Question

The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without the filter is. The dc component of the output voltage is:
A. $15\sqrt 2 V$
B. $\dfrac{{15}}{\Pi }V$
C. 15V
D. $\dfrac{{30}}{\Pi }V$

Answer 1

Hint: We know that ${V_{dc}} = \dfrac{{{V_{PP}}}}{\Pi }$; and by substituting ${V_{PP}}$ i.e. peak to peak voltage we get ${V_{dc}} = $ the DC component of voltage. But in a half wave rectifier ${V_{PP}}$ is nothing but ${V_P}$ i.e., the peak voltage = 15V

Complete Answer:
We know: ${V_{dc}} = \dfrac{{{V_{PP}}}}{\Pi }$; where ${V_{PP}}$ is same as ${V_P}$.
Substituting ${V_P}$ in the above formula we get,
 ${V_{dc}} = \dfrac{{{V_{PP}}}}{\Pi } = \dfrac{{{V_P}}}{\Pi } = \dfrac{{15}}{\Pi }V$
Hence, option (B) is correct.

Additional Information: :
${V_P} = {V_{rms}}\sqrt 2 $ for all rectifiers.
And for Half wave rectifier ${f_{in}} = {f_{out}}$; where f is frequency of the wave
For full wave rectifier and Bridge rectifier \[2{f_{in}} = {f_{out}}\].

Note: Here ${V_{dc}} = \dfrac{{{V_{PP}}}}{\Pi }$ i.e., peak to peak voltage and in a half-wave rectifier peak to peak voltage is nothing but peak voltage. Whereas in a full-wave rectifier peak to peak voltage is twice that of the peak voltage. Here, we have approximated the diode as an ideal approximation which means the voltage across the diode is zero volts. Whereas in the first approximation there is (for silicon diode) a drop across the diode and in the second approximation we consider the voltage falling across the resistance of the diode.