DC-DC conversion multi-power output device

Abstract

The invention relates to the technical field of DC power conversion, in particular to a DC-DC conversion multi-power output device. The DC-DC converter comprises at least one DC-DC converter and an inductor connected in series. The inductor is connected in series with a middle point of a DC-DC conversion circuit formed by the DC-DC converters; the pulse amplitude of a middle point is adjusted through series connection of inductors, and then needed voltage reduction output is obtained through rectification and filtering; or, through the switching pulse output by the DC-DC converter, the series capacitor is charged, the pulse voltage offset after capacitor coupling is adjusted, and the coupled pulse is rectified to obtain the required negative voltage or boost output. According to the invention, the intermediate point pulse amplitude is adjusted through series connection of inductors, and then needed voltage reduction output is obtained through rectification and filtering; the switching pulse output by the DC-DC converter is utilized to charge the series capacitor, the pulse voltage offset after capacitor coupling is adjusted, and the needed negative voltage or boost output is obtained after the coupled pulse is rectified.

Description

Technical Field

[0001] The present invention relates to the technical field of direct current power conversion, and in particular to a DC-DC conversion multi-power output device. Background Art

[0002] Many electronic devices use multiple DC voltage power supplies, and each additional working voltage requires a corresponding conversion circuit to achieve this. Not only is the circuit complex, the occupied area increases, but the cost is also increased. Sometimes, the requirements for the additional working voltage are not high, the current is small, and the contribution to the power consumption of the whole machine is relatively small. Under such requirements, it is usually hoped that a simpler method can be used to obtain the required working voltage, which can not only reduce costs, but also make the circuit more compact.

[0003] The main power supply voltage of the device is usually higher than most working power supply voltages, and a DC-DC converter is needed to step down the higher voltage to various working voltages, such as 3.3V for digital logic circuits or RF circuits, 1.1V for processor cores, 1.8V for SOC analog circuits, and 5V for analog amplifiers. In addition, some devices also require negative voltages for op amps or comparators.

[0004] There are two main methods for voltage conversion: LDO and DC-DC conversion. LDO is used to add a voltage slightly lower than the existing working voltage. The added voltage is achieved by directly stepping down the voltage. It can also be achieved with a voltage regulator diode. If the difference between the added voltage and the existing working voltage is large, a DC-DC converter is needed. DC-DC can achieve boost, buck, and reverse polarity conversion. The most basic step-down DC-DC conversion circuit contains the following components: input filter capacitor, DC-DC converter, energy storage inductor, output smoothing capacitor, feedback voltage divider resistor, see Figure 1 As shown in the figure, under normal circumstances, if a DC-DC converter is used to output multiple voltages, a multi-winding transformer is required, and each winding can output one voltage. Among them, the one with the highest voltage requirement stabilizes the output voltage through negative feedback, and the other windings are open-loop outputs, and their voltage levels are affected by changes in load current.

[0005] Most devices use independent DC-DC or LDO to obtain multi-voltage power supply. The advantage is that the converted output voltage is stable and is basically not affected by load changes. But the disadvantages are also obvious. When the circuit has low requirements for voltage stability and the load current is small, using a separate DC-DC converter to obtain a new working voltage is not flexible enough, the circuit is complex, the cost is increased, and each additional DC-DC conversion will increase an EMI interference frequency.

[0006] Based on this, a DC-DC conversion multi-power output device is proposed, which adopts a simpler method to obtain multiple working power supply voltages, reduce circuit complexity and reduce costs. Summary of the invention

[0007] The purpose of the present invention is to solve the problems existing in the background technology and propose a DC-DC conversion multi-power output device; the present invention can generate multiple working voltages with only one DC-DC converter, and the circuit is simple; compared with the existing solutions, it occupies a small area and has low cost.

[0008] The technical solution of the present invention is a DC-DC conversion multi-power output device, comprising at least one DC-DC converter and an inductor connected in series;

[0009] The inductor is connected in series at the middle point of the DC-DC conversion circuit composed of the DC-DC converter;

[0010] By connecting the inductor in series, the pulse amplitude at the midpoint is adjusted, and then the required step-down output is obtained after rectification and filtering;

[0011] Alternatively, the series capacitor is charged by the switching pulse output by the DC-DC converter, the pulse voltage offset after capacitor coupling is adjusted, and the coupled pulse is rectified to obtain the required negative voltage or boost output.

