## Chapter 2

## 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*).

## Step-by-Step

## 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*)