# Clipper and Clamper Help

Discussion in 'Electronic Basics' started by Kartik Aiyer, Jul 2, 2004.

1. ### Kartik AiyerGuest

Need some help trying to figure this out

Consider a clipper circuit with only one diode.

-----Res-------------
| +
+ D (+)
Vi i (-) Vo
- |
|
D (+ Vb) -
C (-)
----------------------

I am sorry but its the best I could do. So Di is a diode (Vt cut it
Voltage) and DC is a dc source (Vb). Now i believe the diode stays
off when Vi < Vb+Vt
and the output follows the input. (Res is a resistor R)

When Vi> Vb+ Vt the diode turns on and the Vo is clipped at Vb+Vt.

I dont understand why the diode is off when Vi<Vb+Vt ( i cant figure
what the voltage is across the diode) and why the diode is on when Vi
Also in a clamper circuit with one capacitor and one diode, when we
use an AC sinusoidal input, I understand that on the rising of the
signal the output voltage (measured across the diode) follows the
input (assuming zero forward resistance on the diode). Why is it when
the signal begins to decrease the diode is reverse biased.

I would appreciate any help whatsoever.

2. ### dBGuest

Are you sure that you understand diodes?