Here are some questions about the ground insulation detection circuit

西里古1992

Here are some questions about the ground insulation detection circuit [Copy link]

 

The detection circuit diagram is posted here. The DC voltage value is 500V. R20, R26, R27, R28 and Rx form a sampling resistor between DC+, GND and DC-. The purpose is to collect the voltage across R20 as an input signal and output a value through the differential amplifier circuit. The operational amplifier is rail-to-rail, with a range of 0~3.3V. Rx is an equivalent analog resistance value between DC+ and GND. According to theoretical calculation, it should be 500/((390K×3)+200K+0.51K)×0.51K. The input value should be about 180mV, and the voltage value of R26, R27, R28 should be about 142V, while the voltage of Rx is about 70V at most. The current test situation is that the voltage across Rx reaches 477V, and the voltage of R26-R28 is only about ten volts. Is there an equivalent resistor between DC- and GND? The control circuit signals in my circuit are all normal, and the power supply is output by 3 independent voltage regulators.

maychang

This circuit cannot be used.

Your DC power supply (i.e. the source of the MOS tube) has a negative voltage relative to the ground, and the voltage of the two input terminals of the op amp relative to the ground (GND) is very high.

西里古1992

maychang posted on 2019-11-13 22:18 This circuit cannot be used. Your DC power supply (i.e. the source of the MOS tube) is a negative voltage to the ground, and the voltage of the two input terminals of the op amp to the ground (GND) is very high.

The resistor between the two input terminals is only 510o.

西里古1992

maychang posted on 2019-11-13 22:18 This circuit cannot be used. Your DC power supply (i.e. the source of the MOS tube) is a negative voltage to the ground, and the voltage of the two input terminals of the op amp to the ground (GND) is very high.

Teacher, don't those three 390k resistors already reduce the voltage a lot?

西里古1992

maychang posted on 2019-11-13 22:18 This circuit cannot be used. Your DC power supply (i.e. the source of the MOS tube) is a negative voltage to the ground, and the voltage of the two input terminals of the op amp to the ground (GND) is very high.

Maybe the common mode voltage of the op amp +- to ground is too high. Should we replace the three 390K resistors to the top of R20?

maychang

Sirigu1992 posted on 2019-11-14 10:21 The common mode voltage of the op amp +- to ground is too high. Should I replace the three 390K resistors on R20?

"It's just that the common-mode voltage of the op amp +- to ground is too high."

Yes. It is precisely because the common-mode voltage exceeds the allowable range of the op amp that I said this circuit cannot be used.

maychang

Sirigu1992 posted on 2019-11-14 10:21 The common mode voltage of the op amp +- to ground is too high. Should I replace the three 390K resistors on R20?

"Change those three 390K resistors to the top of R20?"

Maybe it works.

I don't know what the purpose of the first post circuit is, so I'm not sure.

The title of the first post says that this circuit is "an insulation detection circuit to ground", and the text says that the DC voltage value is 500V, and that "Rx is the equivalent analog resistance value between DC+ and GND". However, it does not explain the relationship between DC+, DC-, GND and DC (500V). Generally speaking, measuring insulation resistance is to measure the leakage current through the insulation, but in this circuit, it is not clear how the op amp measures the leakage current.

西里古1992

maychang posted on 2019-11-14 10:35 "It's just that the common-mode voltage of the op amp +- to the ground is too high." Yes. It is precisely because the common-mode voltage exceeds the allowable range of the op amp that I say this...

I thought the op amp was placed in a differential circuit and only needed to consider the differential signal.

gmchen

The formula “500/((390K×3)+200K+0.51K)×0.51K” is totally wrong!

The reason for the error is that the resistor string is connected to the ground! The potential of that ground point is 0!

西里古1992

gmchen posted on 2019-11-15 19:24 The formula "500/((390K×3)+200K+0.51K)×0.51K" is totally wrong! The reason for the error is that string of electricity...

There are so many resistors between DC+-500V. It is always right to calculate the current by dividing the voltage by the resistance. I calculate the voltage at both ends by multiplying the current by the resistance.

gmchen

Sirigu1992 posted on 2019-11-20 11:01 There are so many resistors between DC+-500V. It is always correct to calculate the current by dividing the voltage by the resistance. When I calculate the voltage at both ends, I multiply the current by the resistance.

