Feedback type pixel circuit
By adding feedback transistors to the pixel circuits of OLED and LED displays, the brightness uneven problem caused by inconsistent characteristics of different pixel drive tubes is solved, and better brightness uniformity is achieved, especially significantly improved at low grayscale.
Patent Information
- Application Number
- CN202510445285.8
- Authority / Receiving Office
- CN · China
- Patent Type
- Applications(China)
- Current Assignee / Owner
- Filing Date
- 2025-04-10
- Publication Date
- 2025-06-10
AI Technical Summary
The active driving circuits of existing OLED and LED displays have inconsistent characteristics of different pixels due to manufacturing deviations, resulting in uneven luminance luminance, especially at low gray levels.
A feedback transistor is added to a conventional pixel circuit. Specifically, the feedback transistor is in a linear region when emitting light, and the feedback transistor is specifically between the driver tube source and the feedback tube, or between the driver tube source and the light emitting device.
Through the feedback suppression mechanism, the luminous current of different pixels is adjusted, the brightness difference between pixels is reduced, and the brightness uniformity is improved, especially at low grayscales.
Smart Images
Figure CN120126418A_ABST
Abstract
Description
Technical Field
[0001] The present invention relates to a display pixel circuit, especially for the active driving power supply lines of OLED and LED, belonging to the field of display technology. Background Art
[0002] An active driving circuit is generally used to drive a display of array pixels, and its feature is that an active device such as a transistor is provided for each pixel. For a pixel self-luminous display such as OLED and micro-LED, its active driving is generally current-type driving, that is, the light-emitting device of each pixel controls the current through the active device to achieve driving. The most typical pixel structure of the active driving circuit is 2T1C, that is, each pixel is provided with 2 transistors and 1 capacitor. Generally, through the row scanning transistor 1, the data (voltage) is written into the capacitor, and then the driving transistor (transistor 2) connected in series with the light-emitting device is controlled through the capacitor, thereby realizing the adjustment of the current and the light-emitting brightness of the light-emitting device.
[0003] However, due to manufacturing deviations, the characteristics of the driving transistors of different pixels are not consistent, resulting in the problem of uneven light-emitting brightness of different pixels.
[0004] The 7T1C architecture (7 transistors + 1 capacitor) proposed by Samsung can solve the problem of uneven threshold values (Vth) of the driving transistor characteristics through compensation, and its principle operation is divided into three steps. As shown in Figure 1.
[0005] In the first reset stage, the N-1 signal is low, the switching transistors T1 and T2 are turned on, the negative voltage Vini is written into the anode a point of the OLED through the switching transistor T1, and the negative voltage Vini is written into the gate DrG of the driving transistor through the switching transistor T2.
[0006] In the second data writing and threshold compensation stage, the N signal is low, the switching transistors T3 and T4 are turned on, the data information passes through the switching transistor T3 and reaches the gate DrG of the driving transistor through the switching transistor T4. At this time, the gate voltage of the driving transistor is Vdata + Vth, and Vth is negative.
[0007] In the third light-emitting stage, the Em signal is low, the switching transistors T5 and T6 are turned on, Vdd is applied to the source of the driving transistor, the gate-source voltage |Vgs| of the driving transistor = |Vdata + Vth - Vdd|, and the overdrive voltage |Vgs| - |Vth| of the driving transistor = |Vdata - Vdd|. The light-emitting current formula is I=(1 / 2)*u*Cox*(W / L)*(Vgs-Vth) 2 =(1 / 2)*u*Cox*(W / L)*(Vdata-Vdd) 2 .
[0008] The emission current forms a closed-loop circuit by flowing from Vdd through the switching transistor 5, the driving transistor DrT, the switching transistor 6, and the light-emitting device to Vss.