[0012] A DC-DC converter and a series inductor form a buck, negative voltage or boost DC-DC conversion circuit.

[0013] Preferably, in the step-down DC-DC conversion circuit, Vin is input to the power input port of the DC-DC converter; the switch port of the DC-DC converter is connected to the inductor L1;

[0014] The first output terminal of the inductor L1 is connected in series with the anode of the Schottky diode D1 and then connected to the second output terminal Vout2;

[0015] The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1;

[0016] The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1;

[0017] One end of the capacitor C1 is connected to Vin, and the other end is grounded; one end of the capacitor C2 is connected to the output end of the inductor L2, and the other end is grounded; one end of the capacitor C3 is connected to the output end of the Schottky diode D1, and the other end is grounded.

[0018] When the load of Vout2 is light enough, much smaller than the load current of Vout1, its current is ignored; then the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is:

[0019] V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2);

[0020] Switch output voltage V SW The peak value is equal to Vin, and Schottky diode D1 rectifies the positive half-cycle of the pulse. The output voltage is approximately equal to the peak voltage of the positive half-cycle of the pulse minus the conduction voltage V of diode D1. DIODE ,Right now

[0021] Vout2≈Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE

[0022] When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the output voltage of Vout2 to decrease. Since Vout2 works in an open loop, its voltage is affected by the load current.

[0023] Preferably, in the negative voltage DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1;

[0024] The first output terminal of the inductor L1 is connected in series with the capacitor C4 and the cathode of the Schottky diode D1 in sequence, and then connected to the second output terminal Vout2;

[0025] The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1;

[0026] The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1;

[0027] One end of capacitor C1 is connected to Vin, and the other end is grounded; one end of capacitor C2 is connected to the output end of inductor L2, and the other end is grounded; one end of capacitor C3 is connected to the anode of Schottky diode D1, and the other end is grounded; the anode of Schottky diode D2 is connected to the output end of capacitor C4, and the cathode is grounded.

[0028] When the load of Vout2 is light enough, much smaller than the load current of Vout1, and its current is ignored, the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is:

[0029] V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)

[0030] The voltage across capacitor C4 is approximately the peak voltage of the positive half-cycle of the pulse at point A minus the conduction voltage V of D2. DIODE , that is, V C4 =Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE；

[0031] The voltage at point B is V B =V A -V C4 ,Right now:

[0032] V B =V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)-[Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE ]

[0033] =(V SW -Vin)*L2 / (L1+L2)+V DIODE

[0034] Schottky diode D1 rectifies the negative half cycle of the pulse, and the output voltage is approximately equal to the peak voltage of the negative half cycle of the pulse plus the diode conduction voltage V DIODE；

[0035] V SW The peak voltage of the negative half cycle of -Vin is equal to -Vin, so

[0036] Vout2≈-Vin*L2 / (L1+L2)+2*V DIODE

[0037] When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the negative voltage amplitude of Vout2 to change; since Vout2 works in an open loop, its voltage is affected by the load current.

[0038] Preferably, in the boost DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1;

[0039] The first output terminal of the inductor L1 is connected in series with the capacitor C4 and the anode of the Schottky diode D1 in sequence, and then connected to the second output terminal Vout2;

[0040] The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1;

[0041] The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1;

[0042] One end of capacitor C1 is connected to Vin, and the other end is grounded; one end of capacitor C2 is connected to the output end of inductor L2, and the other end is grounded; one end of capacitor C3 is connected to the cathode of Schottky diode D1, and the other end is grounded; the anode of Schottky diode D2 is connected to Vin, and the cathode is connected to the output end of capacitor C4 and the anode of D1.

[0043] If the load of Vout2 is light enough, much smaller than the load current of Vout1, and its current is ignored, the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is:

[0044] V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)

[0045] V SW The negative half-cycle voltage is 0, and the voltage across capacitor C4 is approximately the peak voltage of the pulse negative half-cycle at point A minus the conduction voltage V of Vin and diode D2. DIODE The difference, V C4 =Vout1*L1 / (L1+L2)-(Vin-V DIODE );

[0046] The voltage at point B is V B =V A -V C4 ,Right now:

[0047] V B =V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)-[Vout1*L1 / (L1+L2)-Vin+V DIODE ]

[0048] =V SW *L2 / (L1+L2)+Vin-V DIODE

[0049] Schottky diode D1 rectifies the positive half cycle of the pulse, and the output voltage is approximately equal to the peak voltage of the positive half cycle of the pulse minus the diode conduction voltage V DIODE； The VSW positive half cycle voltage is equal to Vin, so

[0050] Vout2≈Vin*L2 / (L1+L2)+Vin-V DIODE

[0051] When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the output voltage of Vout2 to decrease. Since Vout2 works in an open loop, its voltage is affected by the load.

