Question 6 of 7: Full-Wave Bridge Rectifier and RC Low-Pass Filter Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 04-BS-4 Electric Circuits and Power — undated sitting (internal
evidence places it at May 2019). 3 hours, closed book, one
double-sided aid sheet, approved Casio/Sharp calculator only. The exam instructs
"any five questions constitute a complete paper" — all seven are solved below as a
complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits
(7th ed.) — Ch. 2–4 (resistive circuits, node/mesh analysis, Thévenin/Norton,
maximum power transfer), Ch. 7 (first-order RC/RL transients), Ch. 9–10 (sinusoidal
steady state, phasors, AC power); Chapman, Electric Machinery Fundamentals
(5th ed.) — Ch. 1 (magnetic circuits, reluctance, fringing, force on an armature);
Sadiku, Elements of Electromagnetics — Ch. 5, 8 (magnetic energy and force);
Mano & Ciletti, Digital Design — Ch. 2–3 (Boolean algebra, combinational
logic design).
Approach. Peak input voltage sets the ideal peak output; average of a rectified sinusoid gives $I_{avg}$; each pair of conducting diodes subtracts its forward drop from the peak; and the single-pole low-pass gain expression $|H(j\omega)|=1/\sqrt{1+(\omega RC)^2}$ is inverted at $f=100\text{ Hz}$ for the required attenuation.
Part (a) — schematic and ideal waveforms. Figure 6a shows the four diodes arranged so that D1/D4 conduct on one AC half-cycle and D2/D3 on the other, always steering current through $R_L$ in the same direction; the output voltage and current are therefore both strictly non-negative, tracing $|v_{in}(t)|$ (Figure 6b). Each diode carries the full load current for exactly one half-cycle and blocks (carries none) for the other.
Part (b) — peak and average load current. The ideal peak (amplitude) of the input is $$V_p=V_{RMS}\sqrt2=(20)(1.41421)=28.28\text{ V}$$ which appears unattenuated across $R_L$ at every voltage peak (ideal diodes), so $$I_p=\dfrac{V_p}{R_L}=\dfrac{28.28\text{ V}}{50\,000\ \Omega}=\boxed{0.5657\text{ mA}}$$ A full-wave rectified sine has average value $2/\pi$ times its peak: $$I_{avg}=\dfrac{2I_p}{\pi}=\dfrac{2(0.5657\text{ mA})}{\pi}=\boxed{0.3601\text{ mA}}$$
Part (c) — with a 0.4 V diode drop. At every instant, two diodes conduct in series with the load, so the peak output is reduced by twice the single-diode drop: $$V_{p,\text{out}}=V_p-2(0.4\text{ V})=28.28-0.8=\boxed{27.48\text{ V}}$$ giving $I_{p,\text{out}}=27.48\text{ V}/50\,000\ \Omega=0.5497\text{ mA}$. The waveform (Figure 6c) is the same rectified-sine shape but flattened slightly at the top and, more importantly, pinched to zero a little earlier/later around each zero-crossing (the diodes only conduct once $|v_{in}|$ exceeds $0.8\text{ V}$), producing brief dead-zones the ideal waveform does not have.
Part (d) — RC low-pass filter design. A one-pole RC low-pass has DC gain $1$ ($0\text{ dB}$) and $$|H(j\omega)|=\dfrac{1}{\sqrt{1+(\omega RC)^2}}.$$ Requiring $-20\text{ dB}$ (a factor of $10^{-20/20}=0.1$ in amplitude) at $f=100\text{ Hz}$: $$0.1=\dfrac{1}{\sqrt{1+(\omega RC)^2}}\ \Rightarrow\ \omega RC=\sqrt{\dfrac{1}{0.1^2}-1}=\sqrt{99}=9.950$$ With $R=50\ \Omega$ and $\omega=2\pi(100)=628.3\text{ rad/s}$: $$C=\dfrac{9.950}{\omega R}=\dfrac{9.950}{(628.3)(50)}=\boxed{316.7\ \mu\text{F}}$$ giving a corner frequency $f_c=1/(2\pi RC)=10.05\text{ Hz}$, comfortably below the $100\text{ Hz}$ ripple to be attenuated and confirming $-20.0\text{ dB}$ exactly at $100\text{ Hz}$ when substituted back into $|H(j\omega)|$.
Figure 6b — input voltage (dashed) and ideal full-wave output (solid).
Figure 6c — output voltage with a 0.4 V on-state drop per conducting diode pair.
