The circuit given in the figure is:

If V_i > V₁  then D₁ is ON

V₀ = V₁

If V_i < -V₁ then D₂ is ON

V₀ = -V₁

If –V₁ < V_i < V₁ both diodes will be off:

 \({V_0} = \frac{{{V_i} \times {R_L}}}{{R + {R_L}}}\)

So it will work like dual clipper.

Output waveform:

1. Positive Clipper: A positive clipper removes that part of i/p signal which lies above reference voltage

Eg.

2. Negative Clipper: It removes that part of i/p signal which lies below reference voltage.

Eg.

3) Dual Clipper: Circuit which removes that part of i/p signal which lies above greater reference voltage and also which lies before smaller resistance.