[0052] Preferably, the switch tube of the DC-DC converter performs pulse width modulation on the input voltage Vin under PWM control, and the duty cycle η of PWM directly corresponds to the DC-DC conversion output voltage, Vin*η=Vout1.

[0053] Compared with the prior art, the present invention has the following beneficial technical effects:

[0054] 1. The present invention adjusts the pulse amplitude at the midpoint by connecting an inductor in series, and then obtains the required step-down output through rectification and filtering;

[0055] 2. The present invention utilizes the switching pulse output by the DC-DC converter to charge the series capacitor, adjusts the pulse voltage offset after capacitor coupling, and obtains the required negative voltage or boost output after rectifying the coupled pulse.

[0056] 3. The present invention can generate multiple working voltages with only one DC-DC converter, and the circuit is simple; the occupied area is small and the cost is low. BRIEF DESCRIPTION OF THE DRAWINGS

[0057] Figure 1 It is a basic step-down DC-DC conversion circuit diagram in the prior art;

[0058] Figure 2 A simplified circuit diagram of a step-down multi-voltage output solution in an embodiment of the present invention;

[0059] Figure 3 A simplified circuit diagram of a negative voltage multi-voltage output solution in an embodiment of the present invention;

[0060] Figure 4 A simplified circuit diagram of a boost-type multi-voltage output solution in an embodiment of the present invention;

[0061] Figure 5 This is a waveform diagram of the output voltage of the switch tube of the DC-DC converter in an embodiment of the present invention;

[0062] Figure 6 This is a diagram showing an implementation example of a buck-type multi-voltage output in an embodiment of the present invention;

[0063] Figure 7 This is a diagram showing an implementation example of a negative pressure type multi-voltage output in an embodiment of the present invention;

[0064] Figure 8 This is a diagram showing an implementation example of a boost type multi-voltage output in an embodiment of the present invention. DETAILED DESCRIPTION

[0065] The present invention provides a DC-DC conversion multi-power output device, comprising at least one DC-DC converter and an inductor connected in series;

[0066] The inductor is connected in series at the middle point of the DC-DC conversion circuit composed of the DC-DC converter;

[0067] By connecting the inductor in series, the pulse amplitude at the midpoint is adjusted, and then the required step-down output is obtained after rectification and filtering;

[0068] Alternatively, the series capacitor is charged by the switching pulse output by the DC-DC converter, the pulse voltage offset after capacitor coupling is adjusted, and the coupled pulse is rectified to obtain the required negative voltage or boost output.

[0069] A DC-DC converter and a series inductor form a buck, negative or boost DC-DC conversion circuit; the switch tube of the DC-DC converter performs pulse width modulation on the input voltage Vin under PWM control, and the output voltage waveform is as follows: Figure 5 As shown, the duty cycle η of PWM directly corresponds to the DC-DC conversion output voltage, Vin*η=Vout1.

[0070] Example 1

[0071] Step-down DC-DC converter circuit

[0072] like Figure 2 As shown, in the step-down DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1;

[0073] The first output terminal of the inductor L1 is connected in series with the anode of the Schottky diode D1 and then connected to the second output terminal Vout2;

[0074] The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1;

[0075] The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1;

[0076] One end of the capacitor C1 is connected to Vin, and the other end is grounded; one end of the capacitor C2 is connected to the output end of the inductor L2, and the other end is grounded; one end of the capacitor C3 is connected to the output end of the Schottky diode D1, and the other end is grounded.

[0077] When the load of Vout2 is light enough, much smaller than the load current of Vout1, its current is ignored; then the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is:

[0078] V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2);

[0079] Switch output voltage V SW The peak value is equal to Vin, and Schottky diode D1 rectifies the positive half-cycle of the pulse. The output voltage is approximately equal to the peak voltage of the positive half-cycle of the pulse minus the conduction voltage V of diode D1. DIODE ,Right now

[0080] Vout2≈Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE

[0081] When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the output voltage of Vout2 to decrease. Since Vout2 works in an open loop, its voltage is affected by the load current.

