How to improve the power of voltage doubler rectifier circuit

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How to improve the power of voltage doubler rectifier circuit [Copy link]

 

This is a post I made before about static electricity generators.

Is there an error in the schematic diagram of the electrostatic generator? - Power Technology - Electronic Engineering World - Forum (eeworld.com.cn)

There is no problem with the mainboard circuit, and the waveforms are normal.

R31-R39 in the "voltage doubler rectifier circuit" divides the voltage and sends it to the microcontroller to detect the output voltage value

question:

When R31-R39 is not connected to the circuit (R39 1M resistor is connected to the ground and disconnected), the output high voltage is 25KV, but after R31-R39 is connected, the voltage drops directly to 17KV

So it should be that the output power is too low, causing the output voltage to be pulled down

How to increase the output power?

Increase the voltage divider resistor (I think it's big enough)?

Parallel voltage doubling capacitors?

maychang

In the figure, P5 is marked as "secondary side of the high-voltage package", so what device drives the primary side of the high-voltage package?

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maychang posted on 2021-7-9 15:08 In the figure, P5 is marked as "secondary side of the high-voltage package", so what device drives the primary side of the high-voltage package?

is a half-bridge driver made by IR2104

maychang

Will the program be published on 2021-7-9 15:13 It is a half-bridge driver made by IR2104

IRF830 withstands 500V and has a maximum continuous current of 4.5A. This circuit can output hundreds of watts if used with a proper rectifier and filter circuit. Your R31-R38 consumes only a few tenths of a watt, so there should be no problem. But in fact, the output voltage drops a lot after connecting R31-R38, so the problem must be in the voltage doubler rectification.

maychang

Will the program be published on 2021-7-9 15:13 It is a half-bridge driver made by IR2104

In the first post, C23D24 is unnecessary and should be removed (open circuit). Removing it will not affect the circuit operation.

maychang

Will the program be published on 2021-7-9 15:13 It is a half-bridge driver made by IR2104

According to the first post, the capacitance of capacitors C19 to C27 is 10nF. With this capacitance, there should be no problem in outputting several mA current. Now the output voltage drops a lot when a slight load is added. Check whether the capacitance of each capacitor is sufficient.

maychang

Will the program be published on 2021-7-9 15:13 It is a half-bridge driver made by IR2104

In addition, the 2nd pin of U4 in the 3rd floor diagram should be a rectangular wave (output from the 6th pin of the right chip). What are the frequency, duty cycle and amplitude of this rectangular wave?

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maychang posted on 2021-7-9 16:02 In addition, the 2nd pin of U4 in the 3rd floor picture should be a rectangular wave (output from the 6th pin of the chip on the right). What are the frequency, duty cycle and amplitude of this rectangular wave?

PWM 20K Duty cycle 50%

maychang

Will the program be published on 2021-7-9 16:56 PWM 20K Duty Cycle 50%

In the first circuit, the output voltage cannot be changed using PWM (pulse width modulation).

20kHz, 50% duty cycle. This should be no problem.

What about the amplitude?

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maychang published on 2021-7-9 17:08 The first circuit, using PWM (pulse width modulation) cannot change the output voltage. 20kHz, duty cycle 50%. This should not be a problem. ...

The amplitude is +-12.5V. I have tried changing the duty cycle of the external switching power supply
from 10% to 50%, and the output voltage can be changed.

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Will the program be published on 2021-7-9 18:53 The amplitude is +-12.5v. I have tried the external switching power supply duty cycle from 10% to 50%, and the output voltage can be changed

Switching power supply 25v

maychang

Will the program be published on 2021-7-9 18:53 The amplitude is +-12.5v, and the duty cycle of the external switching power supply is from 10% to 50%. I have tried it, and the output voltage can be changed

The output of IR2104, HO and LO are complementary, and HO changes with IN. That is, when IN is high, HO is high and LO is low, when IN is low, HO is low and LO is low. The duty cycle of IN only changes the ratio of the conduction time of the upper and lower MOS, and cannot be used to adjust the output voltage.

In addition, for this type of voltage doubler rectifier output, the output voltage has little to do with the conduction time of the upper and lower tubes, and is mainly determined by the HV voltage in the figure.

maychang

Will the program be published on 2021-7-9 18:53 The amplitude is +-12.5v, and the duty cycle of the external switching power supply is from 10% to 50%. I have tried it, and the output voltage can be changed

『The amplitude is +-12.5v, powered by an external switching power supply』

The power supply voltage marked in the red box is 12V, and there is no negative power supply. How can the "amplitude be +-12.5v"?

