayscale value is calculated for all pixels P, and the first driving circuit 30 terminates the program.

[0131] When k=1, k is less than a predetermined value. In this case ("No" in step S6), the first driving circuit 30 calculates the light transmission ratio for each selected pixel P in step S7. First, the first driving circuit 30 calculates the first reduction ratio (the degree of scattering of the first light) and the second reduction ratio (the degree of scattering of the second light), which will be described below.

[0132] As described above, light propagating in pixel P is scattered by components such as the signal line Lb, pixel electrode PE, and switching element SW on the display panel 10, and thus consumed. Furthermore, light propagating in pixel P is absorbed and consumed by metal parts such as the switching element SW and glass parts such as the first substrate 11. Based on these factors, the first reduction ratio corresponds to the proportion of light reduced when propagating in pixel P regardless of whether each pixel P is displayed or not, or its position. That is, the first reduction ratio is determined by the structure of pixel P, and on the other hand, it is a constant value independent of the position of pixel P. In this embodiment, the first reduction ratio is determined as a first predetermined ratio (specifically 1%), which is equal to each other among the plurality of pixels P. That is, the first reduction ratio for each pixel P in the first row is 0.01 (1%).

[0133] On the other hand, when a pixel P is in a display state, i.e., the liquid crystal layer 13 in that pixel P is in a scattering state, the light propagating in that pixel P is scattered and consumed by the liquid crystal layer 13. The second reduction ratio corresponds to the reduction in the amount of light propagating in pixel P due to the scattering state of the liquid crystal layer 13 corresponding to that pixel P. The second reduction ratio changes accordingly with the degree of scattering by the liquid crystal layer 13, i.e., the output grayscale value. Specifically, the greater the degree of scattering by the liquid crystal layer 13 in a pixel P, i.e., the greater the output grayscale value, the greater the second reduction ratio. In this embodiment, the first driving circuit 30 reduces the amount of light propagating in pixel P by adjusting the output grayscale value ( Figure 12 The second reduction ratio is calculated by multiplying the value by a second predetermined ratio (specifically 5%). Consequently, the second reduction ratio increases accordingly, resulting in a larger output grayscale value. This second predetermined ratio is a constant value regardless of the position of pixel P.

[0134] Specifically, in the first row of pixels P, the second reduction ratio of pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is zero (=0×0.05(5%)). That is, the liquid crystal layer 13 corresponding to these pixels P with output grayscale values ​​of zero and not displayed is in a transparent state, and no scattering of light in the liquid crystal layer 13 occurs, so the second reduction ratio is zero. On the other hand, the second reduction ratio of pixels P(1, i+3) and (1, i+4) is 0.025 (=0.5000×0.05(5%)). That is, the liquid crystal layer 13 corresponding to pixels P with output grayscale values ​​of 0.5 and displayed is in a scattering state, and scattering of light in the liquid crystal layer 13 occurs, so the second reduction ratio is greater than 0. Furthermore, the first predetermined ratio and the second predetermined ratio are derived in advance through experiments and stored in the storage unit of the first driving circuit 30. In addition, the first predetermined ratio and the second predetermined ratio are of course not limited to the values ​​described above.

[0135] Furthermore, the first driving circuit 30 calculates the transmission ratio by subtracting a first reduction ratio and a second reduction ratio from 1. That is, in the pixels P of the first row, the transmission ratio of pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is 0.99 (=1-0.01-0). In addition, the transmission ratio of pixels P(1, i+3) and (1, i+4) is 0.965 (=1-0.01-0.025).

[0136] Furthermore, in step S8, the first driving circuit 30 calculates the emitted light I for each tilt angle for each selected pixel P. nThe amount. As described above, in this embodiment, light passing through a pixel P propagates in seven directions (tilt angles of -30°, -20°, -10°, 0°, +10°, +20°, and +30°). Hereinafter, each direction (each tilt angle) will be described separately.

[0137] Figure 13 This diagram illustrates the amount of incident light J0 and emitted light I0 in each of the pixels P in the first row, and the amount of incident light J0 in each of the pixels P in the second row, when the tilt angle is 0° (n = 0). In the pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), and (1, i+6) of the first row, the amount of incident light J0 is calculated in step S3 above and is 0.3664.

[0138] In each pixel P of the first row, the first driving circuit 30 calculates the amount of emitted light I0 by multiplying the amount of incident light J0 by the aforementioned transmission ratio. Specifically, the amount of emitted light I0 for pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is 0.3627 (=0.3664×0.99). Furthermore, the amount of emitted light I0 for pixels P(1, i+3) and (1, i+4) is 0.3536 (=0.3664×0.965).

[0139] Figure 14 This shows the incident light J in each of the pixels P in the first row when the tilt angle is +10° (n = +1). +1 The amount and emitted light I +1 The amount of incident light J in each of the pixels P in the second row. +1 The diagram shows the amount of light. Here, it is assumed that the light propagating to pixel P maintains its direction of travel, i.e., propagates in a straight line. That is, it is assumed that light entering at +10° relative to pixel P (incident light J)... +1 The light emitted from pixel P in a direction of +10° constitutes the emitted light (I). +1 The same applies to the description of other lights with the following tilt angles. In the pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), (1, i+6) of the first row, the incident light J +1 The amount is calculated in step S3 above and is 0.3192.

[0140] In each pixel P of the first row, the first driving circuit 30 controls the incident light J. +1 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. +1The quantity. Specifically, the emitted light I from pixels P(1,i+1), (1,i+2), (1,i+5), and (1,i+6) +1 The amount is 0.3160 ​​(=0.3192×0.99). In addition, the amount of emitted light I+1 from pixels P(1,i+3) and (1,i+4) is 0.3080 (=0.3192×0.965).

[0141] Figure 15 This illustrates the incident light J in each of the pixels P in the first row when the tilt angle is -10° (n = -1). -1 The amount and emitted light I -1 The amount of incident light J in each of the pixels P in the second row. -1 The quantity of the graph. In the first row, among pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), (1, i+6), the incident light J -1 The amount is calculated in step S3 above and is 0.3192.

[0142] In each pixel P of the first row, the first driving circuit 30 calculates the outgoing light I by multiplying the amount of incident light J-1 by the aforementioned transmission ratio. -1 The quantity. Specifically, the emitted light I from pixels P(1,i+1), (1,i+2), (1,i+5), and (1,i+6) -1 The amount is 0.3160 ​​(=0.3192×0.99). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4) is... -1 The amount is 0.3080 (=0.3192×0.965).

[0143] Figure 16 This shows the incident light J in each of the pixels P in the first row when the tilt angle is +20° (n = +2). +2 The amount and emitted light I +2 The amount of incident light J in each of the pixels P in the second row. +2 The quantity of the graph. In the first row, among pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), (1, i+6), the incident light J +2 The amount is calculated in step S3 above and is 0.2064.

[0144] In each pixel P of the first row, the first driving circuit 30 controls the incident light J. +2 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. +2The quantity. Specifically, the emitted light I from pixels P(1,i+1), (1,i+2), (1,i+5), and (1,i+6) +2 The amount is 0.2043 (=0.2064×0.99). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4) is... +2 The amount is 0.1992 (=0.2064×0.965).

[0145] Figure 17 This represents the incident light J in each of the pixels P in the first row when the tilt angle is -20° (n = -2). -2 The amount and emitted light I -2 The amount of incident light J in each of the pixels P in the second row. -2 The quantity of the graph. In the first row, at pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), (1, i+6), the incident light J -2 The amount is calculated in step S3 above and is 0.2064.

[0146] In each pixel P of the first row, the first driving circuit 30 controls the incident light J. -2 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. -2 The quantity. Specifically, the emitted light I from pixels P(1,i+1), (1,i+2), (1,i+5), and (1,i+6) -2 The amount is 0.2043 (=0.2064×0.99). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4) is... -2 The amount is 0.1992 (=0.2064×0.965).

[0147] Figure 18 This represents the incident light J in each of the pixels P in the first row when the tilt angle is +30° (n=+3). +3 The amount and emitted light I +3 The amount of incident light J in each of the pixels P in the second row. +3 The quantity of the graph. In the first row, at pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), (1, i+6), the incident light J +3 The amount is calculated in step S3 above and is 0.0912.

[0148] In each pixel P of the first row, the first driving circuit 30 controls the incident light J. +3 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. +3The quantity. Specifically, the emitted light I from pixels P(1,i+1), (1,i+2), (1,i+5), and (1,i+6) +3 The amount is 0.0903 (=0.0912×0.99). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4) is... +3 The amount is 0.0880 (=0.0912×0.965).

[0149] Figure 19 This represents the incident light J in each of the pixels P in the first row when the tilt angle is -30° (n = -3). -3 The amount and emitted light I -3 The amount of incident light J in each of the pixels P in the second row. -3 The quantity of the graph. In the first row, at pixels P(1, i+1), (1, i+2), (1, i+3), (1, i+4), (1, i+5), (1, i+6), the incident light J -3 The amount is calculated in step S3 above and is 0.0912.