[0082] The following uses a specific circuit to introduce this solution in detail:

[0083] like Figure 5 As shown, the switching frequency f of the DC-DC converter SW =1MHz, feedback terminal voltage V FB =0.6V, Schottky diode conduction voltage V DIODE Take 0.25V;

[0084] Calculation: Vout1 = 1.09V, Vout2 = 2.5V.

[0085] Actual: Vout1 = 1.09V, load RL1 = 10Ω, current 109mA; Vout2 = 1.94V, load RL2 = 100Ω, current 19.4mA.

[0086] Example 2

[0087] Negative voltage DC-DC conversion circuit

[0088] like Figure 3 As shown, in the negative voltage DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1;

[0089] The first output terminal of the inductor L1 is connected in series with the capacitor C4 and the cathode of the Schottky diode D1 in sequence, and then connected to the second output terminal Vout2;

[0090] The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1;

[0091] The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1;

[0092] One end of capacitor C1 is connected to Vin, and the other end is grounded; one end of capacitor C2 is connected to the output end of inductor L2, and the other end is grounded; one end of capacitor C3 is connected to the anode of Schottky diode D1, and the other end is grounded; the anode of Schottky diode D2 is connected to the output end of capacitor C4, and the cathode is grounded.

[0093] When the load of Vout2 is light enough, much smaller than the load current of Vout1, and its current is ignored, the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is:

[0094] V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)

[0095] The voltage across capacitor C4 is approximately the peak voltage of the positive half-cycle of the pulse at point A minus the conduction voltage V of D2. DIODE , that is, V C4 =Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE；

[0096] The voltage at point B is V B =V A -V C4 ,Right now:

[0097] V B =V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)-[Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE ]

[0098] =(V SW -Vin)*L2 / (L1+L2)+V DIODE

[0099] Schottky diode D1 rectifies the negative half cycle of the pulse, and the output voltage is approximately equal to the peak voltage of the negative half cycle of the pulse plus the diode conduction voltage V DIODE；

[0100] V SW The peak voltage of the negative half cycle of -Vin is equal to -Vin, so

[0101] Vout2≈-Vin*L2 / (L1+L2)+2*V DIODE

[0102] When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the negative voltage amplitude of Vout2 to change; since Vout2 works in an open loop, its voltage is affected by the load current.

[0103] The following uses a specific circuit to introduce this solution in detail:

[0104] like Figure 6 As shown, the switching frequency f of the DC-DC converter SW =1MHz, feedback terminal voltage V FB =0.6V, Schottky diode conduction voltage V DIODE Take 0.25V;

[0105] Calculation: Vout1 = 1.09V, Vout2 = -2.03V.

[0106] Actual: Vout1 = 1.09V, load RL1 = 10Ω, current 109mA; Vout2 = -1.98V, load RL2 = 200Ω, current 10mA.

[0107] Example 3

[0108] Step-up DC-DC converter circuit

[0109] like Figure 4 As shown, in the boost DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1;

[0110] The first output terminal of the inductor L1 is connected in series with the capacitor C4 and the anode of the Schottky diode D1 in sequence, and then connected to the second output terminal Vout2;

[0111] The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1;

[0112] The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1;

[0113] One end of capacitor C1 is connected to Vin, and the other end is grounded; one end of capacitor C2 is connected to the output end of inductor L2, and the other end is grounded; one end of capacitor C3 is connected to the cathode of Schottky diode D1, and the other end is grounded; the anode of Schottky diode D2 is connected to Vin, and the cathode is connected to the output end of capacitor C4.

