Designing an Active Band Pass Filter

[Lawrenciumbc]

I need to build a band pass filter for a receiver circuit I am building but a couple of things have left me scratching my head:

I decided on a 2 back to back 2 pole Sallen Key filters and my supply voltage is 5V DC from the breadboard rail. Mainly due to the simplicity as I am new to electronics and I think it will be adequate for my circuit.

My center frequency is to be 1.2KHz and I would like this to be as narrow as possible but the main purpose of the filter is to remove ambient noise (the input single is from a photo detector - see datasheet TSL 14S-LF).

- My main question is how do I start when choosing my cut-off frequencies?
- would say a 20% either side frequency gap be sensible and realistic (i.e. 1KHz for the high pass and 1.4KHz for the low pass)?
- Does my capacitor and resistor values bear any relevance to the opposite filter or is it all a ratio to itself? (e.g. could I use pf capacitors in 1 and uf capacitors in the other for instance)

 

[Harald Kapp]

I suggest you use a band-passs filter circuit instead of two back-to-back filters. This allows you to set center frequency and quality factor (or bandwidth) directly. Here are some design notes.

 

[LvW]

- My main question is how do I start when choosing my cut-off frequencies?
- would say a 20% either side frequency gap be sensible and realistic (i.e. 1KHz for the high pass and 1.4KHz for the low pass)?
- Does my capacitor and resistor values bear any relevance to the opposite filter or is it all a ratio to itself? (e.g. could I use pf capacitors in 1 and uf capacitors in the other for instance)

The answer depends on your damping requirements oiutside the pathband.
A lowpass-highpass combination is appropriate if the bandwidth is rather large and if - at the same time - you need a slope of at least 40dB/dec for the transition from the passband to stop band.
A second.order bandpass circuit has an out-of-band attenuation slope of 20 dB/dec only.

 

[Lawrenciumbc]

I suggest you use a band-passs filter circuit instead of two back-to-back filters. This allows you to set center frequency and quality factor (or bandwidth) directly. Here are some design notes.

Ok that's great, i'll try the 'Infinite Gain Multiple Feedback Active Filter' as that appears to be the best performer according to the notes.

With that in mind, so that I get the terminology right, is this type of filter still a '2nd order/2 pole filter' or is it just a single order/1 pole as there is only one instance of a low pass and a high pass embedded within it?

 

[LvW]

Ok that's great, i'll try the 'Infinite Gain Multiple Feedback Active Filter' as that appears to be the best performer according to the notes.

With that in mind, so that I get the terminology right, is this type of filter still a '2nd order/2 pole filter' or is it just a single order/1 pole as there is only one instance of a low pass and a high pass embedded within it?

As I have explained in my former answer - it is a second-order filter (one complex pole pair). However, the transition from the passband to stopband has single-oder properties only (slope 20dB/dec). For this reason, some authors consider such a bandpass filter to be "first-order" only. However, for my opinion, this is not the correct terminology because the denominator is second order.

 

[Lawrenciumbc]

Ok no problem, I have built the filter around my center frequency according to the calculations on the site suggested. (I am using 2 10nf capacitors and 1x10Kohm resistors for R1 and 2x10Kohm for R2 = 1125Hz fc).

However the results on the oscilliscope seem a little strange. I can see the signal clearly attenuating/'flattening' when I go down to 100Hz, but when I increase the frequency of my signal to 10KHz and 100KHz, the output on the oscilliscope actually seems to be quite clear. At 200KHz the signal appears heavily attenuated again.

Does this suggest that my center frequency is shifted somehow to around 100KHz or have I done the maths wrong?! Or, is it just because it is a low order filter and I should expect a relatively wide-ish bandwidth?

I can upload some images of my oscilliscope results if you want me to?

 

[LvW]

I did not check your dimensioning in detail - however, I think the resistor spread must be much larger (larger resistor ratio) for two equal capacitors.

In the mentioned Paper the example is for Q=0.707.
That means: bandwidth B=1.4*Fo. This is a "bandpass" that does not deserve this name.
For a center frequency of 1kHz the bandwidth is 1,4 kHz.
Recommendation: Select the resistor ratio around Q=10.