[0009] As can be seen from the above analysis, although the non-uniformity caused by the threshold Vth of the driving transistor is eliminated, the influence of the non-uniformity of u, Cox, and W / L still exists. This leads to display non-uniformity. Especially when the required emission current value is small at low gray levels, the non-uniformity is more serious. Summary of the Invention
[0010] In the present invention, a feedback transistor is added to the source electrode of the driving transistor in a traditional pixel circuit. For a traditional pixel circuit composed of P-type transistors, during light emission, the feedback transistor is between Vdd and the source electrode of the driving transistor, as Figure 2 . For a traditional pixel circuit composed of N-type transistors, during light emission, the feedback transistor is between the source electrode of the driving transistor and the light-emitting device, as Figure 3 , or between the light-emitting device and Vss, as Figure 4 .
[0011] During the light-emitting stage of the light-emitting device, the feedback transistor is in the linear region, that is, the gate-source voltage |Vgs| > |Vth| and the gate-drain voltage |Vgd| > |Vth| of the feedback transistor. The equivalent resistance Ron of the feedback transistor when linearly conducting is Ron = Vds / Ion. The emission current Ion is usually at the NA level, and Vds is usually small at the mV level. In this way, Ron will have a very large value; while the sheet resistance of the conductor trace is about 3500Ω per square; under the same spatial area, the equivalent resistance Ron of the feedback transistor is much larger than the conventional resistance made by the conductor trace, and the value of the equivalent resistance Ron of the transistor can be adjusted by |Vg| after fabrication, but after the conductor trace is fabricated, the resistance value cannot be adjusted. Therefore, the feedback transistor plays a major feedback suppression role.
[0012] Next, analyze the principle of feedback suppression after using this feedback transistor.
[0013] Take the difference in the mobility u of the driving transistor as an example; the analysis principle of the differences in Cox and (W / L) is the same as the analysis process of u.
[0014] With other parameters being the same, only the analysis operation process for the difference in u: a. The mobility u of the driving transistor, in the B pixel circuit is greater than that in the A pixel circuit, and the current I is proportional to the mobility u. b. The original emission current I output by the driving transistor, in the B pixel circuit is greater than that in the A pixel circuit. c. When there is a feedback transistor, during light emission, the source-drain voltage difference |Vfsd| = |Vfs - Vfd| of the feedback transistor, in the B pixel circuit is greater than that in the A pixel circuit. d. When there is a feedback transistor, the overdrive voltage of the driving transistor during light emission |Vgs| - |Vth| = |Vdd - Vfsd - Vdrg|, and the B pixel circuit is smaller than the A pixel circuit. e. The new current output by the driving transistor is proportional to |Vgs| - |Vth|, I = (1 / 2) * u * Cox * (W / L) * (Vgs - Vth) 2 , and the feedback transistor suppresses the light emission current I. The B pixel circuit is more suppressed, and the A pixel is less suppressed.
[0015] It can be seen that after adding the feedback transistor, due to the source-drain voltage Vfsd regulation function of the feedback transistor, the B pixel with a large original current is more suppressed, and the A pixel with a small original current is less suppressed. It can be seen that the difference between the new currents generated by the A and B pixels becomes smaller, making the brightness of the A and B pixels more uniform.
[0016] For traditional pixels, such as Samsung's 7T1C circuit, the overdrive voltage of the driving transistor |Vgs| - |Vth| = |Vdd - Vdrg|. Without the regulation of the source-drain voltage difference |Vfsd| of the feedback transistor, the difference in mobility between the driving transistors of the A and B pixels is directly reflected in the current and brightness, resulting in uneven brightness between the two, that is, traditional pixels have no ability to suppress the difference in driving transistor mobility.
