Question 6 of 7: Full-Bridge Rectifier and RC Low-Pass Filter
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, 04-BS-4 Electric Circuits and Power — May 2016. Closed book;
one double-sided aid sheet permitted; any five of the seven questions constitute a
complete paper (all seven are answered here as a complete study resource).
Reference texts: Sadiku & Alexander, Fundamentals of Electric
Circuits, 6th ed. (circuit analysis, transients, AC steady state, magnetic circuits);
Boylestad & Nashelsky, Electronic Devices and Circuit Theory (rectifiers);
Mano, Digital Design (combinational logic).
Check: the source labels this item "Problem 6" rather than
"Question 6" — a header-format quirk of this exam; it is an
entirely ordinary, fully-printed question and is answered here as Question 6.
Given. Ideal AC source, 60 Hz, 20 VRMS, feeding a full-wave (Graetz) bridge of four diodes into a 50 kΩ resistive load.
Given data
Quantity
Value
Source frequency
60 Hz
VRMS
20 V
RL (load)
50 kΩ
Diode on-state drop (part c)
0.5 V each
Filter series R (part d)
100 Ω
Target attenuation
−20 dB at 120 Hz
Find. The bridge schematic and waveforms; peak and average load current; the waveforms with diode drop included; and the capacitor value for the specified RC filter.
Figure 6b: Input/output voltage and diode-pair conduction current (each diode pair conducts on alternate half-cycles)
Approach. In a full bridge, exactly two diodes conduct during each half-cycle (D1&D4 on the positive half, D2&D3 on the negative half), producing a full-wave rectified $|\sin|$ output at TWICE the line frequency; peak/average follow directly, the diode drop subtracts $2V_d$ from the peak and creates a dead zone near each zero-crossing, and the RC filter's magnitude response at the known ripple frequency (120 Hz, not 60 Hz) fixes C.
Schematic and waveforms (part a). D1 and D4 conduct together whenever
the source's left terminal is positive, routing current through RL from top to bottom; D2 and D3 conduct
together on the opposite half-cycle, routing current through RL in the same direction —
hence a unipolar, full-wave-rectified output at twice the input frequency (Figure 6, lower panel). Each
diode pair carries current only every other half-cycle, as plotted.
Peak and average load current (part b).
$$ V_p = V_{RMS}\sqrt2 = 20\sqrt2 = \boxed{28.28\text{ V}}, \qquad
I_p = \frac{V_p}{R_L} = \frac{28.28}{50000} = \boxed{0.5657\text{ mA}}. $$
For a full-wave rectified sinusoid, the average is $2/\pi$ times the peak:
$$ I_{avg} = \frac{2}{\pi}I_p = \boxed{0.3601\text{ mA}}. $$
Waveforms with 0.5 V diode drop (part c). Two diodes conduct in
series at every instant, so the peak output is reduced by $2V_d$, and the output stays at zero
whenever the instantaneous input magnitude is below $2V_d$ (a dead zone around each zero-crossing,
red trace in Figure 6):
$$ V_{p,drop} = V_p-2V_d = 28.28-1.0 = \boxed{27.28\text{ V}}. $$
RC low-pass filter design (part d). For $H(j\omega)=1/(1+j\omega RC)$
the DC gain is 0 dB; requiring $-20$ dB (a factor of 0.1) at the 120 Hz ripple frequency:
$$ |H(j\omega)| = \frac1{\sqrt{1+(\omega RC)^2}} = 0.1 \;\Rightarrow\; \omega RC=\sqrt{99}=9.95, \qquad
\omega = 2\pi(120)=753.98\text{ rad/s}. $$
$$ RC = \frac{9.95}{753.98} = 13.20\text{ ms} \;\Rightarrow\; C=\frac{RC}{R}=\frac{0.01320}{100} =
\boxed{132\ \mu\text{F}}. $$
Check: the resulting corner frequency $f_c=1/(2\pi RC)\approx12.1$ Hz sits a decade below the
120 Hz ripple, consistent with the target $-20$ dB/decade attenuation.
