Diodes and Applications
Revision (02)
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Summary of power supply rectifiers
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Example 1
An ac supply of 230 V is applied to a half-wave rectifier circuit through a
transformer of turn ratio 10:1. Assume the diode to be ideal. Find
(i) the output dc voltage
(ii) the peak inverse voltage.
Solution
Primary to secondary turns is N1/N2 = 10
rms of the primary voltage = 230 V
∴ Max primary voltage Vpm = 2 × 230 = 325.3 V
∴ Max secondary voltage Vsm = Vpm x (N2/N1) = 325.3 x (1/10 ) = 32.53 V 3
(i) the output dc voltage
(ii) the peak inverse voltage.
• The maximum secondary voltage appears across the diode.
• ∴ Peak inverse voltage Vsm = 32.53 V
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Example 2
A half-wave rectifier is used to supply 50V dc to a resistive load of 800 Ω.
The diode has a resistance of 25 Ω. Calculate ac voltage required.
Solution
Output dc voltage, Vdc = 50 V
Diode resistance, rf = 25 Ω
Load resistance, RL = 800 Ω
Let Vm be the maximum value of ac
voltage required.
∴ Vdc = Idc × RL
Hence, ac voltage of maximum value
162 V is required. 5
Example 3
A full-wave rectifier uses two diodes, the internal resistance of each diode may
be assumed constant at 20 Ω. The transformer rms secondary voltage from
center tap to each end of secondary is 50 V and load resistance is 980 Ω. Find:
(i) the average load current, (ii) the rms value of load current
Solution
Max. ac voltage
Max. load current
(i) average load current
(ii) RMS value of load current is
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Example 4
In the center-tap circuit shown, the diodes are assumed to be ideal i.e.
having zero internal resistance. Find: (i) dc output voltage (ii) peak
inverse voltage.
Solution
Primary to secondary turns, N1/N2 = 5
RMS primary voltage = 230 V
∴ RMS secondary voltage = 230×(1/5) = 46V
Maximum voltage across secondary = 46 x 2 = 65V
Maximum voltage across half secondary winding is Vm =65/2 = 32.5V
(i) dc output voltage
2Vm
Vdc = VAVG= 2 x 32.5 / 3.14 = 20.7V
(ii) peak inverse voltage.
The peak inverse voltage is equal to the
maximum secondary voltage, i.e.
PIV = 65 V
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Example 5
In the bridge type circuit shown, the diodes are assumed to be ideal.
Assume primary to secondary turns to be 4. Find:
(i) dc output voltage
(ii) peak inverse voltage
(iii) output frequency.
Solution
Primary/secondary turns, N1/N2 = 4
RMS primary voltage = 230 V
∴ RMS secondary voltage = 230 (N2/N1) = 230 × (1/4) = 57.5 V
Maximum voltage across secondary is
Vm = 57.5 × 2 = 81.3V
Average output voltage, 2Vm
(i) ∴ dc output voltage, Vdc = VAVG = 2 × 81.3 / 3.14 = 52V
(ii) peak inverse voltage (PIV = 81.3V)
(iii) In full wave rectification, there are two output pulses for each
complete cycle of the input ac voltage. Therefore, the output
frequency is twice that of the ac supply frequency i.e.
fout = 2 × fin = 2 × 50 = 100Hz 10
More problem to be solved by your self
(1) Figures show the center-tap and bridge type circuits having the same load resistance
and transformer turn ratio. The primary of each is connected to 230 V, 50 Hz supply.
Assume the diodes to be ideal. (i) Find the dc voltage in each case. (ii) PIV for each case
for the same dc output.
(2) The four diodes used in a bridge rectifier circuit have forward resistances which may
be considered constant at 1Ω and infinite reverse resistance. The alternating supply
voltage is 240 V rms and load resistance is 480 Ω. Calculate (i) average load current and
(ii) power dissipated in each diode. 11
