Question 6 of 7: Full-Wave Bridge Rectifier and DC-Side Filter
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2015 — 04-BS-4 Electric Circuits and Power. Closed book; one aid sheet (both sides) and a Casio/Sharp approved calculator permitted; 3 hours. Seven questions; any five constitute a complete paper (only the first five in the answer book are marked). All seven are solved here as a study resource. Marks: each question 20 (four 5-mark parts).
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (KCL/KVL, Thévenin, first-order transients, AC power); R. Boylestad & L. Nashelsky, Electronic Devices and Circuit Theory (bridge rectifier, RC filter); S. J. Chapman, Electric Machinery Fundamentals (magnetic circuits); M. M. Mano & M. Ciletti, Digital Design (combinational logic). Canadian frame: 60 Hz mains, SI units.
Question 6: Full-Wave Bridge Rectifier and DC-Side Filter (20 marks)
Given. A four-diode (Graetz) bridge, ideal 60 Hz / 20 V rms source, R_load = 50 kΩ.
Given data
Source
V_rms
f
R_load
Diode drop (c)
ideal AC
20 V
60 Hz
50 kΩ
0.5 V each
Find. (a) schematic + waveforms; (b) peak and average load current; (c) waveforms with 0.5 V diode drops; (d) an RC filter (R = 100 Ω) giving 20 dB attenuation at 120 Hz.
a) Bridge (Graetz) rectifier. On each half-cycle two diagonally-opposite diodes conduct (D1–D4, then D3–D2), steering both half-cycles the same way through the 50 kΩ load.
a/c) Input (dashed) and output (solid) voltages. Ideal diodes give the full-wave rectified |V_in| with peak 28.3 V; the output current has the same shape scaled by 1/R. With 0.5 V diode drops, two diodes conduct in series, so the output peak falls to 27.3 V and clips to zero while |V_in| < 1.0 V (see part c).
Approach. Peak from $V_{pk}=\sqrt2\,V_{rms}$; full-wave average $=2V_{pk}/\pi$; currents by Ohm's law into 50 kΩ. For the filter, size C so a series-R / shunt-C low pass has magnitude 0.1 (−20 dB) at 120 Hz.
a) Operation and waveforms. Positive half-cycle: D1 and D4 conduct; negative half-cycle: D3 and D2 conduct — the load always sees the same polarity, giving a full-wave rectified output (top diagram). Each diode carries the load current for one half-cycle (half-wave blocks of period 1/60 s) and blocks the other.
b) Peak and average load current. $V_{pk}=\sqrt2(20)=28.28$ V, so
$$I_{pk}=\frac{V_{pk}}{R}=\frac{28.28}{50\,000}=0.566\text{ mA},\qquad I_{avg}=\frac{2V_{pk}}{\pi R}=\frac{2(28.28)}{\pi(50\,000)}=0.360\text{ mA}.$$
$$\boxed{I_{pk}=0.566\text{ mA},\qquad I_{avg}=0.360\text{ mA}}$$
(The average output voltage is $2V_{pk}/\pi = 18.0$ V.)
c) With 0.5 V diode drops. Two diodes conduct in series each half-cycle, so the output is $v_o=|v_{in}|-2(0.5)=|v_{in}|-1.0$ V when $|v_{in}|>1.0$ V, and zero otherwise. The peak output falls to
$$V_{o,pk}=28.28-1.0=27.28\text{ V},$$
with a small dead-zone near each zero crossing where $|v_{in}|<1$ V (shown clipped in the waveform figure).
d) RC low-pass filter (R = 100 Ω). For a series-R, shunt-C low pass the DC gain is 1 and $|H(f)|=1/\sqrt{1+(\omega RC)^2}$. A 20 dB attenuation means $|H|=0.1$:
$$\frac{1}{\sqrt{1+(\omega RC)^2}}=0.1\;\Rightarrow\;\omega RC=\sqrt{99}=9.95,\quad \omega=2\pi(120)=754\text{ rad/s}.$$
$$C=\frac{\sqrt{99}}{\omega R}=\frac{9.95}{754(100)}\;\Rightarrow\;\boxed{C \approx 132\ \mu\text{F}}$$
Check: $20\log_{10}|H(120\text{ Hz})| = -20.0$ dB. (Corner frequency $f_c=1/(2\pi RC)=12.1$ Hz.)