[0150] In each pixel P of the first row, the first driving circuit 30 controls the incident light J. -3 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. -3 The quantity. Specifically, the emitted light I from pixels P(1,i+1), (1,i+2), (1,i+5), and (1,i+6) -3 The amount is 0.0903 (=0.0912×0.99). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4) is... -3 The amount is 0.0880 (=0.0912×0.965).

[0151] As described above, the first driving circuit 30 calculates the incident light J for each pixel P in the first row. n The amount, output grayscale value and emitted light I n The amount.

[0152] Next, in step S9, the first driving circuit 30 increments k by 1 (i.e., k = k + 1). When k = 1, the first driving circuit 30 sets k = 2 (= 1 + 1). Then, the first driving circuit 30 causes the program to return to step S2.

[0153] In step S2, the first driving circuit 30 selects pixel P in the k-th row. When k=2, the first driving circuit 30 selects pixel P located in the second row.

[0154] Furthermore, in step S3, the first driving circuit 30 calculates the incident light J for each tilt angle for each selected pixel P.n The amount. The incident light J incident on each pixel P in the second row and thereafter. n The amount is determined by the emitted light I in the (k-1)th row. n The amount is determined by the angle of inclination, and the calculation method varies for each inclination angle.

[0155] Specifically, the outgoing light I with a tilt angle of 0° (n=0) n That is, the emitted light I0 is along the Y direction, and the emitted light I0 of the pixel P in the (k-1)th row is only incident on the pixel P in the kth row that is adjacent to the +Y side in the same column along the Y direction.

[0156] On the other hand, the emitted light I at tilt angles different from 0° n In the diagram, the emitted light I from pixel P in the (k-1)th row... +1 I +2 I +3 I -1 I -2 I -3 Injected into two pixels P in the k-th row.

[0157] Figure 20 This shows the emitted light I tilted in a direction that is positive (+) relative to the Y direction. +1 I +2 I +3 The image of the incident pixel P. As mentioned above, pixel P is a square, and the length of one side of pixel P is 1. θ is the tilt angle.

[0158] The emitted light I from pixel Pt (specifically, pixel P(k-1,i)) n (Where n = +1, +2, +3) is incident on pixel P(k, i) and its neighboring pixel P(k, i+1). In this case, the outgoing light I... n The amount is based on the allocation ratio r0 and the allocation ratio r + And are assigned to pixel P(k,i) and pixel P(k,i+1).

[0159] Specifically, assuming the emitted light I n It has a width equal to the pixel width (length of one side of pixel P: 1), and is positioned relative to the dividing line L1 that divides the (k-1)th row and the outgoing light I. n The ratio of the length D of the overlapping portion to the length d0 of the portion corresponding to pixel P(k,i) is equivalent to the allocation ratio r0 (=d0 / D). Furthermore, relative to the dividing line L1 and the emitted light I... n The length D of the overlapping portion refers to the length d of the portion corresponding to pixel P(k, i+1). + The ratio is equivalent to the allocation ratio r+ (=d + / D). Furthermore, of course d0+d + =D.

[0160] Figure 21 This shows the emitted light I tilted in a direction negative (-) relative to the Y direction. -1 I -2 Image of pixel P incident on I-3.

[0161] The emitted light I of pixel Pt n (Where n = -1, -2, -3) is incident on pixel P(k, i) and its neighboring pixel P(k, i-1). In this case, the outgoing light I... n The amount is based on the allocation ratio r0 and the allocation ratio r - They are then assigned to pixel P(k, i) and pixel P(k, i-1).

[0162] Specifically, assuming the emitted light I n It has a width equal to the pixel width, and is positioned relative to the dividing line L1 that divides the (k-1)th row and the pixel P in the kth row, and the outgoing light I. n The ratio of the length D of the overlapping portion to the length d0 of the portion corresponding to pixel P(k,i) is equivalent to the allocation ratio r0 (=d0 / D). Furthermore, relative to the dividing line L1 and the emitted light I... n The length D of the overlapping portion refers to the length d of the portion corresponding to pixel P(k, i-1). - The ratio is equivalent to the allocation ratio r - (=d_ / D). Furthermore, of course d0+d - =D.

[0163] Allocation ratio r - r0, r + The value is stored in the second table shown in Table 2. The second table is stored in the storage section of the first drive circuit 30. Allocation ratio r - r0, r + The value of changes accordingly with the magnitude of the tilt angle. Specifically, the allocation ratio r - r + The value corresponds to the value of tanθ, and the value of the allocation ratio r0 is the difference between D and the allocation ratio r. - r + The resulting value. For example, in the case of n = +1 (θ = 10°), the allocation ratio r + =0.176 (=tan10°), distribution ratio r0 = 0.824 (=D(=1)-r) +(=0.176)). Furthermore, when the tilt angle is zero (n=0), as described above, the emitted light I0 is not allocated, and the allocation ratio r0 is 1. Additionally, “N” in the second table indicates a non-existent value.

[0164] [Table 2]

[0165] (Table 2)

[0166] n -3 -2 -1 0 +1 +2 +3 θ -30° -20° -10° 0° 10° 20° 30° <![CDATA[r0]]> 0.423 0.636 0.824 1 0.824 0.636 0.423 <![CDATA[r + ]]> N N N N 0.176 0.364 0.577 <![CDATA[r - ]]> 0.577 0.364 0.176 N N N N

[0167] also, Figure 20 as well as Figure 21 The pixel Pt shown corresponds to the "first pixel", pixel P(k, i) corresponds to the "second pixel", and pixels P(k, i+1) and P(k, i-1) correspond to the "third pixel". Thus, the emitted light I from the first pixel Pt, i.e., pixel Pt, along a direction inclined relative to the Y direction (the second direction), is... n The light from the light source SL (light source device 20) is incident on one of the following: the second pixel P(k, i) adjacent to pixel Pt along the Y direction (first direction), and the third pixel P(k, i+1) and pixel P(k, i-1) adjacent to pixel P(k, i) along the X direction (third direction). Additionally, as described later, the emitted light I from the first pixel Pt... n The amount of light (the amount of light from the light source SL (light source device 20)) is allocated as the incident light J incident on the second pixel, i.e., pixel P(k,i), based on the tilt angle relative to the Y direction (first direction) and the direction (second direction). n The amount of light (light quantity of the light source SL (light source device 20)) and the incident light J incident on one of the third pixel, namely pixel P(k, i+1) and pixel P(k, i-1). n The amount.

[0168] When k≥2, the first driving circuit 30 uses the emitted light I from the (k-1)th row in step S3. n The quantity and the second table, for each pixel P in the k-th row, calculate the incident light J for each tilt angle. n The amount. That is, when k=2, the first driving circuit 30 uses the emitted light I from the first row (k=1). n The quantity and the second table, for each pixel P in the second row, calculate the incident light J for each tilt angle. n The amount.

[0169] When the tilt angle is 0° (n=0), the emitted light I0 of a pixel P is incident only on the pixel P adjacent to the pixel P on the +Y side. For example, the emitted light I0 of pixel P(1, i+2) is incident on pixel P(2, i+2).

[0170] Therefore, when the tilt angle is 0° (n=0), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle. Figure 13 The amount of emitted light I0 of each pixel P in the first row (k=1) is multiplied by the allocation ratio r0 (specifically 1) corresponding to n=0 in the second table to calculate the amount of incident light J0 of each pixel P in the second row (k=2).

[0171] Specifically, such as Figure 13 As shown, in the second row (k=2) of pixel P, the amount of incident light J0 for pixels P(2,i+1), (2,i+2), (2,i+5), and (2,i+6) is 0.3627 (=0.3627×1). Additionally, the amount of incident light J0 for pixels P(2,i+3) and (2,i+4) is 0.3536 (=0.3536×1).

[0172] With a tilt angle of +10° (n=+1), the emitted light I of a pixel P +1 The light is incident on a pixel P adjacent to the pixel P on the +Y side and on the pixel P adjacent to the pixel P on the +X side. For example, the emitted light I from pixel P(1, i+2) +1 Injected into pixels P(2, i+2) and P(2, i+3).

[0173] Therefore, when the tilt angle is +10° (n=+1), the first drive circuit 30, through the... Figure 14 The emitted light I of each pixel P in the first row (k=1) shown +1 The quantity is multiplied by the allocation ratio r0 (specifically 0.824) corresponding to n = +1 in the second table and the allocation ratio r + (Specifically 0.176), to allocate the emitted light I to each pixel P in the first row (k=1). +1 The amount.

[0174] Specifically, such as Figure 14 As shown, the emitted light I from pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is... +1 The quantities are allocated as 0.2604 (=0.3160×0.824) and 0.0556 (=0.3160×0.176). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4)... +1 The amounts were allocated as 0.2538 (=0.3080×0.824) and 0.0542 (=0.3080×0.176).

