Single Phase Rectifiers

Half wave Rectifier:


Figure gives the circuit for the halfwave-rectifier using a pn diode, resistor RL and a
transformer to get the input in the secondary. Let vi be the input voltage obtained the
transformer secondary.

Vi = Vm sinwt = Vm sin a ---(5.1.1)

Where, Vm is the amplitude of the input voltage(v),w is the angular frequency in
radians/second. a is in radians.

 

Let the amplitude Vm >> Vv where,Vv is the cutin voltage for the pn diode.
Assuming Vv = 0. Let the diode be ideal with resistance Rf in the ON state and Rf =& in the OFF state.The current through the diode and the load resistor RL  is given
by
i = Im sina for 0<= a <= p
i = 0 for p<=a<=2 p ----(5.1.2)
where Im is the peak current given by
Im = Vm ---(5.1.3)
Rf + RL

Current flowing through RL will produce a output voltage vo given by

 vo = i x RL ---(5.1.4)

DC or Average Current Idc
2p
Idc = 1 œ L G DOSKD ---(5.1.5)
2 p 0
where i is given by the equation (5.1.2).
Current i = 0 when p < t< 2 p
p
Idc = 1 œL G DOSKD Im ---(5.1.6)
2 p 0 p
and Idc = Vm ---(5.1.7)
p(Rf+RL)
DC Output Voltage Vdc
Vdc = Idc . RL = Im . RL ---(5.1.8)
= Vm ---(5.1.9)
p(1 + Rf/RL)
The Diode Voltage: It is the instantaneous voltage across the diode ( i . R , where R
is the diode resistance. = Rf when ON and =OO When OFF). The wave form is shown in
the figure below

Merits of Half-wave Rectifier:


(i) simple circuit

(ii) low cost.

Disadvantage of Halfwave:

(i) Low rectifier efficiency

(ii) High ripple factor

(iii) Low TUF

(iv) DC saturation of the transformer core due to flow of current in the secondary of
the power transformer. The DC saturation of the core increases the magnetising
current and the hysterisis losses and produces harmonics in the output of the
secondary.