[0114] If the load of Vout2 is light enough, much smaller than the load current of Vout1, and its current is ignored, the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is:

[0115] V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)

[0116] V SW The negative half-cycle voltage is 0, and the voltage across capacitor C4 is approximately the peak voltage of the pulse negative half-cycle at point A minus the conduction voltage V of Vin and diode D2. DIODE The difference, V C4 =Vout1*L1 / (L1+L2)-(Vin-V DIODE );

[0117] The voltage at point B is V B =V A -V C4 ,Right now:

[0118] V B =V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)-[Vout1*L1 / (L1+L2)-Vin+V DIODE ]

[0119] =V SW *L2 / (L1+L2)+Vin-V DIODE

[0120] Schottky diode D1 rectifies the positive half cycle of the pulse, and the output voltage is approximately equal to the peak voltage of the positive half cycle of the pulse minus the diode conduction voltage V DIODE ; VSW positive half cycle voltage is equal to Vin, so

[0121] Vout2≈Vin*L2 / (L1+L2)+Vin-V DIODE

[0122] When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the output voltage of Vout2 to decrease. Since Vout2 works in an open loop, its voltage is affected by the load.

[0123] The following uses a specific circuit to introduce this solution in detail:

[0124] like Figure 7 As shown, the switching frequency f of the DC-DC converter SW =1MHz, feedback terminal voltage V FB =0.6V, Schottky diode conduction voltage V DIODE Take 0.25V;

[0125] Calculation: Vout1 = 1.09V, Vout2 = 5.53V.

[0126] Actual: Vout1 = 1.09V, load RL1 = 10Ω, current 109mA; Vout2 = 4.91V, load RL2 = 400Ω, current 12mA.

[0127] In this embodiment, only one DC-DC converter is needed to generate multiple working voltages, and the circuit is simple; 2) small occupied area and low cost.

[0128] The embodiments of the present invention are described in detail above with reference to the accompanying drawings, but the present invention is not limited thereto, and various changes can be made within the knowledge scope of technicians in the relevant technical field without departing from the purpose of the present invention.

Claims

1. A DC-DC conversion multi-power output device, characterized in that: comprising at least one DC-DC converter and an inductor connected in series; The inductor is connected in series at the middle point of the DC-DC conversion circuit composed of the DC-DC converter; By connecting the inductor in series, the pulse amplitude at the midpoint is adjusted, and then the required step-down output is obtained after rectification and filtering; Alternatively, the series capacitor is charged by the switching pulse output by the DC-DC converter, the pulse voltage offset after capacitor coupling is adjusted, and the coupled pulse is rectified to obtain the required negative voltage or boost output.

2. A DC-DC conversion multi-power output device according to claim 1, characterized in that: A DC-DC converter and a series inductor form a buck, negative voltage or boost DC-DC conversion circuit.

3. A DC-DC conversion multi-power output device according to claim 2, characterized in that: Vin is input to the input port of the DC-DC converter in the step-down DC-DC conversion circuit; the switch port of the DC-DC converter is connected to the inductor L1; The first output terminal of the inductor L1 is connected in series with the anode of the Schottky diode D1 and then connected to the second output terminal Vout2; The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1; The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1; One end of the capacitor C1 is connected to Vin, and the other end is grounded; one end of the capacitor C2 is connected to the output end of the inductor L2, and the other end is grounded; one end of the capacitor C3 is connected to the cathode of the Schottky diode D1, and the other end is grounded.

4. A DC-DC conversion multi-power output device according to claim 3, characterized in that: When the load of Vout2 is light enough, much smaller than the load current of Vout1, its current is ignored; then the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is: V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2); Switch output voltage V SW The peak value is equal to Vin, and Schottky diode D1 rectifies the positive half-cycle of the pulse. The output voltage is approximately equal to the peak voltage of the positive half-cycle of the pulse minus the conduction voltage V of diode D1. DIODE ,Right now Vout2≈Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the output voltage of Vout2 to decrease. Since Vout2 works in an open loop, its voltage is affected by the load current.

5. The DC-DC conversion multi-power output device according to claim 2, characterized in that: In the negative voltage DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1; The first output terminal of the inductor L1 is connected in series with the capacitor C4 and the cathode of the Schottky diode D1 in sequence, and then connected to the second output terminal Vout2; The second output terminal of the inductor L1 is connected to the series inductor L2 and then connected to the first output terminal Vout1; The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1; One end of capacitor C1 is connected to Vin, and the other end is grounded; one end of capacitor C2 is connected to the output end of inductor L2, and the other end is grounded; one end of capacitor C3 is connected to the anode of Schottky diode D1, and the other end is grounded; the anode of Schottky diode D2 is connected to the output end of capacitor C4, and the cathode is grounded.

