TUF for a half-wave rectifier and centre tapped full wave rectifier is ____________ and ____________, respectively.

Transformer utilization factor

The transformer utilization factor (TUF) of a rectifier circuit is defined as the ratio of the DC power available at the load resistor to the kVA rating of the secondary coil of a transformer.

% TUF = \({V_{o(avg)}\times I_{o(avg)}\over V_{s(rms) \times I_{s(rms)}}}\)

For half-wave rectifier

The waveform of a half-wave rectifier is:

The parameters of half wave rectifier are

\(V_{o(avg)}={V_m\over \pi}\) and \(I_{o(avg)}={V_m\over \pi R}\)

\(V_{s(rms)}={V_m\over {2}}\) and \(I_{o(rms)}=I_{s(rms)}={V_m\over 2R}\)

% TUF = \({{V_m\over \pi}\times{V_m\over \pi R}\over {V_m\over \sqrt{2}}\times {V_m\over {2}R}}\)

% TUF = 28.6%

For full-wave rectifier

The waveform of a full-wave rectifier is:

The parameters of the full-wave rectifier are

\(V_{o(avg)}={2V_m\over \pi}\) and \(I_{o(avg)}={2V_m\over \pi R}\)

\(V_{s(rms)}={V_m\over \sqrt{2}}\) and \(I_{s(rms)}=V_m\)

% TUF = \({{2V_m\over \pi}\times{2V_m\over \pi R}\over {V_m\over \sqrt{2}}\times {V_m}}\)

% TUF = 57.2%

Mistake Points The rectification efficiency of the half-wave rectifier is 40.6%

The rectification efficiency of a full-wave rectifier is 81.2%