Given. An ideal $60\text{ Hz}$, $20\text{ V}_{rms}$ AC source drives a 4-diode full-wave bridge into a $50\text{ k}\Omega$ resistive load.
Given data
Quantity
Value
$V_{rms}$
$20\text{ V}$
$f$
$60\text{ Hz}$
$R_L$
$50\text{ k}\Omega$
Diode on-state drop (part c/d)
$0.5\text{ V}$
Filter resistance (part d)
$100\,\Omega$
Find. The rectifier schematic and waveforms; peak/average load current; the effect of diode drop on input/output voltage; and an RC low-pass filter design.
Approach. For an ideal bridge, the output is the full-wave-rectified input ($v_{out}=|v_{in}|$); with real diodes, two diodes conduct at a time and each contributes its own on-state drop. The output ripple repeats at twice the line frequency, so the RC filter's corner is set against $2f=120\text{ Hz}$.
a) Schematic and waveforms. The bridge (Figure 6) routes current through diagonal diode pairs D1-D4 on the positive half-cycle of $v_s(t)$ and D2-D3 on the negative half-cycle, so the load always sees current in the same direction: $v_{out}(t)=|v_s(t)|$ (ideal diodes), a full-wave "humped" waveform at $120\text{ Hz}$; $i_{out}(t)=v_{out}(t)/R_L$ has the same shape; each diode conducts exactly every other half-cycle, carrying the full load current during its ON half and zero during its OFF half.
b) Peak and average load current (ideal diodes). $$V_p = V_{rms}\sqrt2 = 20\sqrt2 = 28.28\text{ V}, \qquad I_{pk} = \dfrac{V_p}{R_L} = \dfrac{28.28}{50{,}000} = \boxed{0.5657\text{ mA}}$$ For a full-wave-rectified sinusoid, the average is $2/\pi$ times the peak: $$I_{avg} = \dfrac{2I_{pk}}{\pi} = \dfrac{2(0.5657)}{\pi} = \boxed{0.3601\text{ mA}}$$
c) Effect of a $0.5\text{ V}$ diode drop. Two diodes conduct in series with the load at every instant, so the output peak is reduced by $2V_D$: $$V_{p,out} = V_p - 2V_D = 28.28-2(0.5) = \boxed{27.28\text{ V}}$$ and the output additionally shows a dead zone near each zero crossing (roughly $|v_s(t)|<2V_D=1.0\text{ V}$) where no diode pair is forward biased and $v_{out}=0$ -- the input remains a clean $28.28\text{ V}$-peak sinusoid throughout, only the output waveform is affected.
d) RC low-pass filter, $20\text{ dB}$ attenuation at $120\text{ Hz}$. A single-pole RC low-pass has DC gain $1$ ($0\text{ dB}$) and $|H(f)|=1/\sqrt{1+(2\pi f RC)^2}$. Requiring $20\text{ dB}$ attenuation means $|H(120\text{ Hz})|=10^{-20/20}=0.1$ relative to DC: $$\sqrt{1+(2\pi(120)RC)^2} = 10 \;\Rightarrow\; 2\pi(120)RC = \sqrt{99} = 9.950 \;\Rightarrow\; RC = \dfrac{9.950}{2\pi(120)} = 0.01320\text{ s}$$ With $R=100\,\Omega$: $$\boxed{C = \dfrac{RC}{R} = \dfrac{0.01320}{100} = 131.96\,\mu\text{F}}$$
