Question 6 of 7: Half-Wave Rectifier and RC Filter Design

Question 6 (Problem 6): Half-Wave Rectifier and RC Filter Design (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. $R_L=50\text{ k}\Omega$, $V_s=20\text{ V}_{rms}$ at 60 Hz (ideal source, part a/b); diode on-state drop $V_D=0.6\text{ V}$ (part c); filter resistor $R_f=100\,\Omega$, target attenuation 20 dB at 60 Hz relative to the DC gain (part d).

Find. (a) schematic + waveform sketches; (b) $I_{pk}$, $I_{avg}$; (c) output with diode drop; (d) filter capacitor $C_f$.

Approach. A single diode conducts only while the source is forward-biasing it (positive half-cycle for the orientation shown), clipping the negative half-cycle entirely; peak/average follow directly from $V_{pk}=\sqrt2\,V_{rms}$ and the standard half-wave average $I_{pk}/\pi$. The filter is a first-order RC low-pass whose gain rolls off at $-20\log_{10}\sqrt{1+(\omega RC)^2}$ dB relative to its DC (0 Hz) gain of unity.

1. a) Schematic and waveforms. The diode D conducts only while $v_s(t)\gt 0$ (forward-biased), clamping $v_{out}$ to $v_s(t)$ during that half-cycle and to zero otherwise (reverse-biased, no current); $i_{out}(t)=v_{out}(t)/R_L$ follows the same clipped shape.
2. b) Peak and average load current. $$V_{pk}=\sqrt2\,V_{rms}=\sqrt2(20)=28.28\text{ V} \qquad I_{pk}=\frac{V_{pk}}{R_L}=\frac{28.28}{50{,}000}=\boxed{0.5657\text{ mA}}$$ For a half-wave rectified sinusoid the average (DC) value is $I_{pk}/\pi$: $$I_{avg}=\frac{I_{pk}}{\pi}=\boxed{0.1801\text{ mA}}$$
3. c) With diode on-state drop $V_D=0.6\text{ V}$. While conducting, $v_{out}=v_s(t)-V_D$, so the clipped output peak is reduced: $$V_{pk,out}=V_{pk}-V_D=28.28-0.6=\boxed{27.68\text{ V}}$$ the waveform shape is otherwise identical (still clipped to zero on the reverse-biased half-cycle), only the conducting-half peak is shifted down by $0.6\text{ V}$ and the diode does not begin conducting until $v_s(t)$ exceeds $0.6\text{ V}$.
4. d) RC low-pass filter design. The filter's transfer function magnitude relative to its DC gain is $|H(j\omega)|=1/\sqrt{1+(\omega R_fC_f)^2}$. Requiring $20\log_{10}|H(j\omega)|=-20\text{ dB}$ at $\omega=2\pi(60)$ means $|H|=10^{-20/20}=0.1$: $$\sqrt{1+(\omega R_fC_f)^2}=10 \ \Rightarrow\ \omega R_fC_f=\sqrt{99}=9.9499$$ $$C_f=\frac{9.9499}{(2\pi\cdot60)(100)}=\boxed{263.9\,\mu\text{F}}$$ This places the filter's corner (−3 dB) frequency at $f_c=1/(2\pi R_fC_f)=6.03\text{ Hz}$, well below the 60 Hz ripple, as expected for 20 dB of ripple attenuation.