abbeyromy said:
@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:
abbeyromy said:
@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.