A Sallen-Key 3-Pole Butterworth Active Highpass Filter - Design Sheet (DS2)

by John-Paul Bedinger

Of the various topologies you can select for making active filters, the Sallen-Key uses the least number of filter components. Furthermore, a 3-pole response (18 db/oct) is possible using only 1 op-amp. Below is a brief mathematical description on how to compute the component values for a Butterworth (steepest response with no ripple) 3-pole highpass filter with selectable output gain. Diagram 1: A 3-pole Sallen-Key highpass filter with output gain.

Looking at node 3b (where v3 = v3a = v3b) , the node voltage equation can be re-written to be: where and M defines the AC gain of the circuit in the passband (Eq. 1 and 2).

The rest of the node voltage equations can be written: (Eq. 3) (Eq. 4) (Eq. 5)

When solved for the filter transfer function H(s) = Vout/Vin, we get: (Eq. 6)

Note that the standard form of a 3-pole Butterworth highpass filter at cutoff frequency 1 rad/sec is: (Eq.7)   where  Kac is the AC gain of the filter, the same as our M.

Equating like terms in Equations 7 and 6 gives the solve block: (Eq.8) (Eq.9) (Eq.10) (Eq.11)

We choose our output gain: Kac = M = 4 (12 dB)

Also, we choose values for components:  C1= 100 uF, C2= 100uF,  C3= 100uF, R5= 10 k-ohms

Solving now for R1, R2, R3,and R4 gives (exact):

R1= 5 k-ohms, R2= 20 k-ohms, R3= 10 k-ohms, R4= 30 k-ohms

Rounding R1-R4 to standard EIA 1% tolerance decade values gives:

R1= 4.99 k-ohms, R2= 20 k-ohms, R3= 10 k-ohms, R4= 30 k-ohms Diagram 2: A 3-pole Sallen-Key Butterworth highpass filter with cutoff at 1 rad/sec and a gain of 4 (12dB).

Practical Notes:

• For different gain values you can refer to my table below, or resolve the solve block with a different value of Kac. Use solutions only with all real positive roots. If you have Mathcad(TM), you can download this zipped Mathcad worksheet to help you.

• C1, C2, and C3 from the table can be scaled together by a factor y, which should be done so that the source impedance Rin is much less than the magnitude of the impedance (Rin + C1) at the cutoff frequency Fc. This will set the cutoff frequency to 1/y rad/sec.

• R1, R2, and R3 from the table can be scaled together by a factor x, which will set the cutoff to 1/(x *y) rad/sec, or:

Fc = 1/(2*3.1416*x*y) Hz        (Eq. 12)

 M 0dB 6dB 12dB 18dB 24dB 30dB 36dB R1(ohms) 7180.57 5862.23 5000 4223 3531.2 2933.18 2425.61 R2(ohms) 2819.43 11533.14 20000 31821.84 49433.83 76312.84 117808 R3(ohms) 49394.66 14790.74 10000 7441.38 5728.67 4467.49 3499.48 R4(ohms) 0 10000 30000 70000 15000 31000 63000 R5(ohms) infinite 10000 10000 10000 1000 1000 1000 C1(Farads) 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 C2(Farads) 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 C3(Farads) 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6

 M 42dB 48dB 54dB 60dB 66dB 72dB 78dB R1(ohms) 1998.55 1640.85 1342.31 1094.1 888.6151 719.3089 580.4752 R2(ohms) 182295.4 282964.2 440633.6 688213.9 1077780 1691800 2660920 R3(ohms) 2744.79 2153.77 1690.71 1328.07 1044.13 821.7427 647.4178 R4(ohms) 127000 25500 51100 102300 20470 40950 81910 R5(ohms) 1000 100 100 100 10 10 10 C1(Farads) 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 C2(Farads) 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 C3(Farads) 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6 100.E-6

Table 1: Prototype component values for a Butterworth highpass filter response at 1 rad/sec.

Example:

We want Fc= 80 Hz, C1= 0.1uF, C2= 0.1uF, and C3= 0.1uF, and a gain of 30 dB. The source resistance is 220 ohms or less.

Use Table 1 for 30dB prototype values, then scale y for the correct capacitor range:

The scale factor y is 0.1uF/100uF, or y = 0.001. Thus, 80 Hz = 1/(2*3.1416*x*0.001). Solving for x gives: x = 1.989

Scaling R1, R2, R3,and R4 by x gives:

R1= 5834.1 ohms, R2= 151786 ohms, R3= 8885.8 ohms, R4= 31 k-ohms,  R5=1 k-ohm

Rounding R1-R4 to standard EIA 1% tolerance decade values gives:

R1= 5.9 k-ohms, R2= 150 k-ohms, R3= 8.87 k-ohms, R4= 30.9 k-ohms, R5=1 k-ohm

The magnitude of (C1 + Rin) at 80Hz is:

(1/(2*3.14*80*0.1e-6)^2+220^2)^.5 = 19.9 k-ohm at 80 Hz.

Since the combined impedance of 19.9 k-ohm at the cutoff frequency is much greater than the 220 ohm source resistance by itself, the value for C1 should work well.

Change Log:

v.1.0 Initial release.

Questions or feedback? E-mail me at jpbedinger@hotmail.com

Do not duplicate, distribute, or modify without my expressed written permission.

Disclaimer: The author is not responsible for any damages resulting from the content or application of this document. Use at your own risk.