Question 5 of 5: Bridge Rectifier with Capacitor Filter
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 07-Elec-A5 Electronics — December 2016.
Closed-book; any non-communicating calculator permitted; three hours. Five questions,
20 marks each — all five constitute a complete paper, so every question is solved
below. Per the paper’s notes, op-amps are ideal with supply rails of
±15 V unless a question states otherwise, and ground/chassis are common.
Reference texts (22-Elec-A5 Electronics): A. S. Sedra
& K. C. Smith, Microelectronic Circuits, 8th ed. (Oxford)
— diode circuits and rectifiers (Ch. 4), MOSFET biasing and amplifiers
(Ch. 5–7), op-amp circuits and offsets (Ch. 2/9), BJT amplifiers
(Ch. 6–7); R. C. Jaeger & T. N. Blalock,
Microelectronic Circuit Design (McGraw-Hill).
Question 5: Bridge Rectifier with Capacitor Filter (20 marks)
Given. A four-diode bridge feeds a filter capacitor $C$ in parallel with
$R=100$ Ω. Source: 100 Hz ($T=10$ ms), 50% duty, levels 0 and
$+10$ V (unipolar). Diode on-voltage 0.7 V; target ripple $V_r=0.5$ V.
Find. (a) output waveform and average $v_O$; (b) minimum $C$; (c) the
$i_{D1}$ waveform and the charging interval; (d) the average current in $D_3$.
Figure 5. Bridge rectifier with $C\|R$ filter.
With a unipolar 0/+10 V input, only $D_1$ and $D_2$ are ever forward biased;
$D_3$ and $D_4$ (the negative-half diodes) never conduct.
Approach. On each $+10$ V half the source pushes current through the
$D_1$–load–$D_2$ path (two diode drops), charging $C$; on each 0 V half all
diodes are off and $C$ discharges into $R$. The input never reverses, so the other diode
pair never turns on.
Peak output (two diode drops). During a $+10$ V half, current flows
$v_{IN}\!\to\!D_1\!\to$ load $\to\!D_2\!\to$ back, so two on-voltages are lost:
$$V_{pk}=10-2(0.7)=\boxed{8.6\ \text{V}}.$$
(a) Average output. The output holds at $8.6$ V during each
$+10$ V half and droops by $V_r=0.5$ V during each 0 V half. Taking the usual
mid-ripple average,
$$v_{O,\text{avg}}\approx V_{pk}-\tfrac12V_r=8.6-0.25=\boxed{8.35\ \text{V}}.$$
The waveform is sketched below (flat at 8.6 V, then a sawtooth droop to 8.1 V).
(b) Minimum capacitor. The capacitor discharges through $R$ for the
whole 0 V half, $t_{dis}=T/2=5$ ms. For a small ripple,
$$V_r\approx V_{pk}\frac{t_{dis}}{RC}\;\Rightarrow\;
C\ge\frac{V_{pk}\,t_{dis}}{R\,V_r}=\frac{8.6(5\times10^{-3})}{100(0.5)}
=\boxed{860\ \mu\text{F}}.$$
(c) $D_1$ current waveform. $D_1$ conducts only during the
$+10$ V halves. At each leading edge it delivers a narrow, tall recharge spike that
restores the charge $\Delta Q=CV_r=430\ \mu$C the cap lost, then it carries the steady load
current $i_O=V_{pk}/R=86$ mA for the rest of that half; during the 0 V halves
$i_{D1}=0$. The capacitor is charging only during that brief leading-edge spike of
each $+10$ V half (essentially impulsive for ideal diodes and a square input); the rest
of the time it is either idle (held) or discharging. The sketch is below.
(d) Average current in $D_3$. A bridge rectifies both input polarities,
but this source is unipolar ($v_{IN}\ge0$ always). $D_3$ and $D_4$ only conduct on
negative half-cycles, which never occur, so
$$\boxed{\overline{i_{D3}}=0\ \text{A}}.$$
Only $D_1$ and $D_2$ carry current; the bridge behaves as a half-wave rectifier in time, and
its average diode current in $D_1$ (or $D_2$) equals the average load current,
$\overline{i_{D1}}=\overline{i_O}=v_{O,\text{avg}}/R\approx 84$ mA.
Figure 5a. Input square wave (0 / +10 V,
100 Hz).
Figure 5b. Output $v_O$: held at
$V_{pk}=8.6$ V during each $+10$ V half, drooping by $V_r=0.5$ V (to
8.1 V) during each 0 V half.
Figure 5c. $i_{D1}$: a recharge spike at the
start of each $+10$ V half (the only interval the cap is charging), then the steady load
current; zero during the 0 V halves.