6. A DC-DC conversion multi-power output device according to claim 5, characterized in that: When the load of Vout2 is light enough, much smaller than the load current of Vout1, and its current is ignored, the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is: V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2) The voltage across capacitor C4 is approximately the peak voltage of the positive half-cycle of the pulse at point A minus the conduction voltage V of D2. DIODE , that is, V C4 =Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V DIODE ; The voltage at point B is V B =V A -V C4 ,Right now: <h2 style=";text-align:left;direction:ltr">V<h2 style=";text-align:left;direction:ltr"> B <h2 style=";text-align:left;direction:ltr"> =V<h2 style=";text-align:left;direction:ltr"> SW <h2 style=";text-align:left;direction:ltr"> *L2 / (L1+L2)+Vout1*L1 / (L1+L2)-[Vin*L2 / (L1+L2)+Vout1*L1 / (L1+L2)-V<h2 style=";text-align:left;direction:ltr"> DIODE <h2 style=";text-align:left;direction:ltr"> ] (V) SW -Vin)*L2 / (L1+L2)+V DIODE Schottky diode D1 rectifies the negative half cycle of the pulse, and the output voltage is approximately equal to the peak voltage of the negative half cycle of the pulse plus the diode conduction voltage V DIODE； V SW The peak voltage of the negative half cycle of -Vin is equal to -Vin, so Vout2≈-Vin*L2 / (L1+L2)+2*V DIODE When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the negative voltage amplitude of Vout2 to change; since Vout2 works in an open loop, its voltage is affected by the load current.

7. The DC-DC conversion multi-power output device according to claim 2, characterized in that: In the boost DC-DC conversion circuit, Vin is input to the power input terminal of the DC-DC converter; the switch terminal of the DC-DC converter is connected to the inductor L1; The first output terminal of the inductor L1 is connected to the capacitor C4 and the anode of the Schottky diode D1 in sequence, and then connected to the second output terminal Vout2; The second output terminal of the inductor L1 is connected to the inductor L2 and then connected to the first output terminal Vout1; The feedback terminal of the DC-DC converter is connected to one end of the resistor R1 and the resistor R2, the other end of R2 is grounded, and the other end of the resistor R1 is connected to the first output terminal Vout1; One end of capacitor C1 is connected to Vin, and the other end is grounded; one end of capacitor C2 is connected to the output end of inductor L2, and the other end is grounded; one end of capacitor C3 is connected to the cathode of Schottky diode D1, and the other end is grounded; the anode of Schottky diode D2 is connected to Vin, and the cathode is connected to the output end of capacitor C4 and the anode of D1.

8. The DC-DC conversion multi-power output device according to claim 7, characterized in that: If the load of Vout2 is light enough, much smaller than the load current of Vout1, and its current is ignored, the pulse amplitude at point A is Vin*L2 / (L1+L2), and the voltage at A is: V A =(V SW -Vout1)*L2 / (L1+L2)+Vout1=V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2) V SW The negative half-cycle voltage is 0, and the voltage across capacitor C4 is approximately the peak voltage of the negative half-cycle of the pulse at point A minus the conduction voltage V of Vin and diode D2. DIODE The difference, V C4 =Vout1*L1 / (L1+L2)-(Vin-V DIODE ); The voltage at point B is V B =V A -V C4 ,Right now: V B =V SW *L2 / (L1+L2)+Vout1*L1 / (L1+L2)-[Vout1*L1 / (L1+L2)-Vin+V DIODE ] =V SW *L2 / (L1+L2)+Vin-V DIODE Schottky diode D1 rectifies the positive half cycle of the pulse, and the output voltage is approximately equal to the peak voltage of the positive half cycle of the pulse minus the diode conduction voltage V DIODE ; VSW positive half cycle voltage is equal to Vin, so Vout2≈Vin*L2 / (L1+L2)+Vin-V DIODE When Vout2 has a certain load current, the voltage at point A will drop due to the shunting of Vout2, causing the output voltage of Vout2 to decrease. Since Vout2 works in an open loop, its voltage is affected by the load.

9. The DC-DC conversion multi-power output device according to claim 1, characterized in that: The switch tube of the DC-DC converter performs pulse width modulation on the input voltage Vin under PWM control.

10. The DC-DC conversion multi-power output device according to claim 9, characterized in that: The duty cycle η of PWM directly corresponds to the DC-DC conversion output voltage, Vin*η=Vout1.

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