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I'm using a 555 timer to generate a triangle waveform with a DC offset. I would like to feed the waveform into a comparator along with an AC signal with zero offset to get a PWM signal, so I need the average voltage of both waveforms to be equal. I'm trying to do this with an DC coupling capacitor to make the triangle waveform centered around 0V, but I'm only getting negative voltage after the capacitor. Why is this happening? How can I get a triangle waveform from the 555 with zero DC offset?

Here is the schematic: 555 Triangle Generator

Here is the voltage after C7. It has a ~-3.7v DC offset for some reason: enter image description here

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  • \$\begingroup\$ If need a low impedance then a biploar supply is an alternative \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 4 '18 at 15:58
  • \$\begingroup\$ That looks like a simulator, not a real steady state trace. Try again 200KHz with a 0.16 Hz HPF and a simulation after 10us is still got a DC offset. Think! (steady state) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 4 '18 at 16:19
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C7 and R5 make an RC high pass filter with a cutoff frequency of .16 Hz and an RC time constant of ~1s. In order to have no DC offset, the capacitor must charge which takes \$5\tau\$ to reach 99% charge, or 5 seconds in your case.

To see the removal of the DC offset, you either need to simulate your circuit for 5s, or use a filter with a higher cutoff frequency. Since it looks like your circuit has an output frequency of ~200 kHz, a .16 Hz high pass filter is not really suitable (and probably wouldn't work well with an electrolytic cap), and you really should use a filter with a much higher cutoff frequency, say 20 kHz.

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C10 might well be an electrolytic which is leaky. Try shunting your output with roughly 10kΩ or so, wtich will reduce the amplitude and very likely the offset. Agreed that you don't need such a long time constant. Try connecting your output to a mA or even μA range of an analog multimeter, if you have access to one.

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    \$\begingroup\$ The simulated capacitor is ideal, so there's no leakage. He just has too large a time constant. \$\endgroup\$ – C_Elegans Apr 4 '18 at 16:19
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    \$\begingroup\$ Try simulating after 1 second \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 4 '18 at 16:29

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