Question 2 of 5: Bridge Rectifier with a Failed Diode

Question 2: Bridge Rectifier with a Failed Diode (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A four-diode bridge feeding a parallel $C\parallel R$ filter, with $D_4$ open-circuit. Ideal diodes ($V_\gamma=0$), $RC=5\text{ ms}$. Input $v_s(t)$: triangular, $f=1\text{ kHz}$ ($T=1\text{ ms}$), peak $\pm 10\text{ V}$, reaching $+10\text{ V}$ at $t=0.25\text{ ms}$.

Find. The steady-state output waveform, its peak $V_p$ and ripple $V_r$, the DC average $V_{dc}$, and the per-period conduction interval $t_{on}$.

Approach. Establish what the open diode does to the rectifier (full-wave → half-wave), then apply the standard capacitor-input-filter ripple estimate with the correct discharge interval.

In the intact bridge the source sits on the top–bottom diagonal and the load on the left–right diagonal. Current reaches the output through opposite arms: the pair $\{D_2,D_3\}$ conducts on the half-cycle when the top terminal is negative, and the pair $\{D_1,D_4\}$ conducts on the other half. With $D_4$ open, the $\{D_1,D_4\}$ path is broken, so one half-cycle is lost entirely and the circuit becomes a half-wave rectifier that recharges the capacitor only once per input period.

1. Peak output voltage. The diodes are ideal (zero drop) and two of them are in series in the conducting path, but with $V_\gamma=0$ there is no forward loss, so the capacitor charges to the full input peak: $$\boxed{V_p=10\text{ V}}$$
2. Ripple voltage. Between recharges the capacitor discharges into $R$ for essentially one full period ($T=1\text{ ms}$, since conduction now happens once per cycle). For $T\ll RC$ the standard capacitor-input-filter estimate applies: $$V_r\approx V_p\,\frac{T}{RC}=10\text{ V}\times\frac{1\text{ ms}}{5\text{ ms}}=\boxed{2\text{ V}.}$$ (The exact exponential value is $V_r=V_p(1-e^{-T/RC})=1.81\text{ V}$; the 2 V linear estimate is the standard exam answer and is used below.)
3. Average DC output. The near-triangular ripple rides symmetrically about the mid-point of the recharge/discharge excursion, so $$V_{dc}\approx V_p-\frac{V_r}{2}=10-\frac{2}{2}=\boxed{9\text{ V}.}$$
4. Conduction interval $t_{on}$. The diodes conduct only while the rising input edge climbs the last $V_r$ up to the peak. The triangular input rises from 0 to $+10\text{ V}$ in a quarter period ($0.25\text{ ms}$), a slope $$\left|\frac{dv_s}{dt}\right|=\frac{10\text{ V}}{0.25\text{ ms}}=40\text{ V/ms}.$$ The capacitor is recharged from $V_p-V_r$ back to $V_p$, so $$t_{on}\approx\frac{V_r}{\lvert dv_s/dt\rvert}=\frac{2\text{ V}}{40\text{ V/ms}}=0.05\text{ ms}=\boxed{50\ \mu\text{s}.}$$

The steady-state output is therefore a positive half-wave: a brief recharge to 10 V near each positive peak (once every 1 ms), followed by a slow $RC$ decay of about 2 V until the next peak.