Electronic Devices - Diode Applications - Discussion

In full wave rectifier f(output) = 2f(input).

therefore, f(o) = 2*50 = 100.

The op of the full wave rectifier is twice the input. So the op becomes 100.

Frequency of of a full wave rectifier is the double of a half wave rectifier.

We can justify the answer as C by taking the harmonic components into consideration as the second harmonic is the Ripple function and it is a 2*f(z)

where f(z) is frequency component...

In a fullwave rectifier diodes D1 conducting for +ve half cycle of the input & diode D2 conducting for -ve half cycle of the input, then the output wave consisting of two +ve hail cycles for an i/p wave of time period T. The time period of o/p wave is decreased to T/2 then the o/p frequency will be twice of the i/p signal frequency.