Band pass filter not achieving cut off frequencies

Thread Starter

CydneyC26

Joined Dec 30, 2020
4
I've got to design a second order band pass filter than has cut-off frequencies of 1kHz and 250kHz. I have to use the E12 series for resistors and capacitors so I can't get that exact bandwidth - but as near as.

I initially designed a low pass filter and then a high pass filter individually, which achieved near enough the right frequencies. Upon combining the two stages, my upper cut off frequency has reduced by over half.

Please can someone explain why this might be? My knowledge is very limited, hence my design I've tried to keep easy to understand.
 

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Papabravo

Joined Feb 24, 2006
16,157
The model for a 741 cannot slew fast enough at 250 kHz, thus reducing the high frequency gain. Try a different opamp with a higher GBW.
 

LvW

Joined Jun 13, 2013
1,247
Yes - watch the slew rate.
More than that, you should not use resistors as low as 10 ohms. The reason is as follows: The preferred resistor range is between 1kOhm and 100 kOhms (if possible!). This is because allresistors should be small (large) if compared with the input (output) resistance of the used opams. Normally , these (parasitic) opamp resistances are neglected during the design process.
 

MrAl

Joined Jun 17, 2014
8,249
I've got to design a second order band pass filter than has cut-off frequencies of 1kHz and 250kHz. I have to use the E12 series for resistors and capacitors so I can't get that exact bandwidth - but as near as.

I initially designed a low pass filter and then a high pass filter individually, which achieved near enough the right frequencies. Upon combining the two stages, my upper cut off frequency has reduced by over half.

Please can someone explain why this might be? My knowledge is very limited, hence my design I've tried to keep easy to understand.
When you design two filters and put them together there could be some interaction or the two responses could simply affect each other. This means that you have to analyze the entire filter with both sections at the same time not individually. You can think about them individually to get a rough idea how the entire filter works and therefore make it a little more intuitive, but for the exact analysis you have to describe the entire filter with one function in order to analyze it with more accuracy.

For a second order bandpass there are two cutoff frequencies the low FL and the high FH. The amplitudes are 3db down from the center gain of the filter. Since -3db corresponds to the center gain G divided by the square root of 2 (which is therefore G/sqrt(2)) you can use that as a guide to find the lower and upper cutoff frequencies.

We can break this down into four steps:
1. Find the transfer function sometimes referred to as T(s) and T(w) where w is angular frequency. You then find the amplitude of T(w) which we can call A(w) for now.
2. Find the mid point gain G from the transfer function A(w).
3. Find the lower cutoff frequency from G and A(w).
4. Find the upper cutoff frequency from G and A(w).

The analytic procedure to find the transfer function can be any suitable analysis method such as nodal analysis.
The most general procedure to find G is to take the first derivative of A(w) with respect to w and set it equal to zero, then solve for w and check that you found the maximum of A(w) which will be G. That gives you a numerical value (or analytic value) for G.
Once you have G, divide by the square root of 2 to get the -3db cutoff amplitudes. This of course will be G/sqrt(2) for both which we can call AL and AH. So both are the same.
Next you just form two expressions:
A(w)=AL
and
A(w)=AH

and solve for w in each case that will give you wL and wH which are the two angular cutoff frequencies. Divide both by pi and you get the two frequencies in Hertz.

I hope this makes sense to you. This is the most general way to do it so in theory it will work with every second order bandpass filter. There could be shortcuts too but they are usually specific to one particular type of filter design.

If you need an example i can supply one. It is not nearly as hard to do as it sounds when you use automatic math software. The midband gain G is quite easy to find even though it may turn out to be a fourth order equation in w because math software makes all this so much easier than doing it by hand.
 

Thread Starter

CydneyC26

Joined Dec 30, 2020
4
The model for a 741 cannot slew fast enough at 250 kHz, thus reducing the high frequency gain. Try a different opamp with a higher GBW.
Thanks for the suggestion, I've swapped out the 741 for a OPA197 and I am now achieving the correct cut off frequencies. However, the roll off of upper stopband is much steeper and does not appear to be -40dB - do you know why this might be?
 

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Thread Starter

CydneyC26

Joined Dec 30, 2020
4
When you design two filters and put them together there could be some interaction or the two responses could simply affect each other. This means that you have to analyze the entire filter with both sections at the same time not individually. You can think about them individually to get a rough idea how the entire filter works and therefore make it a little more intuitive, but for the exact analysis you have to describe the entire filter with one function in order to analyze it with more accuracy.

