"SALLEN KEY" ACTIVE FILTER DESIGNING

[abbeyromy]

I was trying to simulate SALLEN KEY ACTIVE FILTER in PSpice(A low-pass filter, with fc=15.9 kHz and Q = 0.5.) but I am not getting results as per formula( fc coming as 10.3K instead of 15.9K). Please see attached circuit, simulation results and the formula I am using. The theory I have taken from Wikipedia.

What is the advantage/disadvantage of Sallaen key and where you find its use in general?

Please guide!
-Hemanshu

 

[LvW]

I was trying to simulate SALLEN KEY ACTIVE FILTER in PSpice(A low-pass filter, with fc=15.9 kHz and Q = 0.5.) but I am not getting results as per formula( fc coming as 10.3K instead of 15.9K). Please see attached circuit, simulation results and the formula I am using. The theory I have taken from Wikipedia.
What is the advantage/disadvantage of Sallaen key and where you find its use in general?
Please guide!
-Hemanshu

1.) The chosen opamp (uA741) has a transit frequency Ft which is too low for Fc=15 kHz. You have to select a faster opamp (opamp gain at 10 kHz only app. 40 dB).
2.) Advantage S+K: less sensitive to opamp gain and phase errors
3.) Disadvantage: relative high sensitivity to part tolerances.

 

[abbeyromy]

Hi LvW,
I tried with few more opamps from PSPICE library but got the same result. When you say faster opamp do you mean higher slew rate and higher 3dB bandwidth?

For some higher slew rate models, I was getting convergence error in pspice.
By any chance can you suggest kind of opamps I can try using here. I will try to search its pspice model.
I was referring ?National website
**broken link removed**

Thanks for the help!
-Hemanshu

 

[abbeyromy]

Can somebody please simulate the results?

 

[hock]

try to use "filter Wiz Pro" software.
it will give you good results.

hock

 

[abbeyromy]

@hock,

This is not a software issue I guess. Can I request you to simulate the same circuit in the software you are suggesting just to be sure? I have some restrictions on this pc for installation of softwares..
Thanks a ton!!

 

[LvW]

@hock,

This is not a software issue I guess. Can I request you to simulate the same circuit in the software you are suggesting just to be sure? I have some restrictions on this pc for installation of softwares..
Thanks a ton!!

My recommendatio: At first, check the element values by simulating the circuit using an ideal opamp model ("OPAMP" in PSpice). If the result is OK, you are sure that all deficiencies are caused by the real opamp. Then, a faster device is required.

 

[abbeyromy]

@LvW

Good idea.. let me try this

Added after 47 minutes:

@LvW
Not much difference on using the OPAMP model in pspice, almost the same result.
So what should I try next? The theory seems right so by formula I should expect 15.9KHZ cutoff..

Please help..can you simulate the circuit n see what is going wrong?

Thanks a lot!

 

[LvW]

Hi LvW,
I tried with few more opamps from PSPICE library but got the same result. When you say faster opamp do you mean higher slew rate and higher 3dB bandwidth?

abbeyromy

There is no problem with your circuit - but with your formulas.
The frequency called "fc" is NOT the corner frequency (3 dB point) but the pole frequency. It is a petty that the book which you have derived the formulas from did not use the name "fp" for the pole frequency. Therefore, this misunderstanding.
It is quite normal that such formulas contain the pole frequency because the bandwidth (3-dB frequency) depends on the filter approximation (Butterworth, Chebyshev,...).
For example:
(a) for Butterworth response (Qp=0.707) : 3dB-frequency f3dB=pole frequency fp
but
(b) for Bessel/Thomson (Qp=0.577) : f3dB=0.577*fp

Therefore, in your case with Qp=0.5 the 3dB frequency also is pretty below the chosen pole frequency of 15.9 kHz. Thats quite normal. But the factor between both frequencies normally cannot be found in books and tables because it is very uncommon to use an active filter for a pole Q as low as QP=0.5.
In this case, you have a real double pole on the negative axis of the s-plane.

Added after 14 minutes:

Addendum: The exact factor between both frequencies is 0.64359.

It can be derived from the filter response with "critical damping".
The formula considers the series connection of n equal first order stages.
The formula for the factor is

√(2exp(1/n)) - 1 (the whole expression is part of the root). For n=2: as given above.
This means, if you decrease the product RC by this factor you will arrive at the correct 3 dB point.

 

[abbeyromy]

@LvW

Oh I was so confused.. Thanks for correcting me.. I am really amazed with the level of knowledge you guys have.. hats off to you man!!

Ok now I will try to simulate butterworth active filter just to make sure I have understood or not. Will post the results and discuss with you again!!
Thanks man!

 

[abbeyromy]

@LvW
I made my Q=0.707 and fc=fp showed it in simulation. You were right as always so full points to you

Now I have attached the circuit a little theory which might help if somebody wants to do this at their end.

