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I have a simple sawtooth oscillating circuit that I would like to center to 0V (remove the DC component.) I have tried AC coupling but the output is still a little to high and not centered. Could someone explain me why is this happening and what should I do to have my oscillation centered to 0V with positive and negative peaks being equal?

Schematic:

enter image description here

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    \$\begingroup\$ But it is centered at 0 V. The average voltmeter on the left indicates <4 pV, i.e. 0 V. \$\endgroup\$
    – swineone
    Commented Feb 16, 2023 at 14:16

2 Answers 2

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Your signal is not symmetrical - the positive peaks are much narrower than the negative peaks.

The coupling capacitor will shift the signal so that the areas of the waveform above and below zero will be equal. With a symmetrical signal like a sine wave, this will result in the positive and negative peak voltages being equal, but with the signal you show, the waveform will have a higher positive peak voltage than negative peak voltage.

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I would like to center to 0V (remove the DC component.)

Centering the signal's amplitude (peaks) around 0V doesn't necessarily remove the DC component. It only does so if the signal's average value - the DC component - is exactly at 50% of the amplitude - i.e. the signal has to be symmetric.

The AC coupling network you devised does in fact remove the DC component - it just doesn't center the amplitude around 0V, because the DC component wasn't 50% of the amplitude.

To get a most generic "amplitude centering" circuit, you need to detect the envelope of the voltage, and then shift the signal accordingly so that the center of the envelope is at 0V.

For example:

schematic

simulate this circuit – Schematic created using CircuitLab

The output signal at B is indeed centered around 0V:

Waveforms at the signal input A, envelope peaks P and N, envelope midpoint MID, offset voltage OFS, and centered output B

A more practical circuit of this kind would use rail-to-rail I/O op-amps, perhaps a bit faster than TL084 to improve accuracy.

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