# Understanding Clipper Waveform

Discussion in 'General Electronics Chat' started by macman, Jan 22, 2015.

1. ### macman Thread Starter New Member

Dec 18, 2014
8
0

Consider that the voltage drop about R[L] is V[Out].

As we can see that V[0ut] gets clipped in the output waveform...now i am trying to model V[out] using a KVL equation. I think I can get the value of V[out] like:

V(in)-V(ref)=V(out)

Am I right or i am completely wrong?
I think i am wrong because suppose V(in) is time dependent and V(ref) a constant so V(out) should be time dependent and change with V(in) but as we can see there's a straight line in the graph so i must be doing something wrong.

All i want to know is that i somewhat get how clipper works but I can't get to understand the value of V(out) as a straight line can be get using a KVL equation?

2. ### Veracohr Well-Known Member

Jan 3, 2011
602
86
It has to be described as a piecewise function.

While Vout<(Vref+Vd), Vout is proportional to Vin.
Once Vin rises to the point that Vout would be greater than (Vref+Vd), the diode conducts and Vout = Vref+Vd.