Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, December 2016 — 04-BS-4 Electric Circuits and Power. Three hours duration, closed book (one aid sheet permitted). Seven questions are printed; any five constitute a complete paper, but all seven are solved below as a complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits, 7th ed. (circuit analysis, transients, AC power, rectifiers – Questions 1–6); Chapman, Electric Machinery Fundamentals, 5th ed. (magnetic circuits – Question 7).
Given. An ideal 50 Hz, 20 Vrms AC source feeding a full-wave (4-diode) bridge rectifier into a 50 kΩ resistive load; part (c) adds a 0.4 V diode on-state drop; part (d) filters the DC output with a 50 Ω series resistor.
Given data
Vrms
20 V
RL
50 kΩ
f
50 Hz
Diode drop
0.4 V
Filter R
50 Ω
Attenuation target
−20 dB @ 100 Hz
Find. (a) the bridge schematic and waveform sketches; (b) peak and average load current; (c) the sketch and output peak with diode drop included; (d) the filter capacitor value.
Figure 6 — full-wave bridge rectifier: D1/D2 conduct on the positive half-cycle, D3/D4 on the negative half-cycle, so RL always sees current in the same direction.
Input $v_{in}(t)$ (dashed) versus rectified output $v_{out}(t)$ (solid) — every half-cycle of the input is folded onto the positive axis; with a diode drop the rectified peaks are flattened by $2V_D$.
Approach. With ideal diodes the bridge simply reflects the negative half-cycles of the input onto the positive axis, so the output is $|v_{in}(t)|$ with peak $V_p=V_{rms}\sqrt2$; the average of a full-wave-rectified sine is the standard $2V_p/\pi$ result. With a real diode drop, two diodes conduct in series at any instant, so the output peak is reduced by $2V_D$. The RC low-pass filter is sized from the standard first-order magnitude response evaluated at the attenuation frequency relative to its (unity) DC gain.
(a) Schematic and waveforms. See Figure 6 for the bridge topology and the sketch above for $v_{in}(t)$ vs.\ $v_{out}(t)$: the output current $i_{out}(t)=v_{out}(t)/R_L$ has the same full-wave-rectified shape; each diode conducts for one half-cycle only, carrying the full load current during its conduction interval and zero otherwise (D1&D2 conduct together on the positive half-cycle, D3&D4 on the negative half-cycle).
(b) Peak and average load current.
$$ V_p=V_{rms}\sqrt2=20\sqrt2=28.284\text{ V}\;\Rightarrow\;I_p=\frac{V_p}{R_L}=\frac{28.284}{50{,}000}=\boxed{0.5657\text{ mA}} $$
$$ I_{avg}=\frac{2I_p}{\pi}=\frac{2(0.5657)}{\pi}=\boxed{0.3601\text{ mA}} $$
(c) Output peak with a 0.4 V diode drop. At the peak of conduction, two diodes are in series in the current path (e.g. D1 and D4), so
$$ V_{p,\text{drop}}=V_p-2V_D=28.284-2(0.4)=\boxed{27.484\text{ V}} $$
The sketched output waveform is identical in shape but its humps are flattened, sitting $2V_D=0.8\text{ V}$ below the ideal curve at every peak (and clipped to zero, rather than following the input, whenever $|v_{in}(t)|\lt2V_D$).
(d) RC low-pass filter design. A series-R, shunt-C low-pass filter has DC gain 1 and magnitude $|H(j\omega)|=1/\sqrt{1+(\omega RC)^2}$. Requiring $-20\text{ dB}$ (i.e. $|H|=10^{-20/20}=0.1$) at $\omega_a=2\pi(100)=628.3\text{ rad/s}$:
$$ 0.1=\frac{1}{\sqrt{1+(\omega_aRC)^2}}\;\Rightarrow\;\omega_aRC=\sqrt{\frac{1}{0.1^2}-1}=\sqrt{99}=9.950 $$
$$ C=\frac{9.950}{\omega_a R}=\frac{9.950}{628.3(50)}=\boxed{316.71\ \mu\text{F}} $$