For a second order bandpass there are two cutoff frequencies the low FL and the high FH. The amplitudes are 3db down from the center gain of the filter. Since -3db corresponds to the center gain G divided by the square root of 2 (which is therefore G/sqrt(2)) you can use that as a guide to find the lower and upper cutoff frequencies.

We can break this down into four steps:
1. Find the transfer function sometimes referred to as T(s) and T(w) where w is angular frequency. You then find the amplitude of T(w) which we can call A(w) for now.
2. Find the mid point gain G from the transfer function A(w).
3. Find the lower cutoff frequency from G and A(w).
4. Find the upper cutoff frequency from G and A(w).

The analytic procedure to find the transfer function can be any suitable analysis method such as nodal analysis.
The most general procedure to find G is to take the first derivative of A(w) with respect to w and set it equal to zero, then solve for w and check that you found the maximum of A(w) which will be G. That gives you a numerical value (or analytic value) for G.
Once you have G, divide by the square root of 2 to get the -3db cutoff amplitudes. This of course will be G/sqrt(2) for both which we can call AL and AH. So both are the same.
Next you just form two expressions:
A(w)=AL
and
A(w)=AH

and solve for w in each case that will give you wL and wH which are the two angular cutoff frequencies. Divide both by pi and you get the two frequencies in Hertz.

I hope this makes sense to you. This is the most general way to do it so in theory it will work with every second order bandpass filter. There could be shortcuts too but they are usually specific to one particular type of filter design.

If you need an example i can supply one. It is not nearly as hard to do as it sounds when you use automatic math software. The midband gain G is quite easy to find even though it may turn out to be a fourth order equation in w because math software makes all this so much easier than doing it by hand.
Thanks very much for putting the time in this response. I'm a little lost though because I have already been told my cut off frequencies are 1kHz and 250kHz - therefore I don't need to calculate these like you mentioned, do I?

I also thought that because of the configuration of my circuit, that the op-amps act as unity gain.
 

LvW

Joined Jun 13, 2013
1,247
Thanks for the suggestion, I've swapped out the 741 for a OPA197 and I am now achieving the correct cut off frequencies. However, the roll off of upper stopband is much steeper and does not appear to be -40dB - do you know why this might be?
Yes - that is an unwanted effect for each Sallen-Key lowpass. The problem is as follows:
For rising frequencies (above the cut-off) , the output of the opamp decreases - and at the same time an even increasing portion of the input arrives DIRECTLY via R4-C4 at the output and produces an unwanted output voltage across the finite (and also increasing) output impedance of the opamp. More than that, due to phase shift effects the opamps output impedance contains an inductive part which - together with C4 - creates a resonant effect (as can be seen directly after cut-off).
Note: For rising frequencies, the output impedance of the unity-gain amplifier increases because the loop gain goes down with frequency.

Reminder: The ac analysis (Bode diagram) is a small-signal analysis only - and you cannot see any slew rate induced distortions.
 
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Thread Starter

CydneyC26

Joined Dec 30, 2020
4
Yes - that is an unwanted effect for each Sallen-Key lowpass. The problem is as follows:
For rising frequencies (above the cut-off) , the output of the opamp decreases - and at the same time an even increasing portion of the input arrives DIRECTLY via R4-C4 at the output and produces an unwanted output voltage across the finite (and also increasing) output impedance of the opamp. More than that, due to phase shift effects the opamps output impedance contains an inductive part which - together with C4 - creates a resonant effect (as can be seen directly after cut-off).
Note: For rising frequencies, the output impedance of the unity-gain amplifier increases because the loop gain goes down with frequency.

Reminder: The ac analysis (Bode diagram) is a small-signal analysis only - and you cannot see any slew rate induced distortions.
So I would be better off designing a filter that has a feedback loop as opposed to unity gain?
 

LvW

Joined Jun 13, 2013
1,247
So I would be better off designing a filter that has a feedback loop as opposed to unity gain?
When this is a problem for your application - YES.
Unless you have an additional opamp available. In this case, you could use it as a buffer between the output and C4.
As an alternative, use a lowpass in MFB-topology (and not Sallen-Key).
 

MrAl

Joined Jun 17, 2014
8,249
Thanks very much for putting the time in this response. I'm a little lost though because I have already been told my cut off frequencies are 1kHz and 250kHz - therefore I don't need to calculate these like you mentioned, do I?

I also thought that because of the configuration of my circuit, that the op-amps act as unity gain.
Well from that you can derive everything else, but also to check your results.
 
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