220V motor control board, discharges very slowly after power failure, and can still shock people after a long time

selfcap_mcu

220V motor control board, discharges very slowly after power failure, and can still shock people after a long time [Copy link]

 

 

Below is the circuit diagram of the power supply part of the board.

The function of the board is that the 300V DC motor rotates through PWM controlled by the microcontroller.

The problem now is that after I cut off the main power supply of the board, there is always a point on the board, and it just won't go down. After ten hours, I tested the voltage across the capacitor and found that there was still 160V.

Where should I modify the circuit?

okhxyyo

Is there energy storage somewhere?

maychang

"Ten hours later, I tested the capacitor and found that there was still 160V voltage across both ends."

There are six capacitors in the picture. Which capacitor do you measure has a voltage of 160V?

maychang

"Ten hours later, I tested the capacitor and found that there was still 160V voltage across both ends."

It is estimated that R9 or R10 is broken. You can remove it and measure it.

selfcap_mcu

maychang posted on 2020-8-26 10:31 "After ten hours, I tested the capacitor and found that there was still 160V voltage across both ends." It is estimated that R9 or R10 has been broken. You can remove it and measure it...

Hi, I just tested it and the resistance is still 120K and there is no short circuit.

selfcap_mcu

maychang published on 2020-8-26 09:56 "Ten hours later, I tested the capacitor and found that there was still 160V voltage across both ends." There are six capacitors in the picture. Which one do you measure with 160V voltage...

I tested the capacitor C5, I removed the capacitor C51, so there was only one capacitor, so I tested it directly on the pin of the C51 capacitor.

selfcap_mcu

okhxyyo posted on 2020-8-26 09:51 Is there any energy storage somewhere?

It should be that the capacitor storage energy of C5 cannot be discharged.

selfcap_mcu

maychang posted on 2020-8-26 10:31 "After ten hours, I tested the capacitor and found that there was still 160V voltage across both ends." It is estimated that R9 or R10 has been broken. You can remove it and measure it...

Is it because the resistance of R9 and R10 is too large, because the two resistors add up to 300K. As a result, the current flowing through is too small and the discharge is slow.

I just replaced both resistors with 51K each. I just tested the capacitor discharge and found that it can be discharged from 300V to below 30V within 30 seconds.

I don't know if this connection is normal, and if the resistance is too small will it affect the circuit.

maychang

selfcap_mcu posted on 2020-8-26 10:50 Is it because the resistance of R9 and R10 is too large, because the two resistors add up to 300K. As a result, the current flowing through is too small, and the discharge is slow. & ...

Now only one capacitor is connected. If R9 and R10 are 150 kΩ, then the time constant is 100uF*300kΩ＝30s. In 5 times 30 seconds, the voltage across the capacitor will drop to about 1% of the original value. But you said that it is still 160V after ten hours, so I suspect that the resistor is broken.

maychang

selfcap_mcu posted on 2020-8-26 10:50 Is it because the resistance of R9 and R10 is too large, because the two resistors add up to 300K. As a result, the current flowing through is too small, and the discharge is slow. & ...

The smaller the resistor, the faster the discharge. Now the resistor is 51 kilo-ohms, and the time constant should be about 10 seconds.

The smaller the resistor, the larger the power dissipated by the resistor. The voltage across the capacitor is 300V, and the power dissipated in each resistor is 0.44W.

selfcap_mcu

maychang posted on 2020-8-26 11:29 Now only one capacitor is connected. If R9 and R10 are 150 kilo-ohms, then the time constant is 100uF*300kΩ＝30s. In 5 times 30 seconds...

The circuit is indeed not short-circuited. I have also had this problem in another board and a fat-shaking machine. I was testing the board at the customer's place and went out for lunch for about 2 hours. When I came back to pick up the board, I was still electrocuted.

Thank you very much for your reply, I feel you are very professional and enthusiastic.

Are there any requirements for the value of this resistor?

1. What should be the minimum power of the resistor so that the resistor will not burn out?

2. The resistance value of this resistor is not the smaller the better. If the smaller the better, then would it be better if there is no resistor? So what should the value of this resistor be? I want it to release all the electricity within 10 seconds. Can you help me summarize it again?

Thank you so much.

selfcap_mcu

maychang posted on 2020-8-26 11:29 Now only one capacitor is connected. If R9 and R10 are 150 kilo-ohms, then the time constant is 100uF*300kΩ＝30s. In 5 times 30 seconds...

In 5 times 30 seconds, the voltage across the capacitor will drop to about one percent of its original value.

