## Q. 2.10

In Fig. 2-13, $v_S = V_m \sin ωt$ and the diode is ideal. Calculate the average value of $v_L$.

## Verified Solution

Only one cycle of $v_S$ need be considered. For the positive half-cycle, $i_D > 0$ and, by voltage division,
$v_L = \frac{R_L}{R_L + R_S} (V_m \sin ωt) ≡ V_{Lm} \sin ωt$

For the negative half-cycle, the diode is reverse-biased, $i_D = 0$, and $v_L = 0$. Hence,
$V_{L0} = \frac{1}{2 \pi} \int_0^{2\pi} v_L (ωt) d(ωt) = \frac{1}{2 \pi} \int_0^{\pi} V_{Lm} \sin ωt d(ωt) = \frac{V_{Lm}}{\pi}$

Although the half-wave rectifier gives a dc output, current flows through $R_L$ only half the time, and the average value of the output voltage is only $1/\pi = 0.318$ times the peak value of the sinusoidal input voltage.    The output voltage can be improved by use of a full-wave rectifier (see Problems 2.28 and 2.50).
When rectifiers are used as dc power supplies, it is desirable that the average value of the output voltage remain nearly constant as the load varies.    The degree of constancy is measured as the voltage regulation,
$\text{Reg} ≡ \frac{(\text{no-load} V_{L0}) – (\text{full-load} V_{L0})}{\text{full-load} V_{L0}}$          (2.6)

which is usually expressed as a percentage.   Note that 0 percent regulation implies a constant output voltage.