Question 6 of 7: Full-Wave Rectifier and RC Low-Pass Filter Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 Electric Circuits and Power — December 2017
National Exams, 3 hours, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; all seven are solved below as a complete study resource.
Reference texts
Sadiku, C.K. Alexander & M.N.O. Sadiku, Fundamentals of Electric Circuits — DC/AC circuit analysis, Thevenin theorem, magnetic circuits (Ch. 13).
S. Chapman, Electric Machinery Fundamentals — magnetic circuits and reluctance.
R. Boylestad, Electronic Devices and Circuit Theory — diode rectifiers and RC filters.
M.M. Mano, Digital Design — combinational logic design.
Given. Centre-tapped secondary, each half $10\text{ V}_{RMS}$ (110/10/10 turns ratio, primary 110 V$_{RMS}$); $R_{LOAD}=50\text{ k}\Omega$; two diodes, one per half-secondary, common cathode at the load.
Given data
$V_{sec,half}$
$R_{LOAD}$
$f$
Diode drop (c)
Filter $R$ (d)
Target attenuation
10 VRMS
50 kΩ
60 Hz
0.5 V
100Ω
20 dB @ 60 Hz
Find. The rectifier schematic and waveform sketches; $I_{peak}$, $I_{avg}$ in the load; the rectified output with diode drop; the RC filter capacitor value.
Approach. Each half-secondary alternately forward-biases its own diode once per AC cycle, so the load sees the rectified magnitude of a single half-secondary sine wave at twice the line frequency (full-wave); average and peak follow the standard full-wave-rectified sine results, and the RC filter is sized from the standard first-order low-pass magnitude response evaluated at the literal 60 Hz source frequency the question specifies.
a) Schematic and waveforms. See Figure 6: the 110 V primary drives a centre-tapped 10-0-10 V secondary; $D_1$ conducts when $v_1(t)\gt0$ (top half positive), $D_2$ conducts when $v_2(t)\gt0$ (bottom half positive, i.e. $v_1\lt0$), so exactly one diode conducts at all times and the load sees a full-wave-rectified sine at twice the source frequency (120 Hz ripple). The output current $i_{LOAD}(t)$ mirrors $|v_{sec}(t)|/R_{LOAD}$, and each diode's current is a half-wave replica of that, active only during its own half-cycle.
b) Peak and average load current. Each half-secondary peak: $V_m=10\sqrt2=14.142$ V (ideal diodes, no drop, for this part).
$$I_{peak}=\frac{V_m}{R_{LOAD}}=\frac{14.142}{50{,}000}=\boxed{2.828\times10^{-4}\text{ A}=0.283\text{ mA}}$$
$$I_{avg}=\frac{2I_{peak}}{\pi}=\frac{2(2.828\times10^{-4})}{\pi}=\boxed{1.801\times10^{-4}\text{ A}=0.180\text{ mA}}$$
c) With diode on-state drop 0.5 V. Each conducting half-cycle now peaks at $V_m-V_D=14.142-0.5=\boxed{13.64\text{ V}}$, and the output additionally shows a brief "dead zone" near each zero-crossing where $|v_{sec}(t)|\lt0.5$ V and neither diode conducts (output = 0). The input waveform (each secondary half) is an undistorted $\pm14.142$ V sine; the output is the full-wave-rectified, diode-drop-reduced, dead-zone-clipped version at 120 Hz.
d) RC low-pass filter design. First-order LPF magnitude: $|H(j\omega)|=1/\sqrt{1+(\omega RC)^2}$, DC gain $=1$ (0 dB). Requiring $-20$ dB at $f=60$ Hz means $|H|=10^{-20/20}=0.1$:
$$\sqrt{1+(\omega RC)^2}=10 \ \Rightarrow\ \omega RC=\sqrt{99}=9.950$$
$$C=\frac{9.950}{\omega R}=\frac{9.950}{2\pi(60)(100)}=\boxed{2.639\times10^{-4}\text{ F} \approx 264\ \mu\text{F}}$$
(Note: the target is the literal 60 Hz source frequency the question specifies, even though the rectifier's own ripple is really at 120 Hz — this problem is testing RC transfer-function sizing, not ripple analysis.)
