The voltage comparator has an excessive current problem at its input.

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background :

I1 and I2 are 4-20mA current sources, which flow through the load RL. A wire is drawn from RL to the comparator. The comparator compares the voltage and outputs it to control a small relay. It is powered by 24V and the comparator is made of the op amp TP2722.

Problem description:

When both I1 and I2 inputs have fixed values (assuming 12mA), there is no problem.

But if one circuit is disconnected, for example, I2 has no output, or I1 has no output, then the current actually output on the load is less than 12mA.

When the comparator input terminal was connected to the ground with a 10pF capacitor, the load current was short of 0.5mA. After removing it, it became short of 0.3mA-0.4mA. After removing the comparator, the load current was normal. It was initially believed that the comparator input terminal might have absorbed the current. However, the input-related parameters written in the manual were all at the pA level. Why would this current go to the chip?

In the past, I only compared the voltage and didn't need to consider this loss, but I never compared the current. If it is a normal phenomenon, how can I reduce the impact on the current? ? ?

ps1: 4-20mA is for signaling and is very sensitive to this kind of loss.

ps2: There is a problem with the original product and there is no condition to make major changes to the circuit.

The basic schematic diagram is shown below.

The detailed schematic diagram is as follows:

Chip manual is here

https://en.eeworld.com/bbs/forum.php?mod=attachment&aid=NDI5Mzg1fDhmZTY3MTE4NTc2YTU2ZWY5NjA0YjYyODc1NTFiYWFkfDE3MzE2MzcyNTM%3D&request=yes&_f=.pdf

If you need a detailed explanation of the circuit function, we can provide it separately

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I posted too much, but I will lose points if I delete my own reply? ? What?

maychang

"For example, I2 has no output, or I1 has no output. At this time, the current actually output to the load is less than 12mA."

I2 has no output, so the current on RL2 is zero. Are you saying that the current on RL1 is less than 12mA?

maychang

What is the voltage comparator model? What is the power supply?

chunyang

How is the current measured? Why not measure the voltage across the sampling resistor?

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chunyang posted on 2019-8-24 19:37 How is the current measured? Why not measure the voltage across the sampling resistor?

The current is the ammeter connected in series with the 96 ohm resistor. The voltage across the 96 ohm resistor will become a little lower, from 1.8V to 1.6V. This resistor is the main load.

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maychang posted on 2019-8-24 14:54 Voltage comparator model? Power supply?

It is written in the background, there is a chip data sheet, 24V DC power supply, and the op amp used for the comparator is TP2272

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maychang posted on 2019-8-24 14:49 "For example, I2 has no output, or I1 has no output. At this time, the actual current output on the load is less than 12mA." I2 ...

Now that you mention it, my description of the simplified diagram is a bit problematic. The simplified diagram should say that when I1 is not input, the current flowing through the 95 ohm resistor of I2 is less than 12mA.

The 95 and 96 ohms in the two pictures are the same resistor. The actual resistor is 96 ohms, which is a manual error.

maychang

Mongoose posted on 2019-8-25 16:40 Now that you say that, my description of the simplified diagram is a bit problematic. The simplified diagram should be the current flowing through the 95 ohm resistor of I2 when there is no input to I1...

As far as I can see, the purpose of this circuit is to compare the two currents I1 and I2. The larger current will be connected to the 96 ohm RL by the relay contacts.

According to your description, when I1 has current (less than 18.75mA) and I2 has current (equal to 18.75mA), the voltage across RL is 1.8V, and when I1 becomes zero, the voltage across RL decreases to 1.6V.

Is that right?

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maychang posted on 2019-8-25 17:02 As far as I can see, the purpose of this circuit is to compare the two currents I1 and I2. The larger current is connected to the 96 ohm RL by the relay contact. According to your description, I1 ...

What you described is correct.

When I mentioned 18.75, I thought of a question. When the voltage is 1.6v, the current on the 96 ohm resistor is about 11.5-11.6mA. If you do not connect this circuit and directly connect the 96 ohm resistor to 4-20mA, the voltage across the resistor is about 1.2v. No matter how you think about it, this phenomenon is very unscientific.

maychang

Mongoose posted on 2019-8-25 17:23 What you described is correct. When it comes to 18.75, I thought of a question. When it is 1.6v, the current on 96 ohms is about 11.5-11.6mA, and then don't connect this...

When the 96 ohm RL is directly connected to the 4-20mA, the voltage across the resistor is 1.2v. No matter how you think about it, this phenomenon is very unscientific.

This is a direct violation of Ohm's law. I doubt whether the voltmeter and ammeter you are using are accurate.

chunyang

Mongoose published on 2019-8-25 16:34 The current is the ammeter connected in series with the 96 ohm resistor. The voltage across the 96 ohm resistor will become a little lower, from 1.8V to 1.6V. This resistor is the main load.

It is recommended that you only measure the voltage. I strongly suspect that the problem is caused by your measurement operation process.

chunyang

In addition, during the test, use another meter to monitor the 24V power supply voltage throughout the process.

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chunyang posted on 2019-8-25 20:22 In addition, during the test, use another meter to monitor the 24V power supply voltage throughout the process.

OK, I'll test it tomorrow.

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maychang posted on 2019-8-25 18:01 When the 96 ohm RL is directly connected to the 4-20mA, the voltage across the resistor is 1.2v. No matter how you think about it, this phenomenon is very unscientific. ...

I'll change the watch tomorrow and test it again

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The issue is resolved.

I changed the meter and measured 99 ohms for the 96 ohm RL. Under normal circumstances, the voltage across the two ends is about 1.91v at 12mA. After disconnecting one signal, it is only 1.81V.

I replaced it with a 358 and the problem was solved. It turned out that there was something wrong with the TP2722. The 3 dollar op amp is not as good as the one that costs ten cents.