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!

  • Susan_B_Good
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    1 year ago

    The standard way of looking at this is to consider a capacitor-resistor series combination going to ground. Connect a 10v (wrt ground) supply to the capacitor and the voltage across the resistor rises to +10v, then decays. Now connect that capacitor to ground and that same resistor gets -10v across it, which then decays. Whatever is connected to the capacitor “top” terminal has to be able to sink current as well as source it.

    That’s what generators in simulators do - they have zero internal impedance (usually). They sink currents as well as source them.