[0175] Next, the first driving circuit 30, based on the allocated emitted light I+1 The quantity is used to calculate the incident light J for each pixel P in the second row (k=2). +1 The amount. Specifically, the incident light J at pixel P(2, i+1). +1 This is equivalent to converting the emitted light I from pixel P(1, i+1) into a single beam. +1 The light corresponding to the allocation ratio r0 and the emitted light I from pixel P(1, i+0) (not shown) on the -X side adjacent to pixel P(1, i+1) +1 The distribution ratio r + The light obtained by corresponding light synthesis. The emitted light I from pixel P(1, i+0) +1 The quantity and distribution ratio r + The corresponding light intensity is 0.0556. That is, the incident light J at pixel P(2, i+1) is... +1 The amount is 0.3160 ​​(=0.2604+0.0556). Furthermore, Figure 14 The dashed arrows shown represent the emitted light I from pixel P(1, i+0). +1 The distribution ratio r + This corresponds to the situation where light is incident on pixel P(2, i+1).

[0176] Additionally, the incident light J at pixel P(2, i+2) +1 This is equivalent to converting the emitted light I from pixel P(1, i+2) into a single beam. +1 The light corresponding to the allocation ratio r0 and the emitted light I of pixel P(1, i+1) +1 The distribution ratio r + The light obtained by corresponding light synthesis. That is, the incident light J of pixel P(2, i+2). +1 The amount is 0.3160 ​​(=0.2604+0.0556).

[0177] Furthermore, the first driving circuit 30 and the incident light J of the aforementioned pixel P(2, i+2) +1 Similarly, the incident light J of pixels P(2, i+3), (2, i+4), (2, i+5), and (2, i+6) is calculated. +1 The quantity. That is, the incident light J of pixel P(2, i+3). +1 The amount is 0.3094 (=0.2538+0.0556), and the incident light J of pixel P(2, i+4) is... +1 The amount is 0.3080 (=0.2538+0.0542). Additionally, the incident light J at pixel P(2, i+5) +1 The amount is 0.3146 (=0.2604+0.0542), and the incident light J of pixel P(2, i+6) is... +1 The amount is 0.3160 ​​(0.2604 + 0.0556).

[0178] With a tilt angle of -10° (n = -1), the emitted light I of a pixel P -1 The light is incident on a pixel P adjacent to the pixel P on the +Y side and on the pixel P adjacent to the pixel P on the -X side. For example, the emitted light I from pixel P(1, i+2) is... -1 Injected into pixels P(2, i+2) and P(2, i+1).

[0179] Therefore, when the tilt angle is -10° (n = -1), the first drive circuit 30, through the... Figure 15 The emitted light I of each pixel P in the first row (k=1) shown -1 The quantity is multiplied by the allocation ratio r0 (specifically 0.824) corresponding to n=-1 in the second table and the allocation ratio r - (Specifically 0.176), to allocate the emitted light I to each pixel P in the first row (k=1). -1 The amount.

[0180] Specifically, such as Figure 15 As shown, the emitted light I from pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is... -1 The quantities are allocated as 0.2604 (=0.3160×0.824) and 0.0556 (=0.3160×0.176). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4)... -1 The amounts were allocated as 0.2538 (=0.3080×0.824) and 0.0542 (=0.3080×0.176).

[0181] Next, the first driving circuit 30, based on the allocated emitted light I -1 The quantity is used to calculate the incident light J for each pixel P in the second row (k=2). -1 The amount. Specifically, the incident light J at pixel P(2, i+6). -1 This is equivalent to converting the emitted light I from pixel P(1, i+6) into a single beam. -1 The light corresponding to the allocation ratio r0 and the emitted light I from pixel P(1, i+7) (not shown) on the +X side adjacent to pixel P(1, i+6) -1 The distribution ratio r - The light obtained by corresponding light synthesis. The emitted light I from pixel P(1, i+7) -1 The quantity and distribution ratio r - The corresponding light intensity is 0.0556. That is, the incident light J at pixel P(2, i+6) is... -1The amount is 0.3160 ​​(=0.2604+0.0556). Furthermore, Figure 15 The dashed arrows shown represent the emitted light I from pixel P(1, i+7). -1 The distribution ratio r - This corresponds to the situation where light is incident on pixel P(2, i+6).

[0182] Additionally, the incident light J at pixel P(2, i+1) -1 This is equivalent to converting the emitted light I from pixel P(1, i+1) into a single beam. -1 The light corresponding to the allocation ratio r0 and the emitted light I of pixel P(1, i+2) -1 The distribution ratio r - The light obtained by corresponding light synthesis. That is, the incident light J of pixel P(2, i+1). -1 The amount is 0.3160 ​​(=0.2604+0.0556).

[0183] Furthermore, the first driving circuit 30 and the incident light J of the aforementioned pixel P(2, i+1) -1 Similarly, the incident light J of pixels P(2, i+2), (2, i+3), (2, i+4), and (2, i+5) is calculated. -1 The quantity. That is, the incident light J of pixel P(2, i+2). -1 The amount is 0.3146 (=0.2604+0.0542), and the incident light J of pixel P(2, i+3) is... -1 The amount is 0.3080 (=0.2538+0.0542). Additionally, the incident light J at pixel P(2, i+4) -1 The amount is 0.3094 (=0.2538+0.0556), and the incident light J of pixel P(2, i+5) is... -1 The amount is 0.3160 ​​(0.2604 + 0.0556).

[0184] With a tilt angle of +20° (n=+2), the emitted light I of a pixel P +2 The light is incident on a pixel P adjacent to the pixel P on the +Y side and on the pixel P adjacent to the pixel P on the +X side. For example, the emitted light I from pixel P(1, i+2) +1 Injected into pixels P(2, i+2) and P(2, i+3).

[0185] Therefore, when the tilt angle is +20° (n=+2), the first drive circuit 30, through the... Figure 16 The emitted light I of each pixel P in the first row (k=1) shown +2The quantity is multiplied by the allocation ratio r0 (specifically 0.636) corresponding to n = +2 in the second table and the allocation ratio r + (Specifically 0.364), to allocate the emitted light I to each pixel P in the first row (k=1). +2 The amount.

[0186] Specifically, such as Figure 16 As shown, the emitted light I from pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is... +2 The quantities are allocated as 0.1300 (=0.2043×0.636) and 0.0744 (=0.2043×0.364). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4)... +2 The amounts were allocated as 0.1267 (=0.1992×0.636) and 0.0725 (=0.1992×0.364).

[0187] Next, the first driving circuit 30, based on the allocated emitted light I +2 The quantity is used to calculate the incident light J for each pixel P in the second row (k=2). +2 The amount. Specifically, the emitted light I of each pixel P in the first row when the tilt angle is +20° (n=+2). +2 The incident light J of each pixel P in the second row +2 The correspondence is the same as the correspondence in the case where the tilt angle is +10° (n=+1). Additionally, the emitted light I from pixel P(1, i+0) +2 The quantity and distribution ratio r + The corresponding light intensity is 0.0744. That is, the incident light J at pixel P(2, i+1) is... +2 The amount is 0.2043 (=0.1300+0.0744). Furthermore, Figure 16 The dashed arrows shown represent the emitted light I from pixel P(1, i+0). +2 The distribution ratio r + This corresponds to the situation where light is incident on pixel P(2, i+1).

[0188] Additionally, the incident light J at pixel P(2, i+2) +2 The amount is 0.2043 (=0.1300+0.0744). The incident light J at pixel P(2, i+3) is... +2 The amount is 0.2011 (=0.1267+0.0744), and the incident light J of pixel P(2, i+4) is... +2 The amount is 0.1922 (=0.1267+0.0725). Additionally, the incident light J at pixel P(2, i+5)+2 The amount is 0.2025 (=0.1300+0.0725), and the incident light J of pixel P(2, i+6) is... +2 The amount is 0.2043 (0.1300 + 0.0744).

[0189] With a tilt angle of -20° (n = -2), the emitted light I of a pixel P -2 The light is incident on a pixel P adjacent to the pixel P on the +Y side and on the pixel P adjacent to the pixel P on the -X side. For example, the emitted light I from pixel P(1,2) -2 Injected into pixels P(2,2) and P(2,1).

[0190] Therefore, when the tilt angle is -20° (n = -2), the first drive circuit 30, through the... Figure 17 The emitted light I of each pixel P in the first row (k=1) shown -2 The quantity is multiplied by the allocation ratio r0 (specifically 0.636) corresponding to n = -2 in the second table and the allocation ratio r - (Specifically 0.364), to allocate the emitted light I to each pixel P in the first row (k=1). -2 The amount.

[0191] Specifically, such as Figure 17 As shown, the emitted light I from pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is... -2 The quantities are allocated as 0.1300 (=0.2043×0.636) and 0.0744 (=0.2043×0.364). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4)... -2 The amounts were allocated as 0.1267 (=0.1992×0.636) and 0.0725 (=0.1992×0.364).

