I need to add a highpass filter between a signal source and an ADC to get rid of high-amplitude low-frequency signals that would cause clipping.

I built a second-order Sallen-Key filter and the circuit is working as expected. However the filter requires a symmetric power supply and I need to run it on a single supply.

This is what I have:

enter image description here

  • The input U_IN is a 0V ... 3.3V signal (the attached sensor has low output impedance).
  • The filter is currently configured to have gain 1.1 (R2=1k, R1=10k) but a unity gain with a voltage follower would work aswell (the exact gain is not critical).
  • The Op-Amp currently is supplied with +3.3V and -3.3V.
  • The cutoff frequency is at 500Hz.

In addition I already have a reference voltage source (voltage divider and buffer) that provides half of the maximum signal voltage (1.65V):

enter image description here

How can I change the filter to output 0V ... 3.3V while supplying the Op-Amp with +3.3V and GND?


1 Answer 1


You'll be kicking yourself to see how easy this is.

enter image description here

But perhaps you should review where is the 0V signal = "ground" and how can you use an INA with shielded twisted pair to eliminate the low frequency (grid) voltage noise from high E or H fields.

  • \$\begingroup\$ Also split R1 with two 20K resistor just as R3 did. Use 0 - 5Vdc supply. Add a series capacitor at the output. and use 1% resistor for the split to minimize any additional DC offset. \$\endgroup\$
    – Louis
    Commented Jan 26, 2022 at 3:32
  • \$\begingroup\$ Thanks, this works fine (I went for unity gain with voltage follower). How do you know the value of R3 and the new resistor? The values need to be equal to center the signal at half the supply voltage. But how do you calculate their exact values (why did you double the 3.3k?). \$\endgroup\$
    – Johannes
    Commented Jan 26, 2022 at 16:18
  • \$\begingroup\$ @KevinWhite: Having the output centered on 1.6V is exactly what I wanted. I need to read the value with an ADC that has the range 0 ... 3.3V. \$\endgroup\$
    – Johannes
    Commented Jan 26, 2022 at 16:19
  • \$\begingroup\$ @Johannes , because these Resistors are effectively in parallel you need exactly twice the values. In series it adds, but in // it is the ratio of product and sum electronics.stackexchange.com/questions/130267/… \$\endgroup\$ Commented Jan 26, 2022 at 16:51
  • \$\begingroup\$ @Johannes - sorry, my confusion. \$\endgroup\$ Commented Jan 26, 2022 at 17:03

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