this post was submitted on 07 Jul 2023
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Hello there, This oscillator is a 0V +10v DC oscillator, which after current passes through the capacitor, it produces a -5v +5v AC on the resistor.

We've all heard that AC removes DC component and let's AC pass by. I understand the dynamics of this circuit in case the oscillator were operating with AC (capacitive reactance), however this oscillator is DC, the voltage across the capacitor never changes polarity (since the other side of circuit is ground), so what gives? And why the 10V DC is split on half +5 -5 volts after the capacitor? Thank you!

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[–] FearlessPhoenix@lemmy.world 7 points 1 year ago* (last edited 1 year ago)

True AC is sort of "balanced" in that it has just as much positive as negative. The positive area of the waveform is the same size as the negative area. For waveforms that are sort of symmetric across the 0V with a time offset, such as a sine or square wave, this means that it is centered along the 0V line. A DC source, on the other hand, never changes voltage.

The 0V to +10V source you have is actually a -5V to +5V square AC plus a +5V DC. The capacitor is getting rid of the DC component leaving just the AC, which happens to be the -5V to +5V AC that you are getting.