The peak voltage in the output of a half wave diode rectifier fed with a sinusoidal signal without filter is 10V. The d. c. component of the output voltage is :-
A
$\dfrac{10}{\pi}$ V
B
10 V
C
$\dfrac{20}{\pi}$ V
D
$\dfrac{10}{\sqrt{2}}$ V
The answer is A.
Solution
In a half wave rectifier, the output is the positive half of the sine wave for one half cycle, and zero for the other half cycle. The d.c. component is the average value over a full cycle: $V_{dc} = \dfrac{1}{T}\displaystyle\int_0^{T/2} V_0\sin\omega t\,dt = \dfrac{V_0}{\pi}$ For the half wave diode rectifier with $V_0 = 10$ V: $V_{dc} = \dfrac{V_0}{\pi} = \dfrac{10}{\pi}$ volts (For a full wave rectifier it would be $\dfrac{2V_0}{\pi}$; the r.m.s. value $\dfrac{V_0}{\sqrt{2}}$ is not the d.c. component.)
ConceptDC (average) output of half wave rectifier
Semiconductor Electronics: Materials, Devices and Simple Circuits in past papers
106 questions from this chapter have appeared across 24 exam years.
