I am trying to understand the output waveform of this diode in series with capacitor circuit and I am using the constant voltage source model for diode.
I will recount what the professor has told in lecture "As the voltage increases , the capacitor is charged with the voltage of = Peak positive voltage of sinusoid input voltage - the diode voltage and will remain so for the rest of positive and negative cycles"
I am having trouble understanding the negative cycles , when the input voltage becomes negative then the diode is an open switch according to the constant source model.
And the negative terminal of capacitor that was previously connected to negative terminal of input voltage is now connected to positive terminal of voltage.
So we have a wire in between positive battery end and the negative capacitor plate ,isn't it against circuit theory rules of wire having no potential differences ?
How would that affect the output waveform ?