[0192] Next, the first driving circuit 30, based on the allocated emitted light I -2 The quantity is used to calculate the incident light J for each pixel P in the second row (k=2). -2 The amount. Specifically, the emitted light I of each pixel P in the first row when the tilt angle is -20° (n=-2). -2 The correspondence between the incident light J-2 of each pixel P in the second row and the corresponding correspondence is the same as that in the case of a tilt angle of -10° (n=-1). Additionally, the outgoing light I of pixel P(1, i+7)... -2 The quantity and distribution ratio r -The corresponding light intensity is 0.0744. That is, the incident light J at pixel P(2, i+6) is... -2 The amount is 0.2043 (=0.1300+0.0744). Furthermore, Figure 17 The dashed arrows shown represent the emitted light I from pixel P(1, i+7). -2 The distribution ratio r - This corresponds to the situation where light is incident on pixel P(2, i+6).

[0193] Additionally, the incident light J at pixel P(2, i+1) -2 The amount is 0.2043 (=0.1300+0.0744). The incident light J at pixel P(2, i+2) is... -2 The amount is 0.2025 (=0.1300+0.0725), and the incident light J of pixel P(2, i+3) is... -2 The amount is 0.1922 (=0.1267+0.0725). Additionally, the incident light J at pixel P(2, i+4) -2 The amount is 0.2011 (=0.1267+0.0744), and the incident light J of pixel P(2, i+5) is... -2 The amount is 0.2043 (0.1300 + 0.0744).

[0194] With a tilt angle of +30° (n=+3), the emitted light I of a pixel P +3 The light is incident on a pixel P adjacent to the pixel P on the +Y side and on the pixel P adjacent to the pixel P on the +X side. For example, the emitted light I from pixel P(1, i+2) +3 Injected into pixels P(2, i+2) and P(2, i+3).

[0195] Therefore, when the tilt angle is +30° (n=+3), the first drive circuit 30, through the... Figure 18 The emitted light I of each pixel P in the first row (k=1) shown +3 The quantity is multiplied by the allocation ratio r0 (specifically 0.423) corresponding to n = +3 in the second table, and the allocation ratio r + (Specifically 0.577), to allocate the emitted light I to each pixel P in the first row (k=1). +3 The amount.

[0196] Specifically, such as Figure 18 As shown, the emitted light I from pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is... +3The quantities are allocated as 0.382 (=0.0903×0.423) and 0.0521 (=0.0903×0.577). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4)... +3 The amounts were allocated as 0.0372 (=0.0880×0.423) and 0.0508 (=0.0880×0.577).

[0197] Next, the first driving circuit 30, based on the allocated emitted light I +3 The quantity is used to calculate the incident light J for each pixel P in the second row (k=2). +3 The amount. Specifically, the emitted light I of each pixel P in the first row when the tilt angle is +30° (n=+3). +3 The incident light J of each pixel P in the second row +3 The correspondence is the same as the correspondence in the case where the tilt angle is +10° (n=+1). Additionally, the emitted light I from pixel P(1, i+0) +3 The quantity and distribution ratio r + The corresponding light intensity is 0.0521. That is, the incident light J at pixel P(2, i+1) is... +3 The amount is 0.0903 (=0.0382+0.0521). Furthermore, Figure 18 The dashed arrows shown represent the emitted light I from pixel P(1, i+0). +3 The distribution ratio r + This corresponds to the case where light is incident on pixel P(2, i+1).

[0198] Additionally, the incident light J at pixel P(2, i+2) +3 The amount is 0.0903 (=0.0382+0.0521). The incident light J at pixel P(2, i+3) is... +3 The amount is 0.0893 (=0.0372+0.0521), and the incident light J of pixel P(2, i+4) is... +3 The amount is 0.0880 (=0.0372+0.0508). Additionally, the incident light J at pixel P(2, i+5) +3 The amount is 0.0890 (=0.0382+0.0508), and the incident light J of pixel P(2, i+6) is... +3 The amount is 0.0903 (0.0382 + 0.0521).

[0199] With a tilt angle of -30° (n = -3), the emitted light I of a pixel P -3The light is incident on a pixel P adjacent to the pixel P on the +Y side and on the pixel P adjacent to the pixel P on the -X side. For example, the emitted light I from pixel P(1, i+2) -3 Injected into pixels P(2, i+2) and P(2, i+1).

[0200] Therefore, when the tilt angle is -30° (n = -3), the first drive circuit 30, through the... Figure 19 The emitted light I of each pixel P in the first row (k=1) shown -3 The quantity is multiplied by the allocation ratio r0 (specifically 0.423) corresponding to n = -3 in the second table and the allocation ratio r - (Specifically 0.577), to allocate the emitted light I to each pixel P in the first row (k=1). -3 The amount.

[0201] Specifically, such as Figure 19 As shown, the emitted light I from pixels P(1, i+1), (1, i+2), (1, i+5), and (1, i+6) is... -3 The quantities are allocated as 0.382 (=0.0903×0.423) and 0.0521 (=0.0903×0.577). Additionally, the emitted light I from pixels P(1, i+3) and (1, i+4)... -3 The amounts were allocated as 0.0372 (=0.0880×0.423) and 0.0508 (=0.0880×0.577).

[0202] Next, the first driving circuit 30, based on the allocated emitted light I -3 The quantity is used to calculate the incident light J for each pixel P in the second row (k=2). -3 The amount. Specifically, the emitted light I of each pixel P in the first row when the tilt angle is -30° (n=-3). -3 The incident light J of each pixel P in the second row -3 The correspondence is the same as the correspondence in the case where the tilt angle is -10° (n=-1). Additionally, the emitted light I from pixel P(1, i+7) -3 The ratio of quantity to distribution r - The corresponding light intensity is 0.0521. That is, the incident light J at pixel P(2, i+6) is... -3 The amount is 0.0903 (=0.0382+0.0521). Furthermore, Figure 17 The dashed arrows shown represent the emitted light I from pixel P(1, i+7). -3 The distribution ratio r - This corresponds to the situation where light is incident on pixel P(2, i+6).

[0203] Additionally, the incident light J at pixel P(2, i+1) -3 The amount is 0.0903 (=0.0382+0.0521). The incident light J at pixel P(2, i+2) is... -3 The amount is 0.0890 (=0.0382+0.0508), and the incident light J of pixel P(2, i+3) is... -3 The amount is 0.0880 (=0.0372+0.0508). The incident light J at pixel P(2, i+4) is... -3 The amount is 0.0893 (=0.0372+0.0521), and the incident light J of pixel P(2, i+5) is... -3 The amount is 0.0903 (0.0382 + 0.0521).

[0204] Next, in step S4, the first driving circuit 30 calculates the incident light J for each selected pixel P. n The total amount. Specifically, the first driving circuit 30 provides incident light J to each pixel P in the second row. -3 J -2 J -1 J0, J +1 J +2 J +3 The quantities are totaled.

[0205] Specifically, such as Figure 11 As shown, the incident light J of pixel P(2, i+1) n (i.e., incident light J) -3 J -2 J -1 J0, J +1 J +2 J +3 The total amount is 1.5840 (=0.0903 () Figure 19 )+0.2043( Figure 17 )+0.3160( Figure 15 )+0.3627( Figure 13 )+0.3160( Figure 14 )+0.2043( Figure 16 )+0.0903( Figure 18 Similarly, the incident light J at pixel P(2, i+6) n The total amount is 1.5840 (=0.0903+0.2043+0.3160+0.3627+0.3160+0.2043+0.0903).

[0206] Additionally, the incident light J at pixels P(2, i+2) and (2, i+5)n The total amount is 1.5794 (=0.0903+0.2043+0.3160+0.3627+0.3146+0.2025+0.0890). Furthermore, the incident light J at pixels P(2,i+3) and (2,i+4) n The total amount is 1.5486 (=0.0893+0.2011+0.3094+0.3536+0.3080+0.1992+0.0880).

[0207] Furthermore, in step S5, the first driving circuit 30 calculates and outputs grayscale values ​​for each selected pixel P. Specifically, the first driving circuit 30 uses... Figure 8 The input grayscale values ​​shown are Figure 11 The incident light J shown n The total amount is calculated by dividing the input grayscale value by the incident light J for each pixel P in the second row. n The total amount of incident light is used to calculate the output grayscale value (=Input grayscale value / Total amount of incident light Jn).

[0208] like Figure 8 As shown, the input grayscale values ​​of pixels P(2, i+1), (2, i+2), (2, i+5), and (2, i+6) are zero, thus... Figure 12 As shown, the output grayscale values ​​of pixels P(2, i+1), (2, i+2), (2, i+5), and (2, i+6) are zero. Additionally, as... Figure 8 As shown, the input grayscale value of pixels P(2, i+3) and (2, i+4) is 0.8, thus... Figure 12 As shown, the output grayscale values ​​of pixels P(2, i+3) and (2, i+4) are 0.5166 (=0.8). Figure 8 ) / 1.5486( Figure 11 )).

[0209] like Figure 11 As shown, since the light from the emitting element SL is consumed in pixel P of the first row, the incident light J of pixel P in the first row is... n Compared to the total amount, the incident light J of pixel P in the second row n The total amount is relatively small.

