A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is 1000 Omega. The forward resistance R_(f) ideal diode is 10 Omega. Calculate.
(i) Peak, average and rms values of load current
(ii) d.c power output
(ii) a.c power input
(iv) % Rectifier efficiency
(v) Ripple factor.
Answers
Given:
A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is 1000 Omega. The forward resistance R_(f) ideal diode is 10 Omega.
To find:
Calculate.
(i) Peak, average and rms values of load current
(ii) d.c power output
(ii) a.c power input
(iv) % Rectifier efficiency
(v) Ripple factor.
Solution:
From the given information, we have the data as follows.
A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is 1000 Omega. The forward resistance R_(f) ideal diode is 10 Omega.
(i) Peak, average and rms values of load current
I_{in} = V_m / (R_f + R_L)
= 25 / (10 + 1000)
= 24.75 A
I_{dc} = I_m/π
= 24.75/3.14
= 7.88 mA
I_{max} = I_m/2
= 24.75/2
= 12.38 mA
(ii) d.c power output
P_{dc} = I²_{dc} × R_L
= (7.88 × 10^-3)² × 10^3
= 63 mW
(ii) a.c power input
P_{ac} = I²_{rms} × (R_f + R_L)
= (12.38 × 10^-3)² ×( 10 + 1000)
= 155 mW
(iv) % Rectifier efficiency
η = P_{dc}/P_{ac} × 100
= 63/155 × 100
= 40 %
(v) Ripple factor.
= √ [(I_{rms}/I_{ac})² - 1]
= √ [(12.38/7.88)² - 1]
= 1.21