Problems of overcurrent and overvoltage protection in voltage doubler rectifier circuit

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Problems of overcurrent and overvoltage protection in voltage doubler rectifier circuit [Copy link]

 

The voltage doubler rectifier circuit is shown in the figure

The driving circuit is IR2153 half-bridge driver

I saw that other people’s electrostatic devices have (overcurrent protection, discharge protection, overvoltage protection, short circuit protection), but mine has nothing and I want to add these in.

My understanding is that overvoltage protection is achieved by sampling the output voltage and comparing it with a reference voltage (a comparator made of an op amp).

Overcurrent protection, discharge protection, and short circuit protection should be achieved by detecting the output current. These three situations will increase the output current, but how to detect the output current? Connect a 1K resistor in series in the high-voltage output circuit to detect the voltage drop above

Is this possible?

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maychang

Will the program be published on 2022-5-8 15:04

Your circuit already has voltage detection.

Output short circuit detection can be achieved by connecting a resistor in series with the high voltage output loop. 1 kilo ohm resistor is too small because your circuit output is already connected in series with a 200 kilo ohm resistor.

maychang

Will the program be published on 2022-5-8 15:04

The overcurrent detection resistor can also be placed on the AC side, that is, connected between E2 and PE. In this case, the resistance cannot be too large, otherwise the high voltage cannot be raised. The advantage of placing the detection resistor between E2 and PE is that the damage of each high-voltage capacitor or diode can also be detected. However, in this arrangement, there are AC pulses at both ends of the detection resistor, and the detection circuit is slightly complicated.

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maychang posted on 2022-5-8 15:24 Your circuit already has voltage detection. Output short circuit detection can be achieved by connecting a resistor in series with the high voltage output loop. 1 kilo ohm resistor is too small...

I plan to remove these two resistors and only connect a 1K current detection resistor in series

maychang

Will the program be published on 2022-5-8 15:04

The voltage doubler rectifier circuit is not afraid of output short circuit. Even if the output is short-circuited, the current will not be as large as that of the transformer boost rectifier output short circuit. Moreover, your circuit has a 200 kΩ resistor in series. So it is worth considering whether your circuit needs output current detection. Adding a function will definitely increase the cost.

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maychang posted on 2022-5-8 15:37 The voltage doubler rectifier circuit is not afraid of output short circuit. Even if the output is short-circuited, the current will not be as large as the transformer boost rectifier output short circuit. Moreover, your...

Yes, but I also want to detect the current and display it

maychang

Will the program be published on 2022-5-8 15:34 I plan to remove these two resistors and only connect a 1K current detection resistor in series

It is better to keep these two resistors.

The 6th floor has already said: the voltage doubler rectifier circuit is not afraid of output short circuit. Even if the output is short-circuited, the AC power supply on the left side of E1E2 will not be damaged, and the rectifier tubes and capacitors will not be damaged. If these two resistors are removed, the capacitors will discharge quickly when the output is short-circuited, generating considerable sparks, which can easily cause fire and other secondary disasters.

Of course, the resistance of these two resistors can be reduced.

maychang

Will the program be published on 2022-5-8 15:43 Yes, but I also want to detect the current and display it

If you want to display the current, you can use the method of connecting a resistor in series in the output loop to sample it. The output current of this voltage doubler rectifier circuit will not exceed a few tenths of a milliampere, and the sampling resistor can be larger than 1 kilo-ohm.

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maychang posted on 2022-5-8 16:08 If you want to display the current, you can use the method of connecting a resistor in series in the output loop to sample. The output current of this voltage doubler rectifier circuit will not exceed a few milliseconds...

Is the output current of this circuit only a few tenths of a milliampere? If I want to output a current of several milliamperes, can I increase the capacity of the voltage multiplier capacitor?

maychang

Will the program be published on 2022-5-8 17:03 Is the output current of this circuit only a few tenths of a milliampere? If I want to output a current of several milliamperes, can I increase the capacity of the voltage multiplier capacitor?

"If I want to output a current of several milliamperes, can I increase the capacity of the voltage-doubling capacitor?"

A few tenths of a milliamp is my estimate.

Judging from the withstand voltage of the capacitor, the output voltage of your circuit may reach 50kV. If the current is 1mA, the DC power is 50W. The instantaneous AC power at your E1 and E2 terminals may reach hundreds of W. Can the AC power supply at E1 and E2 terminals reach such a large instantaneous power?

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maychang posted on 2022-5-8 17:23 "If I want to output a current of several milliamperes, can I increase the capacity of the voltage-doubling capacitor? " A few tenths of a milliampere is my estimated value. ...

R62 is used to detect the output current, take out the voltage at both ends, perform double-ended to single-ended conversion and then perform AD detection. The double-ended to single-ended circuit is as follows (just an example diagram, ignoring various parameters)

The problem now is that although the relative voltage across the sampling resistor is very low, the voltage to ground is very high (tens of KV), which will break down the op amp. How can we improve the circuit diagram?

蛙蛙蛙

The output current of this circuit is only a few tenths of a milliampere?

maychang

Will the program be published on 2022-5-9 21:28 R62 is used to detect the output current, take out the voltage at both ends, convert the double-ended to single-ended, and then perform AD detection. The double-ended to single-ended circuit is as follows...

In these two pictures, the current sampling resistor is placed at the 50kV (guess value) output end, that is, the voltage at the negative output end is 50kV. The input of your differential amplifier is 50kV to ground, which is the common mode voltage. Can your differential amplifier withstand such a high voltage?

maychang

Will the program be published on 2022-5-9 21:28 R62 is used to detect the output current, take out the voltage at both ends, convert the double-ended to single-ended, and then perform AD detection. The double-ended to single-ended circuit is as follows...

You said it yourself: this will break down the op amp. So this circuit cannot be used at all.

In addition, the "ground" of the voltage doubler rectifier circuit and the "ground" of the differential amplifier are not the same symbol. If the two "grounds" are not connected together, the circuit will not work at all.

maychang

Will the program be published on 2022-5-9 21:28 R62 is used to detect the output current, take out the voltage at both ends, convert the double-ended to single-ended, and then perform AD detection. The double-ended to single-ended circuit is as follows...

Improvement method: put the current sampling resistor R62 at the low end of the high voltage output, that is, on the PE on the left, and connect PE to C41D20E2 through the sampling resistor.

But please note: the output of the current sampling resistor is negative at this time.

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maychang posted on 2022-5-10 08:51 Improvement method: put the current sampling resistor R62 at the low end of the high voltage output, that is, put it on the PE on the left, and connect PE to C41D20E2 through the sampling resistor. ...

Should I change it to this? PE and the op amp ground are shorted together.

If it is changed to this, just use a directional proportional circuit, that is, to provide a negative voltage to the op amp

Thanks

maychang

Will the program be published on 2022-5-10 09:45 Change it to this? The ground of PE and the op amp are short-circuited together. If it is changed to this, just use a directional proportional circuit directly, that is, to...

Yes, that's how it's connected.

Placing the sampling resistor at the low end of the output avoids common-mode voltages of tens of kilovolts.

For this resistor R62, try to use a better resistor. You can consider using several resistors in parallel to prevent the resistor from being open-circuited and damaging the amplifier or even the circuit board.

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Wa Wa Wa published on 2022-5-9 23:58 Is the output current of this circuit only a few tenths of a milliampere?

How much current can it output?

秋叶2012

Brother, have you verified the circuit in the end? I have also been doing flow detection recently and would like to learn from it.