There is a peak in the bandpass filter curve. How to eliminate it?

zk643

There is a peak in the bandpass filter curve. How to eliminate it? [Copy link]

 

According to the curves of each frequency band measured by the circuit in the figure, there is a peak at 10HZ and 100HZ. What is the influence of this? Also, the bandwidth is not enough at 100K and 1M frequency bands. I don’t know what’s going on?

gmchen

There is a peak which should be due to the high Q value.

The bandwidth of 100k and 1M cannot be increased. There are many factors. It is difficult to judge accurately without a complete circuit diagram. The general factors are: 1. If the resistance of the two filters is the same as the low frequency and only the capacitance is changed, the capacitance should be 100p and 10p, then the distributed capacitance will have an impact. 2. The frequency response of the op amp will also have an impact, but it should not be as big as the previous one.

zk643

gmchen posted on 2020-7-30 09:31 There is a spike, which should be due to the high Q value. The bandwidth of 100k and 1M cannot be increased, which is related to many factors. It is difficult to judge accurately without a complete circuit diagram. Roughly...

Yes, the capacitors for 100K and 1M are 100p and 10p, is it because the capacitor is too small? Should the capacitor be increased and the resistance value recalculated? How to solve the problem of high Q value?

gmchen

Yes, the capacitor cannot be too small, but the resistor cannot be too small either, so the value should be determined through debugging. As for the Q value, I need to study the transfer function to answer you.

zk643

gmchen posted on 2020-7-31 10:37 Yes, the capacitor cannot be too small, but the resistor cannot be too small either, so the value should be determined through debugging. As for the Q value, we need to study the transmission...

Oh. Thank you for your help and advice. I will calculate the value of the capacitor first.

gmchen

The method to reduce the Q value is as follows: increase the resistance of the 6.93k resistor and reduce the resistance of the 23.3k resistor, but ensure that their product remains unchanged. The amount of increase and decrease can be determined through experiments.

zk643

But there is only one solution to calculate the resistance value according to the equation.

gmchen

zk643 posted on 2020-8-1 12:51 But there is only one solution to calculate the resistance value according to the equation.

You calculated it according to the equations I gave you in another post.

The premise of the equations is that the op amp and the capacitor are ideal. But in fact, neither of them is ideal, so the resistance value obtained by the above equation is likely to be incorrect. This can also be proved from your experimental curve: the actual -3dB cutoff frequency is 13Hz and 120+Hz, which is higher than the expected 10 and 100.

So the next adjustment should be: 1. Increase the value of the two capacitors (parallel capacitors) so that the cutoff frequency is adjusted to 10 and 100 Hz. 2. If the amplitude frequency still has a bump after the cutoff frequency is adjusted, adjust the two resistors according to my previous reply.

zk643

gmchen posted on 2020-8-1 14:43 zk643 posted on 2020-8-1 12:51 But there is only one solution to calculate the resistance value according to the equation. Are you following the one I gave in another post...

Doubling the capacitance means that the cutoff frequency is reduced by half, and 10HZ becomes 5HZ. This is the curve of the measured filter. There is still a peak at the end from 2HZ to 5HZ.

gmchen

zk643 published on 2020-8-5 11:39 Doubling the capacitance means that the cut-off frequency is reduced by half, and 10HZ becomes 5HZ. This is the curve of the measured filter. At the end of 2HZ to 5HZ, there will still be...

Have you tried changing the resistor?

zk643

gmchen posted on 2020-8-5 13:53 zk643 posted on 2020-8-5 11:39 Doubling the capacitance means that the cutoff frequency is reduced by half, and 10HZ becomes 5HZ. This is the measurement of the filter...

gmchen

zk643 Published on 2020-8-5 11:39 Doubling the capacitance means that the cut-off frequency is reduced by half, and 10HZ becomes 5HZ. This is the curve of the measured filter. At the end of 2HZ to 5HZ, there will still be...

Oh, I understand. When I told you to increase the capacitance and connect capacitors in parallel, I didn't mean to connect the same capacitor in parallel! The original capacitor cutoff frequency is about 13Hz, which means that the actual capacity of the capacitor is only about 80% of the nominal value. This situation is very common. The tolerance of many capacitors can reach 20%. So you either choose a capacitor with a small tolerance, or connect a small-capacity capacitor in parallel. In short, let the cutoff frequency reach 10Hz.

zk643

gmchen posted on 2020-8-5 18:50 Oh, I see. When I said I wanted you to increase the capacitance and connect capacitors in parallel, I didn’t mean I wanted you to connect the same capacitor in parallel! The original capacitor cutoff frequency is about...

The problem of increased gain was found. The 20K input resistor of the first op amp was removed and then forgotten to be connected, which led to the subsequent output ratio comparison. After adding this resistor, the three capacitors were still doubled. When the resistor was changed from 6.9K 23.3K to 10.2K 15.8k, there was no spike from 0.1HZ to 5HZ, and the signal began to attenuate after 5HZ.

zk643

gmchen posted on 2020-8-5 18:50 Oh, I see. When I said I wanted you to increase the capacitance and connect capacitors in parallel, I didn’t mean I wanted you to connect the same capacitor in parallel! The original capacitor cutoff frequency is about...

This is the graph I measured again. The blue one is the curve of the previous value, and the red one is the curve measured when the capacitor is increased and the resistance is changed to 10.2K and 15.8K

gmchen

zk643 posted on 2020-8-6 13:42 This graph is the curve of the previous value, and the red one is the curve measured when the capacitor is increased and the resistance is changed to 10.2K and 15.8K

OK, it worked.

zk643

gmchen posted on 2020-8-6 16:25 OK, it worked.

Yes, thanks for the guidance, the next step is to adjust the problem of insufficient bandwidth at high frequency 1M. But there is still one thing I am not clear about, that is, what is the function of the string of resistors and capacitors in the wreath in the picture, and what effect will adjusting them have?

gmchen

zk643 posted on 2020-8-7 09:14 Yes, thanks for the guidance, the next step is to adjust the problem of insufficient bandwidth at high frequency 1M. But there is one thing I am not very clear about, that is, the picture...

This is a high-frequency compensation network. When the signal frequency increases, the capacitance reactance decreases, and the 1k resistor is connected in parallel with R14, which makes the actual resistance of R14 decrease and the gain of the amplifier increase.

By the way, I am a school teacher, not a factory master.

gmchen

The RC product in the red circle determines the turning point frequency of the compensation, and R determines the amount of compensation.

zk643

gmchen published on 2020-8-7 11:34 This is a high-frequency compensation network. When the signal frequency increases, the capacitance reactance decreases, and the 1k resistor is connected in parallel with R14, which makes the actual resistance of R14 decrease, ...

Oh, now I know this, I will call you teacher. So the bandwidth of this circuit design is 0.1HZ--1MHZ, is it mainly to give a 1M signal at the input end and then see if the output is proportionally amplified, if not, then adjust R14 and C3 to make the signal proportionally amplified? So in theory, should the resistance value be larger to maintain the proportion?

gmchen

This is the same as the measurement method you used in the first building. Input a signal, change the frequency, and then look at the ratio of its output to input. Make an amplitude-frequency characteristic curve like the figure in the first building. Its flat area should be 0.1Hz~1MHz. If it is not right, then adjust the RC