The voltage at the positive terminal and the voltage at the negative terminal of the follower are not equal

xiaxingxing

The voltage at the positive terminal and the voltage at the negative terminal of the follower are not equal [Copy link]

 

As shown in the test circuit below, the voltages at points A and B are not equal, with a difference of about 30mV. The voltages at points C and D are not equal, with a difference of about 30mV. The voltages at points E and F are equal. Thank you all.

Note: The IDAC power supply voltage is 5V, the op amp is OPA2145 (see the attached specification sheet), and the power supply is 20V/GND. opa2145.pdf (2.65 MB, downloads: 1)

2023-4-13 15:50 上传

点击文件名下载附件

chunyang

Are you measuring the actual object? Then check whether there is a short circuit between the differential inputs, whether the LED circuit is open or the MOS tube is not conducting. If the LED is lit, then the voltages at the two points EF must be unequal.

chunyang

However, the resistance parameters in the LED circuit are incorrect. The 100K resistor will cause the LED to fail to light up because the current is too small.

chunyang

If the circuit is to work properly, the operating current of the LED should at least be the lower limit of its rated operating current, and then 15V divided by this current can get the upper limit of the current limiting resistor.

maychang

【The positive terminal voltage and the negative terminal voltage of the follower are not equal】

What instrument did you use to measure the voltage? I'm afraid you used an ordinary voltmeter to measure it, but the internal resistance of the voltmeter is not large enough, which leads to measurement errors. The output resistance at point A is much larger than that at point B.

xiaxingxing

chunyang posted on 2023-4-13 17:29 Is it a real object? Then check whether there is a short circuit between the differential inputs, whether the LED circuit is open or the MOS tube is not turned on. If the LED is lit, then...

It is a real object measurement. Actually, there are 4 LEDs connected in series. The purpose of my circuit is to measure the consistency of LED lighting under very small current (this kind of LED will be slightly bright under uA level current, and the purpose of the circuit is to determine the consistency of LED)

xiaxingxing

maychang posted on 2023-4-13 17:35 [The voltage at the positive terminal and the voltage at the negative terminal of the follower are not equal] What instrument did you use to measure the voltage? I'm afraid you used an ordinary voltmeter to measure it. Due to the electric...

Senior, I measured it with an ordinary Fluke multimeter. Do you mean that in reality the voltage at point A is equal to the voltage at point B, and the voltage at point C is equal to the voltage at point D, right? Thank you!

xiaxingxing

chunyang posted on 2023-4-13 17:31 However, the resistance parameters in the LED circuit are incorrect. The 100K resistor will cause the LED to fail to light up because the current is too small.

Sir, I didn't explain the background clearly. This kind of LED can be slightly bright at uA level current. My purpose is to see whether the brightness is consistent (or if there is a bad one, it will not be lit). Thank you.

maychang

xiaxingxing posted on 2023-4-13 17:54 Senior, it is measured with an ordinary Fluke multimeter. You mean that in actual conditions, the voltage at point A is equal to the voltage at point B, and the voltage at point C is equal to the voltage at point D...

In reality, the voltage at point A and point B may not be equal (offset voltage), but the difference will not be as large as 30mV.

I guess you used the same multimeter to measure the voltage between point A and ground, and then measured the voltage between point B and ground. You calculated that the difference between the two measurement readings was 30mV.

led2015

Try to optimize the design of the follower, use temperature compensation technology, and optimize the design of the bias circuit

chunyang

xiaxingxing posted on 2023-4-13 17:52 This is a real-life measurement. In fact, 4 LEDs are connected in series. The purpose of my circuit is to measure the consistency of LED lighting under very small current (this kind of LED is at the uA level...

Then you don't need to use such a complicated circuit, just connect a high-value resistor in series. If you want to know more about parameters such as junction voltage drop and current, use two resistors in series, and the one close to the ground also serves as a sampling resistor. Take a suitable value and send it to the ADC after 1:1 buffering. The design points are just to pay attention to the input impedance of the op amp to be high enough, and the temperature coefficient of the current limit and sampling resistors to be low. The overall cost is much lower than your circuit, and no DAC is needed. You can even use a multimeter to measure directly, and the circuit is not needed. Instead, you need to design and make a tooling to improve the test efficiency for batch production.

chunyang

xiaxingxing posted on 2023-4-13 17:54 Senior, it is measured with an ordinary Fluke multimeter. You mean that in actual conditions, the voltage at point A is equal to the voltage at point B, and the voltage at point C is equal to the voltage at point D...

They should be basically equal because the op amp has an input offset voltage, but it is very small and its upper limit can be found by checking the device manual.

chunyang

led2015 posted on 2023-4-13 19:17 Try to optimize the design of the follower, use temperature compensation technology, and optimize the design of the bias circuit

It has nothing to do with what the OP asked...

xiaxingxing

chunyang posted on 2023-4-13 20:41 Then you don't have to use such a complicated circuit, just connect a high-value resistor in series. If you want to know more about the parameters such as junction voltage drop and current, use two resistors...

OK, I will optimize the circuit according to your suggestion, thank you very much!

xiaxingxing

chunyang posted on 2023-4-13 20:42 They should be basically equal, because the op amp has input offset voltage, but it is very small. You can find its upper limit by checking the device manual.

Hmm, the offset voltage of this op amp is very small, I'll check it out, thank you!

xiaxingxing

maychang posted on 2023-4-13 18:01 In reality, the voltage at point A and the voltage at point B may not be equal (offset voltage), but the difference will not be as much as 30mV. I guess you are using the same multimeter, ...

"I guess you used the same multimeter to measure the voltage between point A and ground, and then measured the voltage between point B and ground. You calculated that the difference between the two readings was 30mV."

----Haha, that's how I measured it.

You said earlier that "the output resistance at point A is much larger than that at point B." How do you understand this sentence? Thank you!

bobde163

Theoretically, when used as a follower, the voltage difference between the two input ends will not be so large. I don't know how the OP calculated this voltage difference. Did you measure the two voltages separately and subtract them, or directly test the voltage value between the two points? You need to consider the influence of the input impedance introduced by the measuring instrument.

maychang

xiaxingxing posted on 2023-4-14 08:54 "I guess you used the same multimeter to measure the voltage between point A and ground, and then measured the voltage between point B and ground. The two readings you calculated differed by 30m...

The output resistance at point B refers to the resistance from the op amp inverting input terminal to the outside, which is connected to the op amp output terminal. The op amp works in a linear state with strong negative feedback, and the output resistance is extremely small. The output resistance at point A refers to the resistance from the op amp inverting input terminal to the outside, which is related to the working current of the LED. If the working current of the LED is very small (such as uA level as you said), then the dynamic resistance of the LED is 20 kilo ohms in parallel, which is at least in the order of tens of kilo ohms.

xiaxingxing

bobde163 posted on 2023-4-14 08:58 Theoretically, when used as a follower, the pressure difference between the two ends of the input will not be so large. I don’t know how the OP calculated this pressure difference. Is it measured separately at the two ends...

The voltage between point A and ground and point B is measured separately, and then the two voltages are subtracted. I just changed the test method and directly measured the voltage difference between points A and B. The voltage difference is 0. Thank you!

xiaxingxing

maychang published on 2023-4-14 09:02 The output resistance at point B refers to the resistance seen from the inverting input terminal of the op amp, which is connected to the output terminal of the op amp, and the op amp works in line...

I just changed the test method and directly measured the voltage difference between points A and B. The voltage difference is 0. What is the input impedance of a common multimeter in ohms? Thank you!