Did you understand this sentence?

maychang

selfcap_mcu posted on 2020-8-26 13:13 In 5 times 30 seconds, the voltage across the capacitor will drop to about 1% of the original value. Do you understand this sentence? &nb ...

After power is cut off, the voltage across the capacitor changes according to the following formula:

In the formula, τ = RC, which is called the time constant. t is time.

When t = τ, the voltage across the capacitor drops to 0.37 times of its original value. When t = 3τ, it drops to about 0.05 times. When t = 5τ, it drops to less than 0.01 times of its original value.

maychang

selfcap_mcu posted on 2020-8-26 13:12 The circuit is indeed not short-circuited. I have also experienced this phenomenon in another board and fat-shaking machine. I was testing the board at the customer's place and went out for dinner in the middle...

"I want it to release all the electricity within 10 seconds"

"Discharging completely" means that the voltage drops to zero, which takes an infinite amount of time and is impossible.

If you want the voltage to drop to, for example, 0.05 times its original value within 10 seconds, calculate according to the formula in the previous post.

selfcap_mcu

I understand it in general. I just tested a 100K resistor and found that it takes about 40 seconds to discharge from 300V to below 5V under a 100uF capacitor. A 300K resistor takes about 120 seconds.

However, we must pay attention to the actual power of the resistor. According to the manual, the rated power of the 2512 package is 1W, and the rated power of the 1206 package is 1/4W.

Therefore, the smaller the resistance, the shorter the discharge time, but the power handling capacity of the resistance must be actually considered.

The problem I encountered for the first time was probably what the friend just mentioned. The resistor was burned out, causing the resistor to be short-circuited, so it could not be discharged. Because the first time I used it was 1206.

In fact, this problem is very simple if you understand it. Of course, I am very grateful to my friend " maychang " for his patient answer. However, for me who lacks experience, I did not pay attention to this problem in the two products I made. I directly used the previous circuit diagram without considering the power problem. As a result, the discharge time of one board exceeded 2 minutes, and the other board did not discharge at all. In fact, the resistor was too large, or the resistor power was not enough and the circuit was broken.

Therefore, to get a better discharge effect, you still need to calculate it. According to my situation, I choose to use 150K, 2512 package resistor. The discharge is about 60S. Because if you choose 100S, although the discharge is about 40S, the power is less than 0.9W, which is close to 1W.

selfcap_mcu

maychang posted on 2020-8-26 14:38 "I want it to release all the electricity within 10 seconds" "Release all" means that the voltage is reduced to zero, which takes an infinite amount of time...

Thank you so much.

maychang

selfcap_mcu posted on 2020-8-26 14:38 I have generally understood it. I just tested a 100K resistor and found that it takes about 40S to discharge from 300V to below 5V under a 100uF capacitor. A 300K resistor takes about...

For long-term use, it is recommended that the power dissipation of the resistor be reduced by half, that is, for a resistor with a nominal power of 1W, only let the resistor dissipate 0.5W of power.

chunyang

The smaller the resistance, the faster the discharge, but the greater the power consumption during normal operation, which is a waste of power. Although using a discharge resistor can solve the problem, this is not a good method. It is better to use the system itself.

First, the power switch should be set on the AC side instead of the DC side. This way, after power failure, the system itself will be a discharge circuit, which is much faster than resistance.

Secondly, when the system cannot be used as a discharge circuit, the power indicator light can be used for discharge. The power indicator light should not be installed on the low-voltage side of the system, but directly on the high-voltage output side of the previous stage. The general working current of LED is mA level, which can quickly discharge the capacitor storage power and provide a residual power indication. People will not touch the high-voltage circuit when the light is on.

chunyang

Regarding the power selection of the resistor, first calculate the power consumption, power consumption = voltage squared / resistance value, and select the power of the resistor twice the power consumption value. When using the LED power indicator to discharge, the power of the current limiting resistor should also be selected with attention. For high voltage, the junction voltage drop of the LED can be ignored, and the power consumption of the current limiting resistor can be directly calculated by multiplying the power supply voltage by the working current of the LED.

selfcap_mcu

chunyang posted on 2020-8-26 15:20 The smaller the resistance, the faster the discharge, but the power consumption during normal work is greater, which wastes power in vain. Although using a discharge resistor can solve the problem, this method is not good...

The general operating current of LED is mA level, which can quickly discharge the stored electricity in the capacitor and provide a remaining power indication. People will not touch the high-voltage circuit when the light is on.

This is a very user-friendly method. Only an experienced engineer like you can come up with such a user-friendly method.

For batch production testing of boards, the bare boards are usually plugged in and unplugged repeatedly for testing. So if there is an LED indicator, you can judge whether it is okay to touch it by hand based on whether the LED indicator is bright.