Confusion about the static voltage of op amp

sfesdm

Confusion about the static voltage of op amp [Copy link]

 

Now we have the op amp circuit shown in Figure 1. After removing the capacitor of the AC part, we measure the static point of the circuit. Since we are not familiar with op amps, we encounter a problem and don’t know how to deal with it. So I would like to ask everyone for advice. Thank you!

Figure 1

Simplify the circuit in Figure 1 into Figure 2

Figure 2

According to the voltage obtained by the voltage division ratio, Vp=0.985V, Vn=Vp, at this time Vo=Vn-0.035V. If the value of R24 is changed to 10K, then Vo=Vn. Why is there such a result? Under the premise that R24=2M is not changed, is there any way to make Vo=Vn?

Thank you!

maychang

"According to the voltage division ratio, Vp=0.985V, Vn=Vp, and Vo=Vn-0.035V" Is this "Vo=Vn-0.035V" a simulation result or a measurement result of an actual circuit? Regardless of simulation or actual circuit, what is the op amp model?

sfesdm

maychang posted on 2018-6-13 19:55 "According to the voltage obtained by the voltage division ratio, Vp=0.985V or so, Vn = Vp, at this time Vo=Vn-0.035V or so" Your "Vo=Vn-0.035V or so...

The simulation results are similar to the actual circuit results. The simulation uses LM324 and the actual circuit uses LM321

gmchen

The voltage drop of the bias current at the input of the op amp across the 2M resistor.

maychang

sfesdm posted on 2018-6-13 20:27 The simulation results are similar to the actual circuit results. The simulation uses LM324 and the actual circuit uses LM321

This type of op amp is a relatively "rough" op amp, with a large input bias current and offset voltage. If the bias current is 0.1uA, there will be a 0.2V voltage on the 2M resistor.

sfesdm

maychang posted on 2018-6-13 21:44 This type of op amp is a relatively "rough" op amp, with a large input bias current and a large offset voltage. If the bias current is 0.1uA, on a 2M resistor...

So it is caused by the bias voltage. There is no other way except to choose an op amp with a small bias voltage. Thank you

sfesdm

gmchen posted on 2018-6-13 20:52 The voltage drop of the bias current at the input of the op amp on the 2M resistor.

It turns out to be a problem of bias voltage. Thank you

maychang

sfesdm posted on 2018-6-13 22:03 So it is caused by the bias voltage. There is no other way except to choose an op amp with a small bias voltage. Thank you

"There is no other way except to choose an op amp with a small bias voltage." From the picture in the first post, there is a capacitor C18 at the input end. This is an AC amplifier (the output end is not fully drawn, so I don't know if there is a series capacitor). So the problem you found has no effect on the operation of the amplifier. It is estimated that there is no need to replace the op amp.

sfesdm

maychang published on 2018-6-14 06:51 "There is no other way except to choose an op amp with a small bias voltage." From the picture in the first post, there is a capacitor C18 at the input end. This is an AC amplifier (output...

Vp and Vo are connected to the comparator for comparison. When static, the values of Vp and Vo are close to equal. When a small AC signal enters, the value of Vo is higher than the value of Vp, generating a comparison result. In fact, the comparison result can be generated now, but the AC signal required is slightly larger, and the effect is not so ideal.

maychang

[quote]sfesdm posted on 2018-6-14 11:36 Vp and Vo are connected to the comparator for comparison. In static state, the values of Vp and Vo are close to the same. When a small AC signal enters, the value of Vo is higher than the value of Vp, resulting in a comparison result. What a weird design. Vp is also connected to one of the input terminals of the comparator? Then the input bias current of the comparator should also be provided by R27 and R7//R26?

maychang

[quote]sfesdm posted on 2018-6-14 11:36 Vp and Vo are connected to a comparator for comparison. In static state, the values of Vp and Vo are close to being equal. When a small AC signal enters, the value of Vo is higher than the value of Vp, producing a comparison result. What is the purpose of connecting Vp and Vo to a comparator? To judge whether an AC signal is present or absent? Then just amplify the AC signal and compare it with a fixed voltage. Why is it necessary to compare it with the input signal?

sfesdm

maychang posted on 2018-6-14 12:00 Vp and Vo are connected to the comparator for comparison. When static, the values of Vp and Vo are almost equal. When a small AC signal enters, the value of Vo...

The reference is this circuit, officially provided by TI, PIR detection. In my circuit, the comparator is built-in to the MCU.

maychang

sfesdm posted on 2018-6-14 12:06 The reference is this circuit, officially provided by TI, PIR detection. In my circuit, the comparator is built-in MCU.

This circuit compares the output signal of U2B with a fixed voltage, not Vp with Vo. This is the case for both comparators U3 and U4.

maychang

sfesdm posted on 2018-6-14 12:06 The reference is this circuit, which is provided by TI, PIR detection. In my circuit, the comparator is built-in MCU.

I didn't find the manual of the U2B op amp model. Judging from the 15M ohms of R8~R11, the input bias current of this op amp must be in the pA level. The typical value of the input bias current of U3U4 is 30pA.

gmchen

gmchen

sfesdm posted on 2018-6-13 22:03 It turns out to be a bias voltage problem, thank you

It turns out to be PIR detection. This problem is easy to solve. Just add a zero adjustment potentiometer to the circuit and adjust the zero point of the op amp output to the required level. Specifically, it is to fine-tune the value of R26.

maychang

sfesdm posted on 2018-6-14 12:06 The reference is this circuit, which is provided by TI, PIR detection. In my circuit, the comparator is the MCU's own.

In this circuit, if R7 uses 2M ohms (R24 in the first post), then U2B can be replaced with a common op amp such as TL082, and R8~R11 can be reduced to less than 1M ohms.

sfesdm

maychang posted on 2018-6-14 16:31 In this circuit, if R7 uses 2M ohms (R24 in the first post), then U2B can be replaced with a common op amp such as TL082, and R8~R11 can be reduced to less than 1M ohms.

OK, I will try to adjust the circuit again, thank you

maychang

sfesdm posted on 2018-6-14 19:18 OK, I will try to adjust the circuit again, thank you

It is recommended to use TL082 because the op amp has JFET input and the input bias current is typically 30pA at room temperature and 200pA at maximum.