## Q. 2.13

Calculate the ripple factor for the half-wave rectifier of Example 2.10   (a) without a filter and   (b) with a shunt capacitor filter as in Fig. 2-15(a).

## Verified Solution

(a)  For the circuit of Example 2.10,
$F_r = \frac{Δv_L}{V_{L0}} = \frac{V_{Lm}}{V_{Lm/π}} = π ≈ 3.14$

(b)  The capacitor in Fig. 2-15(a) stores energy while the diode allows current to flow, and delivers energy to the load when current flow is blocked. The actual load voltage $v_L$ that results with the filter inserted is sketched in Fig. 2-15(b), for which we assume that $v_S = V_{Sm} \sin ωt$ and $D$ is an ideal diode. For $0 < t ≤ t_1, D$ is forward-biased and capacitor $C$ charges to the value $V_{Sm}$. For $t_1 < t ≤ t_2, v_S$ is less than $v_L$, reverse-biasing $D$ and causing it to act as an open circuit. During this interval the capacitor is discharging through the load $R_L$, giving
$v_L = V_{Sm} e^{-(t – t_1)/R_LC} \quad (t_1 < t ≤ t_2)$         (2.9)

Over the interval $t_2 < t ≤ t_2 + δ, v_S$ forward-biases diode $D$ and again charges the capacitor to $V_{Sm}$. Then $v_S$ falls below the value of $v_L$ and another discharge cycle identical to the first occurs.
Obviously, if the time constant $R_{L}C$ is large enough compared to $T$ to result in a decay like that indicated in Fig. 2-15(b), a major reduction in $Δv_L$ and a major increase in $V_{L0}$ will have been achieved, relative to the unfiltered rectifier. The introduction of two quite reasonable approximations leads to simple formulas for $Δ{v_L}$ and $V_{L0}$, and hence for $F_r$, that are sufficiently accurate for design and analysis work:
1. If $Δ{v_L}$ is to be small, then $δ → 0$ in Fig. 2-15(b) and $t_2 – t_1 ≈ T$.
2. If $Δ{v_L}$ is small enough, then (2.9) can be represented over the interval $t_1 < t ≤ t_2$ by a straight line with a slope of magnitude $V_{Sm}/R_LC$.
The dashed line labeled ‘‘Approximate $v_L$’’ in Fig. 2-15(b) implements these two approximations. From right triangle abc,
$\frac{Δ{v_L}}{T} = \frac{V_{Sm}}{R_LC} \quad \text{or} \quad Δ{v_L} = \frac{V_{Sm}}{fR_LC}$

where $f$ is the frequency of $v_S$. Since, under this approximation,
$V_{L0} = V_{Sm} – \frac{1}{2} Δ{v_L}$

and  $R_LC/T = fR_LC$ is presumed large,
$F_r = \frac{Δv_L}{V_{L0}} = \frac{2}{2fR_L C – 1} ≈ \frac{1}{fR_L C}$           (2.10)