[0210] Additionally, in pixels P in the (i+3)th and (i+4)th columns, such as Figure 8 As shown, even if the input grayscale value (0.8) of pixel P in the first row is equal to the input grayscale value (0.8) of pixel P in the second row, such as Figure 12As shown, the output grayscale value of pixel P in the second row (0.5166) is also larger than that of pixel P in the first row (0.500). That is, even if the input grayscale value of pixel P in the first row is equal to that of pixel P in the second row, the first driving circuit 30 increases the output grayscale value of pixel P in the second row based on the amount of light consumed by the light emitter SL in pixel P in the first row.

[0211] Next, in step S6, the first driving circuit 30 determines whether k is above a predetermined value. If k = 2, then k is less than the predetermined value. In this case ("No" in step S6), the first driving circuit 30 calculates the light transmission ratio for each selected pixel P in step S7.

[0212] As described above, the first predetermined ratio (1%) and the second predetermined ratio (5%) are constant values ​​regardless of the position of pixel P. The first reduction ratio is equivalent to the first predetermined ratio.

[0213] In addition, the second reduction ratio and Figure 12 The output grayscale values ​​shown change accordingly. Specifically, in the second row of pixels P, the second reduction ratio of pixels P(2, i+1), (2, i+2), (2, i+5), and (2, i+6) is zero (=0×0.05 (second predetermined ratio: 5%)). Additionally, the second reduction ratio of pixels P(2, i+3) and (2, i+4) is 0.0258 (=0.5166×0.05 (second predetermined ratio: 5%)).

[0214] Therefore, in the second row of pixels P, the transmittance of pixels P(2, i+1), (2, i+2), (2, i+5), and (2, i+6) is 0.99 (=1-0.01-0). Additionally, the transmittance of pixels P(2, i+3) and (2, i+4) is 0.9642 (=1-0.01-0.0258).

[0215] Furthermore, in step S8, the first driving circuit 30 calculates the emitted light I for each tilt angle for each selected pixel P. n The amount.

[0216] Figure 22 This diagram illustrates the amount of incident light J0 and emitted light I0 in each of the pixels P in the second row, and the amount of incident light J0 in each of the pixels P in the third row, when the tilt angle is 0° (n = 0). In the pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), and (2, i+6) of the second row, the amount of incident light J0 is calculated through step S3 described above. Figure 13 ).

[0217] In each pixel P of the second row, the first driving circuit 30 calculates the amount of emitted light I0 by multiplying the amount of incident light J0 by the aforementioned transmission ratio. That is, the amount of emitted light I0 for pixels P(2, i+1), (2, i+2), (2, i+5), and (2, i+6) is 0.3591 (=0.3637×0.99). Furthermore, the amount of emitted light I0 for pixels P(2, i+3) and (2, i+4) is 0.3409 (=0.3536×0.9642).

[0218] Figure 23 This shows the incident light J in each of the pixels P in the second row when the tilt angle is +10° (n = +1). +1 The amount and emitted light I +1 The amount of incident light J in each of the pixels P in the third row. +1 The quantity of the graph. In the second row, for pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), (2, i+6), the incident light J is calculated through the above step S3. +1 The amount ( Figure 14 ).

[0219] In each pixel P of the second row, the first driving circuit 30 controls the incident light J. +1 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. +1 The quantity. That is, the emitted light I from pixels P(2,i+1), (2,i+2), (2,i+3), (2,i+4), (2,i+5), and (2,i+6). +1 The quantities are 0.3128 (=0.3160×0.99), 0.3128 (=0.3160×0.99), 0.2983 (=0.3094×0.9642), 0.2970 (=0.3080×0.9642), 0.3115 (=0.3146×0.99), and 0.3128 (=0.3160×0.99).

[0220] Figure 24 This shows the incident light J in each of the pixels P in the second row when the tilt angle is -10° (n = -1). -1 The amount and emitted light I -1 The amount of incident light J in each of the pixels P in the third row. -1 The quantity of the graph. In the second row, for pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), (2, i+6), the incident light J is calculated through the above step S3. -1 The amount ( Figure 15 ).

[0221] In each pixel P of the second row, the first driving circuit 30 controls the incident light J. -1 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. -1 The quantity. That is, the emitted light I from pixels P(2,i+1), (2,i+2), (2,i+3), (2,i+4), (2,i+5), and (2,i+6). -1 The quantities are 0.3128 (=0.3160×0.99), 0.3115 (=0.3146×0.99), 0.2970 (=0.3080×0.9642), 0.2983 (=0.3094×0.9642), 0.3128 (=0.3160×0.99), and 0.3128 (=0.3160×0.99).

[0222] Figure 25 This shows the incident light J in each of the pixels P in the second row when the tilt angle is +20° (n = +2). +2 The amount and emitted light I +2 The amount of incident light J in each of the pixels P in the third row. +2 The quantity of the graph. In the second row, for pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), (2, i+6), the incident light J is calculated through the above step S3. +2 The amount ( Figure 16 ).

[0223] In each pixel P of the second row, the first driving circuit 30 controls the incident light J. +2 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. +2 The quantity. That is, the emitted light I from pixels P(2,i+1), (2,i+2), (2,i+3), (2,i+4), (2,i+5), and (2,i+6). +2 The quantities are 0.2023 (=0.2043×0.99), 0.2023 (=0.2043×0.99), 0.1939 (=0.2011×0.9642), 0.1920 (=0.1992×0.9642), 0.2004 (=0.2025×0.99), and 0.2023 (=0.2043×0.99).

[0224] Figure 26 This shows the incident light J in each of the pixels P in the second row when the tilt angle is -20° (n = -2). -2 The amount and emitted light I -2 The amount of incident light J in each of the pixels P in the third row.-2 The quantity of the graph. In the second row, for pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), (2, i+6), the incident light J is calculated through the above step S3. -2 The amount ( Figure 17 ).

[0225] In each pixel P of the second row, the first driving circuit 30 controls the incident light J. -2 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. -2 The quantity. That is, the emitted light I from pixels P(2,i+1), (2,i+2), (2,i+3), (2,i+4), (2,i+5), and (2,i+6). -1 The quantities are 0.2023 (=0.2043×0.99), 0.2004 (=0.2025×0.99), 0.1920 (=0.1992×0.9642), 0.1939 (=0.2011×0.9642), 0.2023 (=0.2043×0.99), and 0.2023 (=0.2043×0.99).

[0226] Figure 27 This shows the incident light J in each of the pixels P in the second row when the tilt angle is +30° (n = +3). +3 The amount and emitted light I +3 The amount of incident light J in each of the pixels P in the third row. +3 The quantity of the graph. In the second row, for pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), (2, i+6), the incident light J is calculated through the above step S3. +3 The amount ( Figure 18 ).

[0227] In each pixel P of the second row, the first driving circuit 30 controls the incident light J. +3 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. +3 The quantity. That is, the emitted light I from pixels P(2,i+1), (2,i+2), (2,i+3), (2,i+4), (2,i+5), and (2,i+6). +3 The quantities are 0.0894 (=0.0903×0.99), 0.0894 (=0.0903×0.99), 0.0861 (=0.0893×0.9642), 0.0849 (=0.0880×0.9642), 0.0881 (=0.0890×0.99), and 0.0894 (=0.0903×0.99).

[0228] Figure 28 This shows the incident light J in each of the pixels P in the second row when the tilt angle is -30° (n = -3). -3 The amount and emitted light I -3 The amount of incident light J in each of the pixels P in the third row. -3 The quantity of the graph. In the second row, for pixels P(2, i+1), (2, i+2), (2, i+3), (2, i+4), (2, i+5), (2, i+6), the incident light J is calculated through the above step S3. -3 The amount ( Figure 19 ).

[0229] In each pixel P of the second row, the first driving circuit 30 controls the incident light J. -3 The amount of light emitted is multiplied by the aforementioned transmission ratio to calculate the emitted light I. -3 The quantity. That is, the emitted light I from pixels P(2,i+1), (2,i+2), (2,i+3), (2,i+4), (2,i+5), and (2,i+6). -1 The quantities are 0.0894 (=0.0903×0.99), 0.0881 (=0.0890×0.99), 0.0849 (=0.0880×0.9642), 0.0861 (=0.0893×0.9642), 0.0894 (=0.0903×0.99), and 0.0894 (=0.0903×0.99).

[0230] Next, in step S9, the first driving circuit 30 increments k by 1 (i.e., k = k + 1). When k = 2, the first driving circuit 30 sets k = 3 (= 2 + 1). Then, the first driving circuit 30 causes the program to return to step S2.

[0231] In step S2, the first driving circuit 30 selects pixel P in the k-th row. When k=3, the first driving circuit 30 selects pixel P located in the third row.