I have used a faster opamp- LM118 than ua741. At some point op amps lose voltage gain at some frequency due to their finite bandwidth. I read somewhere that the frequency at which the curve depart from the ideal should be near the Gain Bandwidth Product GBP of the op amp.

Can you help me understand how I can calculate a near range at which this deviation I should be expecting. See the attached document that shows the bending up of the curve at around 1.2MHz. I am also attaching the datasheet for LM118 for your quick reference. If I use uA741 then this thing happens at around 300KHz.

Waiting your expert guidance!!
-Hemanshu

 

[keith1200rs]

I don't know if you have it, but Texas Instruments' Filter Pro software might be worthwhile downloading - it might save you some time. It will tell you the GBW for a given filter and you can design by Q & Fn if you like.

Keith.

 

[LvW]

@LvW
I made my Q=0.707 and fc=fp showed it in simulation. You were right as always so full points to you
Can you help me understand how I can calculate a near range at which this deviation I should be expecting. See the attached document that shows the bending up of the curve at around 1.2MHz.
-Hemanshu

Hi, Hemanshu !
Congratulation!
It is hard to give a final and comprehensive answer to your question, because the real filter curve always will depart from the ideal one due to real opamp properties (gain, phase, impedances,..).
*Of course, the most important influence comes from the finite opamp gain - in particular from its phase deviation. This results in (a) a pole frequency which is somewhat smaller than wanted and (b) a pole Q which is somewhat larger than wanted.
That means, there is no specific frequency at which a deviation from the ideal curve starts. In theory, the deviation starts at zero frequency and becomes more remarkable at higher frequencies. Thus, because of other uncertainties like parts tolerances - nobody expects an ideal filter response.
Therefore, normally a tolerance mask is specified which gives lower and upper limits for the response curve. If the real curve is inside the limits, everything is OK and it is of minor importance, whether some effects are caused by the opamp or by parts tolerances.
*One explanation to the 1.2 MHz-effect:
This is a special property (and a disadvantage) of the Sallen-Key topology and cannot directly attributed to the finite opamp gain. The explanation is as follows:
For higher frequencies the voltage coming out of the opamp becomes lower and lower (low pass effect) - but at the same time there is another unwanted portion which is fed directly via the feedback capacitor to the output of the opamp. And this portion of the input signal creates a voltage across the finite opamp output impedance with a rising characteristic.

Added after 2 hours 13 minutes:

@LvW
I made my Q=0.707 and fc=fp showed it in simulation. You were right as always so full points to you
Now I have attached the circuit a little theory which might help if somebody wants to do this at their end.
-Hemanshu

HEMANSHU, I like to give one additional comment to your design.
Perhaps it helps somebody.
You have found a very simple and good working dimensioning procedure with very nice values:
Two equal resistors and two capacitors with the ratio of two.

But one should be aware of the fact that this holds for the Butterworth response ONLY! For other approximations (Bessel, Chebyshev) the calculation process is somewhat more complex:
* Choose two capacitors which satisfy the equation: C2/C1 ≥4*Qp*Qp.
(In your case: C2/C1=2 with Qp=0.707)
* This leads to a quadratic equation (second order) for the ratio R1/R2 which has two results. Both can be used.
* Further comment: The design of a Sallen-Key filter with a fixed gain of two (instead of one, as in your case) leads to a calculation which is much more simplified (without the necessity to solve a quadratic equation).
The required gain of two can be accomplished by two equal valued feedback resistors .

Perhaps this can be helpful in some cases.

 

[Audioguru]

If the gain of the opamp is 1.6 then equal value resistors and capacitors produce a sharp Butterworth cutoff.

 

[abbeyromy]

@LvW/Keith/Audioguru

Thanks for your guidance and expert advice!

 

[LvW]

If the gain of the opamp is 1.6 then equal value resistors and capacitors produce a sharp Butterworth cutoff.

AUDIOGURU: For my opinion there is only one Butterworth response of 2nd order.
You may call it "sharp" or not.

 

[Audioguru]

I call a Butterworth filter response "sharp".
I call a Bessel filter response "droopy".
I call a Chebychev filter response "all over the place".

 

[LvW]

AUDIOGURU,

I presume, you know what you are speaking about. That's for sure.
But, on the other hand, I know there are some people who perhaps think that a Butterworth 2nd order response can me "more sharp" or "less sharp" - perhaps depending on the gain. Therefore my response.

 

[Audioguru]

I use a "Butterworth" highpass filter with too much gain in my bass extension Circuit. The filter's gain is too high so it forms a peak in the response at the cutoff frequency. If the gain is higher then it oscillates.

 

[LvW]

I use a "Butterworth" highpass filter with too much gain in my bass extension Circuit. The filter's gain is too high so it forms a peak in the response at the cutoff frequency. If the gain is higher then it oscillates.

Just one remark for newcomers:
As soon as the frequency response shows a "peak" in the vicinity of the pole frequency (perhaps as a consequence of tolerances or gain deviations) , the filter has a "Chebyshev" response; it is not a "Butterworth" filter anymore (Butterworth means "maximally flat").