WHAT IS A SALLEN-KEY FILTER?

The Sallen-Key filter can be configured in a number of different ways. We will discuss a low pass filter shown in Fig. 1. (at bottom of post.) This is a second order filter and provides 40db attenuation per decade of frequency above the cut-off frequency. This is a useful circuit for applications where we wish to obtain a fairly rapid roll-off with a one-stage filter.

I won’t go into the transfer function equation here. If you want to study it, you can find the equation on Wikipedia. But to explain my design procedure (which is not found on Wikipedia), we do need to know the equations for the cut-off frequency, fo, and the quality factor, Q, of the circuit which are as follows:

fo = 1 / ( 2 * Pi * (R1*R2*C1*C2)^1/2)

The inner parenthetical quantity is the square-root of the product of the R’s and C’s multiplied together.

Q = ((R1*R2*C1*C2)^1/2) / (C2*(R1+R2))

where again the numerator is the square-root of the product of R’s and C’s. These are the equations we need to design the circuit.

First we combine the two equations to get the product of fo and Q:

Q*fo = 1 / (2*Pi*(R1+R2)*C2)

A Butterworth maximally flat filter would have a Q equal to the reciprocal of the square-root of two. For ease of calculation here, I will set Q = 1 to obtain a simple equation for fo:

fo = 1 / (2*Pi*(R1+R2)*C2)

where in all of the above Pi = 3.14159…

Assume that we want a filter with a 10KHZ cut-off and we are satisfied with a Q = 1 which will provide a reasonably flat response in its pass band. Then we know that the quantity important to the cut-off frequency is

(R1+R2)*C2 = 1 / (2*Pi*fo) = 1.5915 X 10^-5

The units of the above are reciprocal radian frequency. Let C2 be set to a reasonable value of say 1 nano-farad. (Don’t make C2 too large a value because C1 will need to be even larger.) Now we have that

R1+R2 = 15910

So as a quick approximation let us set R1 = R2 = 8K ohms each so the total is 16K ohms. Then we can calculate C1 from the square of our first equation above as

C1 = 1 / (4*Pi^2*R1*R2*fo^2*C2)

C1 = 3.957 X 10^-9 ~ 4 nano-farad

Now let’s check the design by calculating fo and Q:

fo = 1 / (2*Pi*(R1*R2*C1*C2)^1/2) = 9947 Hz

That is close enough to our original goal of a 10KHZ cut-off frequency.

Now for Q:

Q = (R1*R2*C1*C2)^2 / (C2*(R1+R2)) = (8000^2*4X10^-9*1X^-9)^1/2 / (1X^-9*16000) = 1.0

The calculated value of Q agrees with our original decision to set Q = 1.

I have now completed the design of the filter so that we have the following component values:

R1 = R2 =8K

C1 = 1X10^-9

C2= 4X10^-9

Q = 1.0

fo = 10KHZ

In fact I simulated the design using a 1MHZ JFET input op-amp and had the following results:

fo ~ 12.7KHZ (-3db down frequency)

flatness 1.3 db DC to 7KHZ

40 db per decade attenuation above 10KHZ

-54 db at 300KHZ then rising slowly to -46db at 1MHZ.

The response was very smooth with no ring or ripple in the pass-band, with a steady 40db per decade roll-off after cut-off.