[0232] Furthermore, in step S3, the first driving circuit 30 calculates the incident light J for each tilt angle for each selected pixel P. n The amount. As described above, when k≥2, in step S3, the first driving circuit 30 uses the emitted light I from the (k-1)th row. n The quantity and the second table, for each pixel P in the k-th row, calculate the incident light J for each tilt angle. n The amount.

[0233] That is, in step S3, the first driving circuit 30 uses the emitted light I from the second row (k=2). nSimilarly to the case of k=2, the amount and the second table are used to calculate the incident light J for each tilt angle for each pixel P in the third row (k=3). n The amount.

[0234] When the tilt angle is 0° (n=0), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle. Figure 22 The amount of emitted light I0 of each pixel P in the second row (k=2) is multiplied by the allocation ratio r0 (specifically 1) corresponding to n=0 in the second table to calculate the amount of incident light J0 of each pixel P in the third row (k=3).

[0235] Specifically, such as Figure 22 As shown, in the third row of pixels P, the amount of incident light J0 for pixels P(3, i+1), (3, i+2), (3, i+5), and (3, i+6) is 0.3591 (=0.3591×1). Additionally, the amount of incident light J0 for pixels P(3, i+3) and (3, i+4) is 0.3409 (=0.3409×1).

[0236] When the tilt angle is +10° (n=+1), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle of +10° (n=+1). Figure 23 The emitted light I of each pixel P in the second row (k=2) shown is... +1 The quantity is multiplied by the allocation ratio r0 (specifically 0.824) corresponding to n = +1 in the second table and the allocation ratio r + (Specifically 0.176), to allocate the emitted light I to each pixel P in the second row (k=2). +1 The amount.

[0237] Next, the first driving circuit 30, based on the allocated emitted light I +1 The quantity is used to calculate the incident light J for each pixel P in the third row (k=3). +1 The quantity. Specifically, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) +1 The amounts were 0.3128, 0.3128, 0.3009, 0.2972, 0.3089, and 0.3126.

[0238] When the tilt angle is -10° (n = -1), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle. Figure 24 The emitted light I of each pixel P in the second row (k=2) shown is... -1 The quantity is multiplied by the allocation ratio r0 (specifically 0.824) corresponding to n=-1 in the second table and the allocation ratio r -(Specifically 0.176), to allocate the emitted light I to each pixel P in the second row (k=2). -1 The amount.

[0239] Next, the first driving circuit 30, based on the allocated emitted light I -1 The quantity is used to calculate the incident light J for each pixel P in the third row (k=3). -1 The quantity. Specifically, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) -1 The amounts were 0.3126, 0.3089, 0.2972, 0.3009, 0.3128, and 0.3128.

[0240] With a tilt angle of +20° (n = +2), the first drive circuit 30, through... Figure 25 The emitted light I of each pixel P in the second row (k=2) shown is... +2 The quantity is multiplied by the allocation ratio r0 (specifically 0.636) corresponding to n = +2 in the second table and the allocation ratio r + (Specifically 0.364), to allocate the emitted light I to each pixel P in the second row (k=2). +2 The amount.

[0241] Next, the first driving circuit 30, based on the allocated emitted light I +2 The quantity is used to calculate the incident light J for each pixel P in the third row (k=3). + The quantity is 2. Specifically, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) is... +2 The amounts were 0.2023, 0.2023, 0.1969, 0.1927, 0.1974, and 0.2016.

[0242] When the tilt angle is -20° (n = -2), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle of -20° (n = -2). Figure 26 The emitted light I of each pixel P in the second row (k=2) shown is... -2 The quantity is multiplied by the allocation ratio r0 (specifically 0.636) corresponding to n = -2 in the second table and the allocation ratio r - (Specifically 0.364), to allocate the emitted light I to each pixel P in the second row (k=2). -2 The amount.

[0243] Next, the first driving circuit 30, based on the allocated emitted light I -2 The quantity is used to calculate the incident light J for each pixel P in the third row (k=3).-2 The quantity. Specifically, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) -2 The amounts were 0.2016, 0.1974, 0.1927, 0.1969, 0.2023, and 0.2023.

[0244] When the tilt angle is +30° (n=+3), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle of +30° (n=+3). Figure 27 The emitted light I of each pixel P in the second row (k=2) shown is... +3 The quantity is multiplied by the allocation ratio r0 (specifically 0.423) corresponding to n = +3 in the second table, and the allocation ratio r + (Specifically 0.577), to allocate the emitted light I to each pixel P in the second row (k=2). +3 The amount.

[0245] Next, the first driving circuit 30, based on the allocated emitted light I +3 The quantity is used to calculate the incident light J for each pixel P in the third row (k=3). +3 The quantity. Specifically, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) +3 The amounts are 0.0894, 0.0894, 0.0880, 0.0856, 0.0862, and 0.0886.

[0246] When the tilt angle is -30° (n = -3), the first drive circuit 30 drives the first drive circuit 30 by controlling the tilt angle of the first drive circuit 30. Figure 28 The emitted light I of each pixel P in the second row (k=2) shown is... -3 The quantity is multiplied by the allocation ratio r0 (specifically 0.423) corresponding to n = -3 in the second table and the allocation ratio r - (Specifically 0.577), to allocate the emitted light I to each pixel P in the second row (k=2). -3 The amount.

[0247] Next, the first driving circuit 30, based on the allocated emitted light I -3 The quantity is used to calculate the incident light J for each pixel P in the third row (k=3). -3 The quantity. Specifically, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) -3 The amounts are 0.0886, 0.0862, 0.0856, 0.0880, 0.0894, and 0.0894.

[0248] Next, in step S4, the first driving circuit 30 calculates the incident light J for each selected pixel P. n The total amount. Specifically, the first driving circuit 30 provides a total of incident light J for each pixel P in the third row. n (i.e., incident light J) -3 J -2 J -1 J0, J +1 J +2 J +3 ) amount.

[0249] Specifically, such as Figure 11 As shown, the incident light J of pixels P(3,i+1), (3,i+2), (3,i+3), (3,i+4), (3,i+5), (3,i+6) n The total quantities are 1.5665, 1.5561, 1.5022, 1.5022, 1.5561, and 1.5665.

[0250] Furthermore, in step S5, the first driving circuit 30 calculates the output grayscale value (= input grayscale value) for each selected pixel P. Figure 8 ) / Incident light J n The total amount ( Figure 11 )).

[0251] like Figure 8 As shown, the input grayscale values ​​of pixels P(3, i+1), (3, i+2), (3, i+5), and (3, i+6) are zero, thus... Figure 12 As shown, the output grayscale values ​​of pixels P(3, i+1), (3, i+2), (3, i+5), and (3, i+6) are zero. Additionally, as... Figure 8 As shown, the input grayscale value of pixels P(3, i+3) and (3, i+4) is 0.8, thus... Figure 12 As shown, the output grayscale values ​​of pixels P(3, i+3) and (3, i+4) are 0.5325 (=0.8). Figure 8 ) / 1.5022( Figure 11 )).

[0252] Then, the first driving circuit 30 executes steps S6, S7, S8, and S9. In step S2, it selects pixel P in the fourth row (k=4), and executes steps S4 and S5 in the same way as in the cases of k=2 and 3, thereby calculating the incident light J for each pixel P in the fourth row. n The total amount ( Figure 11 ) and output grayscale value (= input grayscale value ( Figure 8 ) / Incident light J nThe total amount ( Figure 11 )).

[0253] like Figure 8 As shown, the input grayscale values ​​of pixels P(4, i+1), (4, i+2), (4, i+5), and (4, i+6) are zero, thus... Figure 12 As shown, the output grayscale values ​​of pixels P(4, i+1), (4, i+2), (4, i+5), and (4, i+6) are zero. Additionally, as... Figure 8 As shown, the input grayscale value of pixels P(4, i+3) and (4, i+4) is 0.8, thus... Figure 12 As shown, the output grayscale values ​​of pixels P(4, i+3) and (4, i+4) are 0.5479 (=0.8). Figure 8 ) / 1.4602( Figure 11 )).

[0254] like Figure 11 As shown, since the light from the emitting element SL is consumed in pixel P, the incident light J of pixel P in the k-th row is... n Compared to the total amount, the incident light J of pixel P in the (k+1)th row n The total amount is relatively small.

[0255] In addition, such as Figure 8 as well as Figure 12 As shown, in the (i+3)th and (i+4)th columns of pixel P, even if the input grayscale value of pixel P in the kth row is equal to the input grayscale value of pixel P in the (k+1)th row (… Figure 8 Compared to the output grayscale value of pixel P in row k+1, the output grayscale value of pixel P in row k+1 is also larger. Figure 12 ).

[0256] That is, the first driving circuit 30 increases the output grayscale value of the pixel P in the (k+1)th row based on the amount of light consumed by the light emitter SL in the pixel P in the kth row. Specifically, when the two input grayscale values ​​corresponding to two pixels P arranged along the Y direction (first direction) are equal, the first driving circuit 30 makes the output grayscale value corresponding to the pixel P on the second side 10d side of the two pixels P greater than the output grayscale value corresponding to the pixel P on the first side 10c side of the two pixels P.

