The intent is probably creating a DC offset AC signal. Looks very unsafe to me.
The large capacitor you can treat as a voltage source as long as R*C (capacitor time constant) >> 1/f_ac_signal.How's that work?
The large capacitor you can treat as a voltage source as long as R*C (capacitor time constant) >> 1/f_ac_signal.
So in your digram capacitor is charge to
Vc = Vdd * R4/(R3+R4). So simply replace capacitor with Vc voltage source.
Well I think that the time constant is equal to resistance seen from the capacitor terminal. t = R * C ---->R = R3||R4 + R1||R2
Simply you must find a equal resistance seeing from the capacitor terminal when the other remaining voltage source replace with a short-circuit.ok, this confused me more, I simply don't understand why R3||R4 and R1||R2, if || here means parallel.
by Duane Benson
by Duane Benson
by Jake Hertz
by Duane Benson