Preparation;
In a half-wave rectifier, as shown in Figure 4-1 , a load resistance (Rl) is connected to an ac source
through a single diode. When the input sine wave is positive , the diode is forward-biased and allows
current to flow in the load resistance. When the input sine wave is negative, the diode is reverse-biased
and prevents current flow in the load resistance. Therefore, only the positive half of the sine wave
voltage is across the load resistance , and the voltage across the load resistance is zero during the
negative half of the input sine wave. This produces a pulsating dc voltage across the load resistance.
The average value of this pulsating half-wave rectified output voltage (V dc) is the value measured by a
dc voltmeter, and is calculated by dividing the peak output voltage (V p) by π. Therefore.
V dc=(V p )/( π)
The diode barrier potential, or forward-bias voltage (V f = 0.7V), causes the peak of the output to be
less than the peak of the input wave. This happens because the output voltage is equal to the input
voltage minus the voltage drop across the diode
The maximum value of the reverse-bias voltage across the diode, called the inverse voltage
(PIV), occurs at the peak of the negative half to the input sine wave. Therefore, the peak inverse
Voltage (PIV) is equal to the peak value of the input sine wave.
The output ripple frequency of this half-wave rectified output is calculated by finding the inverse of
The time period (T) for one complete cycle of the output waveshape. Therefore’
f=1/T
A transformer is often used to reduce or increase the ac line voltage to the level desired for the dc
output. The transformer secondary peak output voltage (V p2) is equal to the transformer turns ratio
times the primary peak input voltage (V p1) Therefore,
V p2 =(N2 )/(N 1) V p1
The peak primary (input) voltage (V p1) can be calculated from the RMS input voltage using the
equation