[0017] Taking the Vdd difference caused by IR-drop as an example, the analysis of the kink effect is the same as the Vdd analysis. a. The power supply voltage Vdd, the proximal B pixel circuit is greater than the distal A pixel circuit. b. The overdrive voltage of the driving transistor during light emission |Vgs| - |Vth| = |Vdd - Vdrg|, and the proximal B pixel circuit is greater than the distal A pixel circuit. c. The original current output by the driving transistor is proportional to |Vgs| - |Vth|, and the original light emission current, the proximal B pixel is greater than the distal A pixel. d. When there is a feedback transistor, the source-drain voltage difference of the feedback transistor during light emission |Vfsd| = |Vfs - Vfd|, and the B pixel circuit is greater than the A pixel circuit. e. When there is a feedback transistor, the overdrive voltage of the driving transistor during light emission |Vgs| - |Vth| = |Vdd - Vfsd - Vdrg|, and the B pixel circuit is smaller than the A pixel circuit. f. The new current output by the driving transistor is proportional to |Vgs| - |Vth|, I = (1 / 2) * u * Cox * (W / L) * (Vgs - Vth) 2 , and the feedback transistor suppresses the light emission current I. The B pixel circuit is more suppressed, and the A pixel is less suppressed.
[0018] It can be seen that after adding the feedback tube, due to the source-drain voltage Vfsd regulation function of the feedback tube, the B pixels with a large original current are more suppressed, and the A pixels with a small original current are less suppressed. It can be seen that the difference between the new currents generated by the AB pixels becomes smaller, making the brightness of the AB pixels more uniform.
[0019] For traditional pixels, such as the 7T1C circuit of Samsung, the overdrive voltage of the driving transistor |Vgs|-|Vth| = |Vdd-Vdrg|, without the regulation of the source-drain voltage difference |Vfsd| of the feedback transistor, the difference between Vdd of the AB pixels is directly reflected in the current and brightness, resulting in uneven brightness of the two, that is, traditional pixels have no ability to suppress the Vdd difference.
[0020] The abbreviated names of the present invention are described as follows: Vdd: Light-emitting power supply, Vss: Cathode power supply of the light-emitting device, Vdata: Data voltage, DrG\DrS\DrD: Gate, source, and drain of the driving transistor, taking Dr for Driver, FG\FS\FD: Gate, source, and drain of the feedback transistor, taking f for feedback, Vfs\Vfd\Vfg: Source, drain, and gate voltages of the feedback transistor, Vfsd: Source-drain voltage of the feedback transistor, Vgs: Gate-source voltage of the transistor, Vds: Source-drain voltage of the transistor, Vgs-Vth: Overdrive voltage of the transistor, Vth is the threshold voltage of the transistor, U\Cox\W\L: Mobility, gate oxide capacitance, width, and length of the transistor, N-1\N: Row scan signal, output by NGOA (gate driver on array), well-known in the industry Em: Row light-emitting switch signal, output by EGOA (gate driver on array), well-known in the industry S: Row scan signal, output by SGOA (gate driver on array), well-known in the industry Description of the Drawings
[0021] Figure 1a . Samsung pixel circuit Figure 1b . Samsung pixel circuit timing Figure 2 . Adding a feedback tube to the source of the P-type driving transistor Figure 3. Add a feedback transistor to the source of the N-type driving transistor Figure 4 . Add a feedback transistor to the source of the N-type driving transistor, and the position of the feedback transistor is between the light-emitting device and Vss Figure 5 . Add a feedback transistor to the traditional pixel circuit 1 Figure 6 . Add a feedback transistor to the source of the P-type driving transistor, which is between the EM switching transistor and the driving transistor Figure 7a . Add a feedback transistor to the source of the P-type driving transistor, and a routing resistor is connected in series at both ends Figure 7b . Add a feedback transistor to the source of the P-type driving transistor, and a routing resistor is connected to the drain Figure 7c . Add a feedback transistor to the source of the P-type driving transistor, and a routing resistor is connected to the source Figure 8a . Add a feedback transistor to the traditional pixel circuit 2 Figure 8b . The corresponding timing sequence Detailed implementation method Embodiment 1
[0022] In the case where the traditional pixel circuit has an EM switching transistor between the driving transistor and VDD, such as Samsung's 7T pixel circuit. Add a feedback transistor to the source of its driving transistor, and the position is between VDD and the Em switching transistor T5. As shown in Figure 5 Therefore.
