Half-Wave Rectifier and RC Smoothing Filter (Problem 6): 04-BS-4, December 2013, Q6

Question 6: Half-Wave Rectifier and RC Smoothing Filter (Problem 6) (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 half-wave diode rectifier feeding a resistive load $R_L=50 kΩ$ from an ideal AC source (60 Hz, 20 V$_{RMS}$).

Given data
$R_L$	frequency	source (RMS)	diode drop $V_D$
50 kΩ	60 Hz	20 V	0.6 V

Find. (a) the schematic and the input/output/current waveforms with the conduction interval marked; (b) peak and average load current; (c) the waveforms with a $0.6$ V diode drop; (d) an RC low-pass filter (using 100 Ω) that attenuates the 60 Hz component by 20 dB relative to DC.

An ideal half-wave rectifier passes only the positive half of each cycle to the load, so the output is a train of half-sinusoids. The peak follows from the RMS rating and the average is the familiar $V_{peak}/\pi$; a finite diode drop simply shifts the conducting threshold and clips the peak.

(a) Schematic and waveforms. The diode conducts while $v_s(t)\gt0$ (ideal), delivering the solid half-sine output below.
Half-wave rectifier schematic: diode D conducts on the positive half-cycle, delivering vL to RL.
Input vs(t) (dashed) and rectified output vL(t) (solid): only positive half-cycles pass.
(b) Peak and average current. $V_{peak}=\sqrt2\,V_{RMS}=28.284$ V, so $$I_{peak}=\frac{V_{peak}}{R_L}=0.566\text{ mA},\qquad I_{avg}=\frac{V_{peak}}{\pi R_L}=0.18\text{ mA}.$$
(c) With a 0.6 V diode drop. Conduction now begins only when $v_s\gt V_D=0.6$ V, and the output peak clips to $V_{peak}-V_D=27.684$ V; the output is the same half-sine shifted down by $V_D$ and slightly narrower in conduction angle.
(d) RC low-pass filter. For a first-order RC section the gain magnitude is $|H|=1/\sqrt{1+(\omega R_fC)^2}$ with DC gain 1. A 20 dB attenuation means $|H|=0.1$, hence $\omega R_fC=\sqrt{99}=9.95$. With $R_f=100 Ω$ at 60 Hz, $$\boxed{C=\frac{9.95}{2\pi f R_f}=263.929\ \mu\text{F}}\quad(f_c=\tfrac{1}{2\pi R_fC}=6.03\text{ Hz}).$$

Final results
Quantity	Value
$V_{peak}$	28.284 V
$I_{peak}$	0.566 mA
$I_{avg}$	0.18 mA
Output peak with $V_D$	27.684 V
Filter $C$ (with $R_f=100 Ω$)	263.929 µF

Topic: Half-wave rectification and first-order RC smoothing.

Key relations: $V_{peak}=\sqrt2V_{RMS}$; half-wave average $V_{dc}=V_{peak}/\pi$; diode threshold $v_s\gt V_D$; RC low-pass $|H|=1/\sqrt{1+(\omega RC)^2}$, corner $f_c=1/(2\pi RC)$; $-20$ dB per decade roll-off.

Why this works: A single diode gates the positive half-cycles; the DC term of that waveform is $V_{peak}/\pi$. Cascading an RC section rejects the ripple fundamental because its reactance falls with frequency, so the ratio of ripple to DC is set purely by $\omega RC$.

Common pitfall: Using $V_{peak}/2$ (full-wave) instead of $V_{peak}/\pi$ for the half-wave average; and forgetting that for a half-wave rectifier the lowest ripple frequency equals the line frequency, not twice it.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed. (filters, Bode); Boylestad & Nashelsky, Electronic Devices and Circuit Theory, Ch. 2 (diode rectifiers).

Question 6 of 7: Half-Wave Rectifier and RC Smoothing Filter (Problem 6)

Question 6: Half-Wave Rectifier and RC Smoothing Filter (Problem 6) (20 marks)