[0257] In addition, the first driving circuit 30 uses the aforementioned transmittance ratio to calculate the emitted light I of the pixel P in the (k-1)th row for each selected pixel P. n The amount is the incident light J of pixel P in the k-th row. nThat is, the first driving circuit 30 uses the degree of scattering (first reduction ratio) of the first light generated by the light emitting element SL (light from the light source device 20) in the pixel P on the first side 10c side irradiating the elements of the display panel 10, to calculate the incident light J incident on the pixel P on the second side 10d side. n The amount of light emitted (the amount of light from the light source SL (light source device 20)). Furthermore, the first driving circuit 30 uses the degree of scattering (second reduction ratio) of the second light in the pixel P on the first side 10c side, which varies according to the output grayscale value, to calculate the incident light J incident on the pixel P on the second side 10d side. n The amount of light emitted by the light source SL (light source device 20).

[0258] By calculating the output grayscale value in this way, the brightness of pixel P in the first row of the display area DA in the (i+3)th and (i+4)th columns is (= (incident light J) n The total amount ( Figure 11 ))×(output grayscale value( Figure 12 The brightness of pixel P in the second row of display area DA is 0.8000 (=1.6×0.500), and the brightness of pixel P in the third row of display area DA is 0.7999 (=1.5022×0.5325), and the brightness of pixel P in the fourth row of display area DA is 0.8000 (=1.4602×0.5479).

[0259] Therefore, in pixels P in the (i+3)th and (i+4)th columns, as follows Figure 8 As shown, the input grayscale values ​​are all equal, such as Figure 11 As shown, even when the amount of light emitted by the emitting element SL decreases along the Y direction from the -Y side to the +Y side, the brightness of the pixel P remains approximately equal and is homogenized.

[0260] If k becomes a predetermined value or higher (step S6: Yes), the first driving circuit 30 calculates the output grayscale value corresponding to all pixels P. Thus, the first driving circuit 30 terminates. Figure 10 The program shown performs inverse gamma correction on the output grayscale value to generate a pixel driving signal.

[0261] Next, the above correction coefficients and the brightness of pixels P in the fifth row of the (i+3)th and (i+4)th columns and thereafter will be explained.

[0262] Figure 29 This illustrates a case where a correction coefficient is used in the display device 1 according to an embodiment of this application to calculate the output grayscale value by the first driving circuit 30. Figure 8 , Figure 11 as well as Figure 12 The graph shows the brightness and grayscale values ​​of pixel P in columns i+3 and i+4. Figure 29 The horizontal axis k represents the row number of pixel P in the (i+3)th and (i+4)th columns. Figure 29 The brightness of the vertical axis represents the incident light J in the pixel P of column i+3, column i+4 and row k. n The total amount and the brightness of pixel P. As mentioned above, the input grayscale value of pixel P in column i+3 and column i+4 is 0.8, which is equal to each other and constant.

[0263] Furthermore, as described above, the second light quantity is calculated by multiplying the first light quantity by a correction factor (1.6 in this embodiment). The correction factor is determined to be such that, when the multiple input gray values ​​corresponding to the multiple pixels P arranged along the Y direction are equal to each other by a predetermined gray value, the pixel P located on the side closest to the second side 10d among the multiple pixels P arranged along the Y direction (i.e., Figure 29 The incident light J in pixel P of pixel k = a predetermined value (1080) n The total amount of light becomes the first light quantity. The correction coefficient varies depending on the size and structure of the display panel 10, and is derived in advance through experiments and stored in the storage section of the first drive circuit 30.

[0264] Furthermore, the value of the correction coefficient can be appropriately changed. In this embodiment, the correction coefficient is set to 1.6, but it can also be set to 1.0 or less depending on the display state and usage state. In this case, the second light intensity is less than the first light intensity, and on the other hand, even for images that are dark as a whole as the display area DA, the brightness difference of the image can be reduced.

[0265] The predetermined grayscale value is a grayscale value determined based on the size and structure of the display panel 10, and in this embodiment it is 0.8. That is, the correction coefficient is determined to be a value such that when the multiple input grayscale values ​​corresponding to the multiple pixels P arranged along the Y direction are equal to each other due to the predetermined grayscale value, the incident light J in the pixel P (k = predetermined value (1080) located on the second side 10d side among the multiple pixels P arranged along the Y direction is adjusted. n The total amount of light becomes the first light quantity (1).

[0266] In other words, by using a correction coefficient, when the multiple input grayscale values ​​corresponding to the multiple pixels P arranged along the Y direction (first direction) are equal to each other, when the light intensity of the light source SL (the light intensity of the light source device 20) is the first light intensity, the light intensity (incident light J) incident on the pixel P located in the first row closest to the first side 10c (equivalent to "the fourth pixel closest to the first side 10c") among the multiple pixels P arranged along the Y direction is the first light intensity. n The total amount of light emitted by the luminescent body SL (when the light intensity of the luminescent body SL is greater than the first light intensity, the light intensity of the luminescent body SL incident on the pixel P (equivalent to "the fourth pixel closest to the second side 10d") of a predetermined row of pixels P arranged along the Y direction when the light intensity of the luminescent body SL is greater than the first light intensity) is equal to the light intensity of the incident light J. n The total amount is equal.

[0267] Therefore, through Figure 10 The flowchart shown illustrates how the first driving circuit 30 calculates the incident light J in pixels P at columns i+3 and i+4. n The total amount of light gradually decreases from the second light amount (1.6) to the first light amount (1) between the pixel P of the first row and the pixel P of the predetermined row.

[0268] Additionally, if the grayscale value of each pixel P in the (i+3)th and (i+4)th columns is equal, then by... Figure 10 The flowchart shown illustrates how the first driving circuit 30 calculates the output grayscale values ​​(= input grayscale value / incident light J) in pixels P of the (i+3)th and (i+4)th columns. n The total amount of the quantity gradually increases from 0.5 to 1 between pixels P in the first row and pixels P in the predetermined row.

[0269] Therefore, the brightness of pixel P in column i+3 and column i+4 (=(total amount of incident light Jn)×(output grayscale value)) is approximately 0.8 between pixel P in the first row and pixel P in the predetermined row, and is thus uniformized. Furthermore, since the brightness of pixel P is approximately equal to the input grayscale value, the desired brightness of pixel P can be obtained.

[0270] That is, by using a correction coefficient, the first driving circuit 30 calculates the output grayscale value, so that the incident light J in the pixel P of the first row is... n The total amount of light becomes the light quantity of the emitting body SL as the second light quantity, thereby enabling the brightness of pixel P to be uniform from pixel P in the first row to pixel P in the predetermined row.

[0271] Therefore, when the display device 1 operates to display the entire display area DA as white, that is, when the input grayscale value of each of the plurality of pixels P is equal to each other and set to 1, white is displayed in a state where the brightness of the pixels P is uniform among the plurality of pixels P and the grayscale is uniform in the display area DA. Furthermore, since the brightness of the plurality of pixels P is uniform, when the display device 1 operates to display a portion of the display area DA as red, the complementary color of red (cyan (blue-green)) is not displayed in the display area DA on the second side 10d side, which is closer to that portion of the display area DA. In other words, the desired color is displayed in the display area DA.

[0272] In this way, in the display device 1 with a display panel 10 where light from the light source device 20 is incident from the side, it is possible to achieve uniformity of brightness of the pixels P when the gray values ​​corresponding to the multiple pixels P are equal to each other.

[0273] The preferred embodiments of this application have been described above, but this application is not limited to such embodiments. The content disclosed in the embodiments is merely an example, and various modifications can be made without departing from the spirit of this application. Appropriate modifications made without departing from the spirit of this application are of course also within the technical scope of this application.

[0274] For example, the light source device 20 may also be configured opposite to one of the fourth XL side 15c, fourth XR side 15d, and fourth YF side 15f of the second substrate 15. Alternatively, the light source device 20 may be configured on the side of any one of the first substrate 11, the second substrate 12, and the first substrate 14.

[0275] In addition, in the incident light J n and emitted light I n In this context, the number of integers corresponding to the subscript n and the tilt angle are not limited to the angles mentioned above. In this case, the values ​​stored in the second table are determined based on the magnitude of the tilt angle.

[0276] Furthermore, if the tilt angle increases, the emitted light I from pixel P in the k-th row will... n The incident pixel P in the (k+1)th row sometimes differs from that in the above-described embodiment. For example, in the case of a tilt angle of +45 degrees, Figure 20 The emitted light I from pixel Pt shown n It only incident on pixel P(k, i+1). Furthermore, when the tilt angle is greater than +45 degrees and less than +90 degrees, the emitted light I from pixel Pt... n Light I is incident on pixels P(k, i+1) and P(k, i+2), and the outgoing light I from pixel Pt is... nThe amount is assigned to pixel P(k, i+1) and pixel P(k, i+2).

