Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up.
Sign up to join this community
simulate this circuit – Schematic created using CircuitLab
I am doing a clamper circuit as shown in the above figure, I am confused regarding its operation when diode is forward biased we have a sinusoidal output, as the circuit is clamped to V2 but what I am not getting is when diode is forward biased shouldn't the output be constant V2, as at node for 1 we get V2 when diode is short circuited. Can you please help?
I think you got it. It is a V2 clamp. Node 1 will follow V1 for small voltages until V1 gets to be larger than 1V + VF(diode) ~ 1.7V. If V1 continues to increase then D1 (and V2) will sink the necessary current to keep node 1 at 1.7V. Hope that helps.
EDIT: After clamping node 1 for the first time, the AC capacitor will charge up. If this charge doesn't leak, node 1 will now have an offset equal to the voltage difference between V1 and the initial clamp voltage. As the other answer states, this is often used as a charge-pump offset generator
The circuit acts kind of like Dickson charge pump. https://en.wikipedia.org/wiki/Voltage_doubler#Dickson_charge_pump
At positive phase of V1 remaining components C1, D2 and V2 will create a voltage divider where C1 will charge to certain voltage level*. Once the sinewave on V1 does negative phase charge doesnt flow anywhere because of diode D1 and Vout is sum of C1 and V1 as they are in series.
Voltage of C1 depends about time constant (RC) and voltage of V1. In steady state its Vc1 = V1-V2-Vf
