Please tell me the role of these two resistors in the circuit

littleshrimp

Please tell me the role of these two resistors in the circuit [Copy link]

 

This is the circuit generated by the ADI analog filter wizard. The amplifier's IN+ is not directly connected to Ref, but is connected through R4A, R3A, R4B, and R3B "voltage divider"

What role do the resistors R4A, R3A, R4B, and R3B play in the circuit?

captzs

This connection method is positive feedback, which can increase the output amplitude and is easy to self-excite, so control R3A<<R4A.

maychang

After voltage division, it is connected to the non-inverting input terminal, which is a positive feedback effect, the purpose of which is to improve the Q value of the filter.

captzs

maychang posted on 2019-8-28 14:37 After the voltage is divided, it is connected to the in-phase input terminal. This is a positive feedback effect, the purpose of which is to improve the Q value of the filter.

I wanted to get a larger filter output, so I tested this form of positive feedback. When the original circuit gain was large, the positive feedback could only be a few thousandths. If it was larger, it would self-excite. The amplitude increased, but I didn't pay attention to calculate the Q value.

maychang

captzs posted on 2019-8-28 15:21 I want to get a larger filter output, so I test this form of positive feedback. When the original circuit gain is large, the positive feedback can only be a few thousandths. If it is larger...

That's right. This positive feedback should not control the Q value, but change the gain. But this approach is prone to self-excitation.

Moreover, the circuit has already changed the gain through R1A, R2A, R1B, R2B, so there is no need to add R3A, R3B, R4A, R4B.

maychang

Teacher gmchen once published "Talking about Active Filters - Bandpass Filters" , in which he talked about multi-channel feedback bandpass filters. Changing the gain is to change the resistors R1A and R2A in the first post.

captzs

maychang posted on 2019-8-28 15:38 Teacher gmchen once published "Talking about Active Filters - Bandpass Filters", in which he talked about multi-feedback bandpass filters. Change...

At first, in order to filter and bootstrap, to overcome the influence of the internal resistance of the signal source on the input signal and suppress noise, two-stage active filtering was required for the phase relationship, but it was difficult to ensure the accuracy of the amplitude after the fourth order, so the first-order method was used instead, but it was unsuccessful. So I had an impression of the circuit posted by the host.

littleshrimp

maychang posted on 2019-8-28 15:31 Yes. This positive feedback should not control the Q value, but change the gain. But this approach is prone to self-excitation. Moreover, the circuit has been passed through R1A ...

This circuit was generated by the ADI analog filter wizard below. I tried it and found that there were no two resistors when the center frequency was below 35KHz. When the frequency was set to above 40KHz, there would be two more resistors.

https://www.analog.com/designtools/cn/filterwizard/

littleshrimp

captzs posted on 2019-8-28 16:05 In order to filter and bootstrap at the beginning, in order to overcome the influence of the internal resistance of the signal source on the input signal and suppress noise, two-stage active filtering is required in terms of phase relationship, but...

This circuit was generated by the ADI analog filter wizard below. I tried it and found that there were no two resistors when the center frequency was below 35KHz. When the frequency was set to above 40KHz, there would be two more resistors.

https://www.analog.com/designtools/cn/filterwizard/

maychang

littleshrimp posted on 2019-8-29 10:16 This circuit was generated by the ADI analog filter wizard below. I tried it and found that the two resistors were not present when the center frequency was below 35KHz...

"I tried it, and when the center frequency is below 35KHz, there are no two resistors. When the frequency is set above 40KHz, there will be two more resistors."

I don't know why there are two more resistors when the frequency is high.

captzs

littleshrimp posted on 2019-8-29 10:16 This circuit was generated by the ADI analog filter wizard below. I tried it and found that there were no two resistors when the center frequency was below 35KHz...

It may be used to adjust the compensation output amplitude when the frequency increases, because it is difficult to achieve the specified gain and keep fc unchanged using R5A/R1A adjustment. Guess.

littleshrimp

captzs posted on 2019-8-29 11:30 It may be used to adjust the compensation output amplitude when the frequency increases, because it is difficult to achieve the specified gain and keep fc unchanged using R5A/R1A adjustment. Guess.

I did some simulation, this circuit design is 40dB 40KHz

When simulating, it was found that the actual gain was only 30dB

In this circuit, using r8, r9, r10, r10 and directly connecting Vref to the positive input of the amplifier have basically the same effect, and has no effect on the gain.

littleshrimp

maychang posted on 2019-8-29 10:28 "I tried it, and when the center frequency is below 35KHz, there are no two resistors. When the frequency is set to above 40KHz, there will be two more resistors...

The difference between the two resistors is not very obvious during simulation.

littleshrimp

captzs posted on 2019-8-29 11:30 It may be used to adjust the compensation output amplitude when the frequency increases, because it is difficult to achieve the specified gain and keep fc unchanged using R5A/R1A adjustment. Guess.

I tried again. The Bode plot did not change after changing the resistance value, but the total number of amplifications on the oscilloscope changed significantly.

captzs

littleshrimp posted on 2019-8-30 14:15 I tried it again. The Bode plot did not change after changing the resistance value, but the total number of amplifications on the oscilloscope changed significantly

My simulation has always been like this. The display of the amplitude-frequency meter and the oscilloscope are different, not to mention the slight difference in frequency. As you said, the amplitude-frequency meter curve cannot be seen with positive feedback and without positive feedback, but the oscilloscope has obvious differences. I have repeatedly considered that the amplitude-frequency meter reflects the circuit characteristics of the circuit itself, and positive feedback has little effect on frequency, so the amplitude-frequency curve has not changed, and it is the same no matter what frequency is input.

meiyao

Voltage divider circuit.

littleshrimp

captzs posted on 2019-8-30 16:14 My simulation has always been like this. The display of the amplitude-frequency meter and the oscilloscope is different. Let’s not talk about the frequency difference. As you said, there is positive feedback and there is no positive feedback...

Are you also using multisim? Are there any other good simulation software? I can compare these two softwares.

gmchen

littleshrimp posted on 2019-8-30 14:15 I tried it again. The Bode plot did not change after changing the resistance value, but the total number of amplifications on the oscilloscope changed significantly

This circuit adds a little positive feedback. Let F = R9/(R8+R9), which is the positive feedback coefficient. The result of the transfer function analysis is as follows:

1. The center frequency has an upward trend, and the increase is approximately (1+3F)/(1-F).

2. The Q value has a strong upward trend, and the increase is about (1+3F)/(1-QF). From this formula, we can also see that if F>=1/Q, the circuit will self-excite.

3. The passband gain has a strong upward trend, and the increase is approximately 1/(1-QF).

The purpose of this positive feedback is probably to compensate for the non-ideal characteristics of the components. It is recommended to try replacing op amps with different gain-bandwidth products during simulation.

littleshrimp

gmchen posted on 2019-9-5 21:46 This circuit adds a little positive feedback. Let F=R9/(R8+R9), which is the positive feedback coefficient. The results of the transfer function analysis are as follows: 1. Center frequency...

Teacher, I have soldered the circuit, but this situation occurred during testing. Can you help me take a look?

https://bbs.eeworld.com.cn/thread-1089632-1-1.html

xiaxingxing

captzs posted on 2019-8-28 15:21 I want to get a larger filter output, so I test this form of positive feedback. When the original circuit gain is large, the positive feedback can only be a few thousandths. If it is larger...

You asked in the first post that R3A<<R4A, so the positive feedback coefficient F is very small. So in the case of R3A<<R4A, isn't the positive feedback gain Av=1/F very large? Doesn't this contradict your statement that "positive feedback can only be a few thousandths"? Thank you!