I am attempting to remove the DC offset from the 1 kHz, ± 2.5 V, AC square wave output pins of an ADS1261 (http://www.ti.com/lit/ds/symlink/ads1261.pdf). However, when measured on the oscilloscope (and modelled in LTSpice), the average DC value is approximately 2 mV.

I have modelled a passive RC high pass filter in LTSpice (similar to Changing a Signal's DC Offset), however, I have found that instead of maintaining the ± 2.5 V and removing the offset, I get 0 - 5 V output instead.

How can I remove the DC offset/average while maintaining the ± 2.5 V square wave?


LtSpice passive RC high pass filter

Result with C1 as 15 nF

LTSpice with C1 as 15 nF

Result with C1 as 100 uF

LTSpice with C1 as 100 uF


1 Answer 1


You tried 2 values 4 orders of magnitude apart.

In the first (15nF) example, the capacitance is too small. Your DC content is being removed, but with a time constant of 150us, the capacitor is almost completely charged before you get to the next edge of the 1KHz square wave. As such, the output is near zero at each edge, and shows the entire 5V P-P transition.

In the second (100uF) example, you have a time constant of about 1s. The cap in the simulation is initialized to 2.5V, and in the 2ms time frame of your scan, it doesn't change much. Therefore, the output remains 2.5V above the input. If you run that simulation for a few seconds, you'll see the waveform you want.

Your challenge is to determine a value that doesn't distort too much, but settles quickly enough for your purposes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.