Question 6 of 7: Full-Bridge Rectifier and RC Low-Pass Filter
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, 04-BS-4 Electric Circuits and Power — May 2017. Closed book;
one double-sided aid sheet permitted; any five of the seven questions constitute a
complete paper (all seven are answered here as a complete study resource).
Reference texts: Sadiku & Alexander, Fundamentals of Electric
Circuits, 6th ed. (circuit analysis, transients, AC steady state, magnetic circuits);
Boylestad & Nashelsky, Electronic Devices and Circuit Theory (rectifiers);
Mano, Digital Design (combinational logic).
Given. Ideal 60 Hz, 20 VRMS AC source feeding a
full-wave (four-diode) bridge rectifier into a 50 kΩ resistive load; part (c) adds a
0.5 V on-state diode drop; part (d) filters the DC output with a 100 Ω series
resistor.
Given data
Quantity
Value
VRMS
20 V
f
60 Hz
RL
50 kΩ
Diode drop (part c)
0.5 V
R (filter, part d)
100 Ω
Ripple frequency
120 Hz
Target attenuation
20 dB
Find. The rectifier schematic and waveforms; peak and average load current;
the waveforms with a 0.5 V diode drop; and the RC low-pass filter capacitor value.
Full-wave bridge rectifier feeding RL
(part a)
Input sinusoid (dashed) and rectified output
(solid) — ideal diodes (part a)
Approach. Ideal-diode peak/average output formulas give parts (a)–(b)
directly; two diodes conduct in series at any instant, so part (c) simply shifts the output peak
down by twice the diode drop with a dead zone near each zero crossing; part (d) is a standard
first-order RC transfer-function magnitude design at the ripple frequency.
Schematic and waveforms (part a). Four diodes (D1–D4) in a bridge feed
RL from the AC source; on each half-cycle one diagonal pair conducts, so the output
voltage/current are both full-wave rectified (always positive) at twice the source frequency
(120 Hz), and each diode conducts for exactly one half-cycle, alternating with the diagonal
pair (see figure).
Peak and average load current (part b). Peak source voltage
$V_{pk}=\sqrt2\,V_{RMS}=\sqrt2(20)=28.28\text{ V}$; with ideal diodes the output peak equals
Vpk:
$$ I_{pk} = \frac{V_{pk}}{R_L} = \frac{28.28}{50{,}000} = \boxed{0.566\text{ mA}}. $$
Full-wave average (DC) output: $V_{dc}=\dfrac{2}{\pi}V_{pk}=\dfrac{2}{\pi}(28.28)=18.0\text{ V}$, so
$$ I_{dc} = \frac{V_{dc}}{R_L} = \frac{18.0}{50{,}000} = \boxed{0.360\text{ mA}}. $$
With a 0.5 V diode drop (part c). Two diodes conduct in series at every
instant, so the output peak is reduced by $2(0.5)=1\text{ V}$, and no conduction occurs while
$|v_{in}|\lt1\text{ V}$ (a short flat dead-zone at each zero crossing):
$$ V_{pk,drop} = V_{pk}-2(0.5) = 28.28-1 = \boxed{27.28\text{ V}}. $$
RC low-pass filter design (part d). For a simple RC low-pass,
$|H(j\omega)|=1/\sqrt{1+(\omega RC)^2}$ with DC gain 1 (0 dB). Requiring 20 dB
attenuation at the 120 Hz ripple means $|H|=10^{-20/20}=0.1$:
$$ \sqrt{1+(\omega RC)^2}=10 \ \Rightarrow\ \omega RC=\sqrt{99}=9.950, \qquad
\omega = 2\pi(120)=754.0\text{ rad/s}. $$
$$ C = \frac{9.950}{\omega R} = \frac{9.950}{(754.0)(100)} = \boxed{132\ \mu\text{F}}. $$