[0277] In addition, the first driving circuit 30 calculates the incident light J for each pixel P selected in step S4. n In the case of the total amount, the incident light J can also be considered. -3 J -2 J -1 J0, J +1 J +2 J +3 The values ​​are multiplied by weighting coefficients. The weighting coefficients vary depending on the size and structure of the pixel P, are derived in advance through experiments, and are stored in the storage section of the first driving circuit 30.

[0278] Alternatively, the display panel 10 can also be a transmissive liquid crystal panel. In this case, the light source device 20 includes a light emitter SL and a light guide plate (not shown), and is a so-called side-lit backlight disposed on the back side 10b of the display panel 10. The surface of the light guide plate faces the back side 10b of the display panel 10. The light emitter SL is disposed opposite the side of the light guide plate parallel to the first side side 10c of the display panel 10. Light from the light emitter SL enters the light guide plate from the side side of the light guide plate and exits from the side side of the light guide plate toward the back side 10b of the display panel 10. The light from the light emitter SL is consumed in the light guide plate, and the amount of light from the light emitter SL decreases as the light from the light emitter SL enters from the side of the light guide plate toward the opposite side. Therefore, in such a display device, similar to the display device 1 of the above embodiment, there is a technical problem that the brightness of the pixel P decreases from the first side side 10c of the display panel 10 toward the second side side 10d. Therefore, by calculating and outputting grayscale values ​​by the first driving circuit 30 as described in the above embodiment, even if the display panel 10 is a transmissive liquid crystal panel and the light source device 20 is a backlight with a light emitter SL arranged on the side of the light guide plate, it is possible to achieve uniformity of multiple pixels P.

[0279] Furthermore, regarding other effects resulting from the methods described in this embodiment, effects known from the description in this specification, or effects that can be reasonably conceived by those skilled in the art, can of course be understood as being brought about by this application.

[0280] Explanation of reference numerals in the attached figures

[0281] 1. Display device; 10. Display panel; 10c. First side of display panel; 10d. Second side of display panel; 13. Liquid crystal layer; 14. First substrate; 20. Light source device; 30. First driving circuit (driving circuit); DA. Display area; P. Pixel.

Claims

1. A display device, characterized in that, have: The display panel has a display area that overlaps with a plurality of pixels arranged in a matrix when viewed from above, and has a first side and a second side opposite to the first side. A light source device is disposed on the first side side, emitting light incident from the first side side to the second side side; as well as The driving circuit calculates the output grayscale value based on the input grayscale value of the image signal, and outputs the pixel driving signal corresponding to the output grayscale value to the corresponding pixel among the plurality of pixels. The plurality of pixels includes a first pixel and a second pixel arranged along a first direction from the first side toward the second side. The second pixel is located between the first pixel and the second side. When the input grayscale value corresponding to the first pixel and the input grayscale value corresponding to the second pixel are equal, the driving circuit makes the output grayscale value corresponding to the second pixel greater than the output grayscale value corresponding to the first pixel. The input grayscale value and the output grayscale value are determined for the color of each pixel. In a frame, an image can be visually confirmed by sequentially emitting light of the corresponding color with a gray level corresponding to the output gray value from the pixels selected as pixels that emit light of the corresponding color.

2. The display device according to claim 1, characterized in that, The light emitted from the light source device propagates within the display panel from the first side towards the second side. The display panel has a liquid crystal layer comprising a polymer-dispersed liquid crystal.

3. The display device according to claim 2, characterized in that, The light emitted by the light source device propagates within the display panel along the first direction and along a second direction inclined relative to the first direction when viewed from above. The driving circuit uses the input grayscale value of the pixel, the amount of light incident on the pixel along the first direction, and the amount of light incident on the pixel along the second direction to calculate the output grayscale value.

4. The display device according to claim 3, characterized in that, The light incident on the first pixel is scattered within the first pixel. The driving circuit calculates the amount of light incident on the second pixel based on the degree of scattering of the light incident on the first pixel.

5. The display device according to claim 4, characterized in that, The first pixel has a switching element. The light incident on the first pixel is scattered when it hits the switching element.

6. The display device according to claim 3, characterized in that, The driving circuit calculates the amount of light incident on the second pixel based on the degree of scattering of the light incident on the first pixel, which changes in accordance with the magnitude of the output gray value.

7. The display device according to claim 3, characterized in that, The plurality of pixels are arranged in a matrix along the first direction and a third direction orthogonal to the first direction when viewed from above. The plurality of pixels includes a third pixel adjacent to the second pixel along the third direction. The light incident on the first pixel exits from the first pixel and is incident on the second pixel and the third pixel.

8. The display device according to claim 7, characterized in that, Based on the tilt angle of the second direction relative to the first direction, the amount of light emitted from the first pixel is allocated into the amount of light incident on the second pixel and the amount of light incident on the third pixel.

9. The display device according to claim 1, characterized in that, When the input grayscale value corresponding to the first pixel and the input grayscale value corresponding to the second pixel are equal, the driving circuit corrects at least one of the first output grayscale value calculated based on the input grayscale value corresponding to the first pixel and the second output grayscale value calculated based on the input grayscale value corresponding to the second pixel. When the input grayscale value corresponding to the first pixel and the input grayscale value corresponding to the second pixel are different, the driving circuit does not correct either the first output grayscale value or the second output grayscale value.

10. The display device according to claim 1, characterized in that, The plurality of pixels includes a plurality of fourth pixels arranged along the first direction. When the input grayscale values ​​corresponding to the plurality of fourth pixels are equal to each other and the light source device emits light of a first emitted amount, light of the first incident amount is incident on the pixel among the plurality of fourth pixels that is closest to the first side. When the input grayscale values ​​corresponding to the plurality of fourth pixels are equal to each other, and the light source device emits light of a second emitted amount greater than the first emitted light amount, light of the second incident amount is incident on the pixel among the plurality of fourth pixels that is closest to the second side. The first incident light intensity is equal to the second incident light intensity.

11. A display device, characterized in that, have: The display panel has a first side; Multiple light-emitting elements are positioned opposite the first side surface; The first pixel is configured on the display panel; The second pixel is disposed on the display panel, located on the side opposite to the first side, separated from the first pixel; as well as The driving circuit is supplied with an image signal having a first input grayscale value corresponding to the first pixel and a second input grayscale value corresponding to the second pixel. The driving circuit outputs a first pixel driving signal having a first output grayscale value corresponding to the first input grayscale value to the first pixel, and outputs a second pixel driving signal having a second output grayscale value corresponding to the second input grayscale value to the second pixel. When the first input grayscale value is equal to the second input grayscale value, the second output grayscale value is greater than the first output grayscale value. The first input grayscale value, the second input grayscale value, the first output grayscale value, and the second output grayscale value are determined for the color of each pixel. In a frame, an image can be visually confirmed by sequentially emitting light of the corresponding color with a gray level corresponding to the first output gray level from the first pixel selected as the pixel emitting light of the corresponding color, and sequentially emitting light of the corresponding color with a gray level corresponding to the second output gray level from the second pixel selected as the pixel emitting light of the corresponding color.

12. The display device according to claim 11, characterized in that, The display panel also has a second side opposite to the first side in a first direction. The first pixel and the second pixel are arranged along the first direction.

13. The display device according to claim 11, characterized in that, The display panel includes: Liquid crystal layers, including polymer-dispersed liquid crystals; The first substrate and the second substrate are separated by the liquid crystal layer; as well as The third substrate is located on the opposite side of the first substrate, separated from the second substrate. The third substrate has the first side surface. The light emitted by the plurality of light emitters enters the third substrate from the first side and exits from the main surface of the third substrate.

14. A display device, characterized in that, have: The display panel includes a liquid crystal layer comprising a polymer-dispersed liquid crystal, a first substrate and a second substrate separated by the liquid crystal layer, and a third substrate, wherein the second substrate is located between the first substrate and the third substrate; Multiple light-emitting elements are positioned opposite the side surface of the third substrate; The first pixel is configured on the display panel; The second pixel is disposed on the display panel and is located at a position arranged with the first pixel in a direction intersecting the side; as well as The driving circuit is supplied with an image signal having a first input grayscale value corresponding to the first pixel and a second input grayscale value corresponding to the second pixel. The light emitted by the plurality of light emitters enters the third substrate from the side and exits from the main surface of the third substrate. The driving circuit outputs a first pixel driving signal having a first output grayscale value corresponding to the first input grayscale value to the first pixel, and outputs a second pixel driving signal having a second output grayscale value corresponding to the second input grayscale value to the second pixel. When the first input grayscale value is equal to the second input grayscale value, the second output grayscale value is greater than the first output grayscale value. The first input grayscale value, the second input grayscale value, the first output grayscale value, and the second output grayscale value are determined for the color of each pixel. In a frame, an image can be visually confirmed by sequentially emitting light of the corresponding color with a gray level corresponding to the first output gray level from the first pixel selected as the pixel emitting light of the corresponding color, and sequentially emitting light of the corresponding color with a gray level corresponding to the second output gray level from the second pixel selected as the pixel emitting light of the corresponding color.

Images