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maychang published on 2021-7-9 19:27 『The amplitude is +-12.5v, powered by an external switching power supply』 The power supply voltage marked in the red box is 12V, and there is no negative power supply. How can it be "amplitude...

Oh, this one you're talking about is 12v! I'm talking about the hv that is supplied by an external switching power supply of 25v! Originally, the hv was going to be supplied by a 220v direct rectifier, but considering that the voltage was too high, a switching power supply was used instead! I experimented that adjusting the duty cycle can adjust the output voltage! I just used an oscilloscope to measure the waveform of the primary of the high-voltage package, and found that the primary drive waveform was not symmetrical up and down. When the duty cycle increased from 10% to 50%, the negative half of the drive waveform slowly changed from close to 0v to become completely symmetrical when the duty cycle reached 50%. Maybe this is the reason for the voltage change! It's strange why it's asymmetrical.

maychang

Will the program be published on 2021-7-9 20:15 Oh, the one you are talking about is 12v! I am talking about hv which is supplied by an external switching power supply of 25v! Originally, hv was going to be supplied by 220v direct rectification, but considering the power...

"Just now I used an oscilloscope to measure the waveform of the primary of the high-voltage transformer and found that the primary drive waveform is not symmetrical up and down."

Not surprising.

I have already said on the 12th floor: The output of IR2104, HO and LO are complementary, HO changes with IN. The change of IN duty cycle is reflected at both ends of the transformer primary (you call it the high-voltage package primary), that is, the square waves at both ends of the primary are asymmetric.

maychang

Will the program be published on 2021-7-9 20:15 Oh, the one you are talking about is 12v! I am talking about hv which is supplied by an external switching power supply of 25v! Originally, hv was going to be supplied by 220v direct rectification, but considering the power...

The square waves at both ends of the transformer primary are asymmetrical. For the transformer, it is not easy to "magnetically reset" in a certain direction, that is, the MOS tube conduction time is long in a certain direction, and the "volt-second product" borne by the primary winding is large, while the reverse MOS tube conduction time is short, and the reverse "volt-second product" borne by the primary winding is small. The iron core must be magnetically reset, so the magnetic reset can only be achieved through the secondary, that is, the voltage at both ends of the secondary in a certain direction is high, and the voltage at both ends of the secondary in another direction is small.

This situation is not conducive to the operation of the transformer core and often causes the core to become saturated in a certain direction.

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maychang published on 2021-7-9 20:33 The square waves at both ends of the transformer primary are asymmetric. For the transformer, it is not easy to "magnetically reset" in a certain direction, that is, the MOS tube in a certain direction...

Oh, I see. It seems that this solution is wrong. If we change it to a flyback type and adjust the PWM duty cycle, we can adjust the output voltage!

Now let's fix the duty cycle at 50% and check why the output power is so low?

maychang

Will the program be published on 2021-7-9 21:02 Oh, I see. It seems that this solution is wrong. If you change it to a flyback type to adjust the PWM duty cycle, you can adjust the output voltage! Now...

In a single-ended flyback circuit, the waveform of the transformer secondary output is asymmetrical, so it is not suitable to use voltage doubler rectification to achieve high DC voltage output.

maychang

Will the program be published on 2021-7-9 21:02 Oh, I see. It seems that this solution is wrong. If you change it to a flyback type to adjust the PWM duty cycle, you can adjust the output voltage! Now...

However, a single-ended flyback circuit can easily achieve high voltage output as long as the transformer secondary has enough turns.

If the transformer secondary peak voltage is 25kV, then only one high-voltage rectifier and one high-voltage capacitor are needed. High-voltage capacitors are quite expensive.

maychang

Will the program be published on 2021-7-9 21:02 Oh, I see. It seems that this solution is wrong. If you change it to a flyback type to adjust the PWM duty cycle, you can adjust the output voltage! Now...

If I were making this power supply, I would choose a single-ended flyback and a common rectifier and filter circuit (a diode and a capacitor).

The difficulty of such a design lies in the winding of the transformer. The transformer has a large number of secondary turns, and the distributed capacitance must be minimized. At the same time, the beginning and end of the winding should be as far apart as possible. Its structure can refer to the high-voltage package of an old-fashioned TV. The rectifier tube can be connected in series with multiple diodes, and the capacity of the capacitor can be quite small (because the output current is very small). The high-voltage filter capacitor of the old-fashioned TV relies entirely on the capacitor formed by the graphite coating on the inner and outer walls of the picture tube.