[0023] The source FS of the feedback transistor is connected to VDD, the drain FD is connected to the source of the transistor T5. At least during the light-emitting stage, the feedback transistor is in the linear region, that is, |Vgs| > |Vth|, |Vgd| > |Vth|. Therefore, its gate FG is connected to a low-voltage signal. If it is connected to a pulse signal, it needs to be a low-voltage signal at least during light emission. So FG can be connected to the a node, or the drain of the driving transistor DrD, or the Em signal, or an additional separate signal. At least during the light-emitting stage, it is required to be at a low voltage and meet the basic requirement that the feedback transistor is in the linear region.
[0024] If connected to the Em signal, it is equivalent to the feedback transistor and the switching transistor T5 being in series, and can be combined into one transistor, but the new transistor (L / W) ≥ 2.
[0025] The working steps are the same as those of the traditional pixel circuit, and the working timing sequence is the same as Figure 1b , except that feedback suppression is added during light emission.
[0026] In the first step, the reset stage T1, the N-1 signal is low, the switching transistors T1 and T2 are turned on, the negative voltage Vini is written into the anode a point of the OLED through the switching transistor T1, and the negative voltage Vini is written into the gate DrG of the driving transistor through the switching transistor T2.
[0027] In the second step, during the data and threshold compensation stage T2, when the N signal is low, the switching transistors T3 and T4 are turned on. The data information passes through the switching transistor T3 and reaches the gate DrG of the driving transistor through the switching transistor T4. At this time, the gate voltage of the driving transistor is Vdata + Vth, where Vth is negative.
[0028] In the third step, during the light-emitting stage, when the Em signal is low, the switching transistors T5 and T6 are turned on. FG is low and the feedback transistor is in the linear state. Vdd is supplied to the source of the driving transistor, and the gate-source voltage of the driving transistor |Vgs| = |Vdata + Vth - (Vdd - Vfsd)|. The overdrive voltage of the driving transistor |Vgs| - |Vth| = |Vdd - Vfsd - Vdata|, where the source-drain voltage of the feedback transistor Vfsd = Vfs - Vfd, which is proportional to the flowing current.
[0029] The light-emitting current formula is I = (1 / 2) * u * Cox * (W / L) * (Vgs - Vth)^2 = (1 / 2) * u * Cox * (W / L) * (Vdd - Vfsd - Vdata)^2.
[0030] The light-emitting current flows from Vdd through the feedback transistor, switching transistor 5, driving transistor DrT, switching transistor 6, and the light-emitting device to Vss, forming a closed-loop circuit.
[0031] If the u, Cox, (W / L), and VDD of pixel B are greater than those of pixel A, then the original light-emitting current Ib of pixel B is greater than the original light-emitting current Ia of pixel A. Then, Vfsdb of pixel B is greater than Vfsda of pixel A, and the overdrive voltage |Vgs| - |Vth| of the driving transistor of pixel B is less than that of the driving transistor of pixel A. Therefore, the light-emitting current of pixel B is more suppressed by Vfsdb, and the light-emitting current of pixel A is less suppressed by Vfsda.
[0032] That is, pixel B with a larger original current is more suppressed, and pixel A with a smaller original current is less suppressed. It can be seen that the new currents generated by pixels A and B are closer to each other, making the brightness of pixels A and B more uniform. That is, the feedback transistor adjusts pixels A and B to be more uniform. Embodiment 2
[0033] Such as Figure 6 , the feedback transistor is between the Em switching transistor and the driving transistor. Similarly, at least during the light-emitting stage, the feedback transistor is in the linear region. For this embodiment, other transistors do not affect the description, so they are omitted.