Please draw all the power supplies (DC+\DC-), otherwise it is difficult to make it clear with just words. Is it a floating 500V power supply (500V connected to DC+/DC-, without ground in between) or two power supplies (+500V, -500V)?

西里古1992

maychang posted on 2019-11-14 10:42 "Replace the three 390K resistors on R20?" Maybe it's feasible. I don't know what the purpose of the first post circuit is, so I dare not...

Teacher, let me briefly explain this insulation detection circuit to you. Figure 1 is the working principle diagram we refer to. In the figure, U is also a high-voltage DC output of about 400V~500V. We just want to design a detection circuit to detect the insulation impedance of U+ to ground and U- to ground. The steps and the final method of calculating the insulation resistance are similar to Figure 1.

Then Figures 2 and 3 are the detection circuits we designed, which are basically similar to the working principle of Figure 1. The MOS tubes of the two bridge arms are turned on alternately, so that the voltage value collected by the sampling resistor is obtained by the differential amplifier circuit to obtain IMD+ and IMD-, and sent to the microcontroller for collection. In this way, the voltage value of the 0.51K resistor in the two bridge arms is calculated in the program through IMD+ and IMD-, so that the formula in Figure 1 can be used to read the resistance values of RX and RY, thereby judging the insulation impedance.

There is no problem with this post before, it is just that the 0.51K and three 390K resistors are switched, but we have tested it in the past few days and have some questions for you:

1) The voltage across the 0.51K resistor of the negative bridge arm should be a negative value, right? According to the formula in Figure 1, RX and RY will always have a negative value. We can see a negative voltage in the programming software window, and the insulation resistance also shows a negative value.

2) I can't figure out how to convert the formula in Figure 1. Isn't it essentially connecting the resistors in series and parallel to find the current, calculating the voltage of the sampling resistors through shunt, that is, U1 and U2 (in Figure 1), and then simplifying the equation to get the expressions for RX and RY?

3) In the differential amplifier circuit, I put the bias voltage of 2V at the non-inverting input terminal. In this way, are the input-output relationship of the differential circuit reversed for the two bridge arms? I calculated the input-output relationship based on the virtual short and virtual break?

maychang

Sirigu1992 posted on 2019-12-13 21:35 Teacher, let me briefly explain this insulation detection circuit to you. Figure 1 is the working principle diagram we refer to. In the figure, U is also high voltage...

I don't think this solution is feasible.

There are two switches in the figure. In fact, an N-channel MOS tube and a P-channel MOS tube are used to replace these two switches. However, the insulation resistance Rx and Ry of the power supply to the ground may vary greatly, ranging from 1 megohm to 1 gigaohm. Therefore, the switch in Figure 1 on the 12th floor may need to withstand the voltage of the power supply U, that is, 500V, when it is turned off.

It may be difficult to find a P-channel MOS tube with a voltage resistance of 500V, at least I have never heard of it.

maychang

Sirigu1992 posted on 2019-12-13 21:35 Teacher, let me briefly explain this insulation detection circuit to you. Figure 1 is the working principle diagram we refer to. In the figure, U is also high voltage...

MOS tubes are closer to ideal switches than bipolar tubes, but they are still not ideal switches. MOS tubes have a certain amount of leakage current when they are turned off.

The above is the characteristics of a 500V withstand voltage N-channel MOS tube. The content in the red box shows: The condition for testing the withstand voltage of this type of tube is a drain current of 250uA, and it is qualified if the drain voltage exceeds 500V. 500V/250uA is 2 megohms. Although this resistance is nonlinear, this current may decrease sharply when the drain-to-source voltage is slightly lower, or this current may basically not change with the drain-to-source voltage.

Regardless of whether the resistance between the drain and source of the MOS tube changes, I said in a previous reply: the insulation resistance variation range is quite large, which may reach 1000 megohms. Even if Rx or Ry is only 5 megohms, the measured result is meaningless when the resistance is 2 megohms when the MOS tube is turned off.

maychang

Sirigu1992 posted on 2019-12-13 21:35 Teacher, let me briefly explain this insulation detection circuit to you. Figure 1 is the working principle diagram we refer to. In the figure, U is also high voltage...

Even if the measurement is carried out according to the scheme in Figure 1 on the 12th floor, the two switches S1 and S2 in the figure must use mechanical switch contacts. Carefully select the relay. It is estimated that ultra-small relays will not work, but slightly larger relays are available.