 

[Lawrenciumbc]

To make a Q=10 then, something like 1Kohm and a 10Kohm should do the trick? with 2x 10nf capacitors, this would give a Q of 10 and a center frequency of 1591Hz (close enough considering the components I have available).

Would I therefore be right in thinking that this should tighten up my bandwidth a little more?

Just out of interest, what would generally be considered a high Q factor? (i.e. how high could you realistically go with the ratio) and is there a likely reason as to why the example chosen in the paper is so low (0.707)?

Many thanks for your help by the way, I've been trying to get my head around this and its much clearer now.

 

[LvW]

I have some experients with filters. For a Q value of 10 and two equal capacitors you need a resistor ratio of 4Q²=400.
In this case, the 3dB-bandwidth is B=fo/Q.
It is up to you to decide which bandwidth meets your requirements.

EDIT: "Just out of interest, what would generally be considered a high Q factor? (i.e. how high could you realistically go with the ratio) and is there a likely reason as to why the example chosen in the paper is so low (0.707)?"
A quality factor Q=0,707 is nonsense!
Don`t ask me why this "obscure" paper has chosen such an example.
By the way, the corresponding homepage dealing with several different analog circuits contains many errors. I would not recommend to rely to it.

 

[Lawrenciumbc]

By the way, the corresponding homepage dealing with several different analog circuits contains many errors. I would not recommend to rely to it.

Ok no problem, thanks for letting me know, I'll keep that in mind.

sorry, for 4Q²=400, do you have a quick example as i'm not sure exactly what you mean? (i.e. how to calculate)

 

[LvW]

Ok no problem, thanks for letting me know, I'll keep that in mind.

sorry, for 4Q²=400, do you have a quick example as i'm not sure exactly what you mean? (i.e. how to calculate)

Sorry, I wrote that the ratio of the two resistors in your circuit should be 4Q²=400 in case of Q=10.
That was an example only.
Therefore, select two resistors having this ratio and calculate the corresponding capacitor value (for two equal capacitors) according to the formula for the center frequency as given in the paper.
This formula is correct.
In case you are interested where this expression 4Q² comes from, it would be necessary to evaluate the corresponding transfer function H(s).

 

[Lawrenciumbc]

Ok I have managed to build a multiple feedback band pass filter (2nd order, one complex pole) which appears to perform well around my center frequency.

My final question (i hope!) off the back of this is even though I know the the topology, how can I work backwards to determine if the response follows a Butterworth/Chebyshev/Bessel?

 

[LvW]

Ok I have managed to build a multiple feedback band pass filter (2nd order, one complex pole) which appears to perform well around my center frequency.

My final question (i hope!) off the back of this is even though I know the the topology, how can I work backwards to determine if the response follows a Butterworth/Chebyshev/Bessel?

A second-order bandpass is a bandpass of lowest order possible (to be compared with a first-order lowpass).
For such circuits we cannot discriminate between various approximations like Butterworth, Chebyshev,...
There is only one single frequency response.

 

[Lawrenciumbc]

I now understand, I was confused as books and websites I was reading describe filter topologies as 'types' such as sallen key and multiple feedback, but then go on to imply that response characteristics are also 'filter types', which didn't make sense to me. I figured that you build to a topology then depending on the exact parameters i.e. your R, C, Vin, Vcc and number of poles, you would then get a 'behaviour' like Butterworth etc... .

As my filter is more like a peak around 1 frequency rather than a defined passband region, it inherently can't actually behave like a Butterworth/Chebyshev etc....

Would that be an accurate interpretation of what's happening and of what you mean LvW?

 

[LvW]

At first, you should not mix different filter topologies (like Sallen-Key, multiple feedback,...) with the various approximations available to shape the filter functions.
However, such a shaping requires more than one single pole pair - that means: It is not applicable to first-order low/highpass functions and second-order bandpass functions.
Therefore, yes your interpretation is correct.