2ND ORDER LOW OR HIGH PASS FILTER

Fig. 1 shows a circuit that can be configured as either a high pass or low pass filter depending on which components are populated. The values shown are for a low pass filter with a cut-off frequency of approximately 200HZ.

In the case shown R2 and R3 are set to 8.1K, and C1 and C4 are set to 0.1UFD. The component parts parallel to these are set to parasitic values (not populated) but would be populated for the high pass version and R2, R3, C1, and C4 would not be populated for the high pass configuration.

We can calculate the cut-off frequency value for either the high pass or low pass filter from the equation

Fo = 1 /( 2 * Pi * ( R2 * R3 * C1 * C4 ) ^ 0.5 ) = 196.48 HZ

R2 and R3 are usually set equal to each other and C1 an C4 are usually set equal to each other. We can then simplify the formula to:

Fo = 1 / ( 2 * Pi * R * C)

which we should recognize as the equation for a simple RC filter, but in this case we have a 2nd filter instead of a 1st order filter with the circuit of Fig. 1.

So if you know the frequency cut-off you want you can set the capacitor values and calculate the R values. For example, say you want a cut-off of 500 HZ and you will set the C values to 0.1UFD as before. The R values are then

R = 1 / ( 2 * Pi * Fo * C ) = 3.183 K

Or if we know the R values and the cut-off frequency, we can calculate the capacitor values;

C = 1 / ( 2 * Pi * Fo * R ) = 1.000 * 10^-7

or 0.1UFD.

The circuit in Fig. 1 was simulated with the components shown using +5V and -5V power supplies on a SPICE simulator and verified to have the calculated cut-off frequency.