[0034] During light emission, the feedback suppression principle is equivalent to that of Embodiment 1 ( Figure 5), that is, the feedback suppression for pixels with a large original current is large, and the feedback suppression for pixels with a small original current is small. Therefore, the process will not be elaborated here.
[0035] As can be seen from Embodiment 1 and Embodiment 2, in the case where there is an EM switch transistor between VDD and the driving transistor in the original pixel circuit, the feedback transistor is between VDD and the Em switch transistor, or between the EM switch transistor and the driving transistor. When emitting light, the principle of the equivalent feedback suppression effect is the same. Embodiment 3
[0036] Such as Figure 7a , Figure 7b , Figure 7c , here the relationship between the feedback transistor and the trace resistance is described. There are trace resistances on both the source and drain sides of the feedback transistor, a trace resistance at the source end of the feedback transistor, and a trace resistance at the drain end of the feedback transistor. For this embodiment, other transistors do not affect the description, so they are omitted. Similarly, at least in the light-emitting stage, the feedback transistor is in the linear region.
[0037] The feedback transistor is in the linear region. When the feedback transistor is linearly conducting, the equivalent resistance Ron = Vds / Ion. The light-emitting current Ion is usually at the NA level, and Vds is usually small at the mV level. In this way, Ron will have a very large value; while the sheet resistance of the conductor trace is about 3500 Ω per square; in the same spatial area, the equivalent resistance Ron of the feedback transistor is much larger than the conventional resistance made by the conductor trace, and the value of the equivalent resistance Ron of the transistor can be adjusted by |Vg| after fabrication, but after the conductor trace is fabricated, the resistance value cannot be adjusted.
[0038] Therefore, the feedback transistor plays a major feedback suppression role.
[0039] When emitting light, the feedback suppression principle is equivalent to that of Embodiment 1 ( Figure 5 ), that is, the feedback suppression for pixels with a large original current is large, and the feedback suppression for pixels with a small original current is small. Therefore, the process will not be elaborated here. Embodiment 4
[0040] In the case where the traditional pixel circuit directly connects VDD and the driving transistor, a feedback transistor is added. In addition to the feedback transistor being in the linear region in the light-emitting stage, according to the requirements of the traditional pixel, in the compensation stage, the feedback transistor also needs to be turned on and be in the linear region, as shown in Figure 8.
[0041] The working steps are the same as those of the traditional circuit, except that feedback suppression is added when emitting light.
[0042] The first step is the reset stage T1. N-1 is low, and the switching transistor T1 and the switching transistor T2 are turned on. Through the transistor T1, Vini is written into the a node to reset the light-emitting device; through the transistor T2, Vini is written into the DrG node to reset the gate of the driving transistor.
[0043] In the second compensation stage T2, N is low, turning on the switching transistor T3; FG is low to turn on the feedback transistor; S is low to turn on the transistor T4; VDD is written to the DrG point through the feedback transistor and the T3 transistor, Vdrg = Vdd + Vth, where Vth is negative. Vofs is written to point b through the data line, Vb = Vofs.
[0044] In the third data writing stage T3, S is low to keep the transistor T4 turned on, and Vdata is written to point b through the data line, Vb = Vdata. At this time, both N-1 and N are high, and DrG is in a floating state. Vdrg = Vdd + Vth + C1*(Vdata - Vofs) / Ctotal, where Vth is negative and Ctotal is the total capacitance of the DrG node. |Vgs| = |(Vdd - Vfsd) - Vdd + |Vth| - C1*(Vdata - Vofs) / Ctotal|, and the source-drain voltage of the feedback transistor |Vfsd| = |Vfs - Vfd|. The overdrive voltage |Vgs| - |Vth| = |C1*(Vofs - Vdata) / Ctotal - Vfsd|.
[0045] In the fourth light-emitting stage T4, FG is low and the feedback transistor is in the linear state, where Vfsd = Vfs - Vfd, which is proportional to the flowing current, and Em is low to turn on the switching transistor T5.
[0046] The light-emitting current I = (1 / 2)*u*Cox*(W / L)*(Vgs - Vth) 2 = (1 / 2)*u*Cox*(W / L)*(C1*(Vofs - Vdata) / Ctotal - Vfsd) 2 . The light-emitting current flows from VDD, through the feedback transistor, the driving transistor, the transistor T5, and the light-emitting device to Vss, forming a closed loop.
[0047] If the u, Cox, (W / L), and VDD of pixel B are greater than those of pixel A, then the original light-emitting current Ib of pixel B is greater than the original light-emitting current Ia of pixel A, then Vfsdb of pixel B is greater than Vfsda of pixel A, then the overdrive voltage |Vgs| - |Vth| of the driving transistor of pixel B is less than that of the driving transistor of pixel A, then the light-emitting current of pixel B is more suppressed by Vfsdb, and the light-emitting current of pixel A is less suppressed by Vfsda.
[0048] That is, pixel B with a large original current is more suppressed, and pixel A with a small original current is less suppressed. It can be seen that the difference between the new currents generated by pixels A and B becomes smaller. This makes the brightness of pixels A and B more uniform. That is, the feedback transistor makes pixels A and B more uniform.
[0049] In the same way, for this type of traditional pixel circuit, there is also a situation where a feedback transistor is paired with a trace resistor, and it is also the feedback transistor that plays the main role in feedback suppression.
[0050] For a traditional pixel circuit composed of N-type transistors, a source feedback transistor is also added to the driving transistor. Similarly, at least during the light-emitting period, the feedback transistor is in the linear region. The feedback transistor can be located between the driving transistor and the light-emitting device, or between the light-emitting device and the cathode Vss. The principle of feedback suppression is the same, that is, the feedback suppression for pixels with a large original current is large, and the feedback suppression for pixels with a small original current is small.
[0051] Those skilled in the art should understand that traditional pixel circuits cannot be enumerated exhaustively, and this patent only gives examples of traditional pixel circuits. The spirit and principle of the present invention lie in adding a feedback transistor to the source of the driving transistor. At least during the light-emitting stage, the feedback transistor is in the linear region, forming feedback suppression on the original current, so as to improve the pixel uniformity. Any modification made based on the spirit and principle of the present invention should be within the protection scope of the present invention.
Claims
1. A feedback pixel circuit, characterized in that: Contains: conventional pixel circuit and feedback transistor; At least in the light emitting stage, the feedback transistor operates in a linear region, the gate-source voltage |Vgs|>threshold |Vth|, the gate-drain voltage |Vgd|>threshold |Vth|; In a traditional pixel circuit composed of P-type transistors, the feedback tube is located between the light-emitting power source Vdd and the source of the driving tube; In a traditional pixel circuit composed of N-type transistors, the feedback tube is located between the source of the driving tube and the cathode Vss; A conventional pixel circuit is a transistor other than a feedback transistor in a pixel circuit.
2. Based on claim 1, the gate FG of the feedback transistor is connected to a signal added outside the traditional pixel circuit.
3. Based on claim 1, in a conventional pixel circuit composed of P-type transistors, the gate FG of the feedback transistor is connected to any node between the drain of the driving tube and the cathode VSS in the conventional pixel circuit.
4. Based on claim 1, in a conventional pixel circuit composed of N-type transistors, the gate FG of the feedback transistor is connected to any node between the light emitting power source Vdd and the drain of the driving tube in the conventional pixel circuit.
5. Based on claim 1, the gate FG of the feedback transistor is connected to the gate control signal (Em signal) of the switching transistor in the light-emitting path of the traditional pixel circuit.
6. Based on claim 5, the feedback transistor and the switch transistor are combined to form a new transistor, and the new transistor is characterized by an aspect ratio greater than or equal to 2.