28 Jun 2025
Rectifier Output, Regulation, Ripple, Efficiency, and Filters
Average and RMS output, ripple factor, efficiency, voltage regulation, and basic smoothing filters.
For a periodic output of period $T_o$, the dc and RMS values are
\[V_{dc}=\frac1{T_o}\int_0^{T_o}v_o(t)\,dt, \qquad V_{rms}=\sqrt{\frac1{T_o}\int_0^{T_o}v_o^2(t)\,dt}.\]For an ideal diode, no filter, and a purely resistive load $R_L$, direct integration gives:
| Rectifier | $V_{dc}$ | $V_{rms}$ | $I_{dc}$ | $I_{rms}$ |
|---|---|---|---|---|
| Half wave | $V_m/\pi$ | $V_m/2$ | $V_m/(\pi R_L)$ | $V_m/(2R_L)$ |
| Full wave | $2V_m/\pi$ | $V_m/\sqrt2$ | $2V_m/(\pi R_L)$ | $V_m/(\sqrt2R_L)$ |
The full-wave $V_m$ is the peak presented to the load: one half-secondary peak for a center-tapped circuit and the complete-secondary peak for a bridge. Constant diode drops reduce all output values and change the conduction interval, so the table is not exact for a real diode.
Ripple factor and rectification efficiency
The RMS value contains dc and ac components:
\[V_{rms}^2=V_{dc}^2+V_{ac,rms}^2.\]Hence the ripple factor is
\[\boxed{r=\frac{V_{ac,rms}}{V_{dc}} =\sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^2-1}}.\]For the unfiltered ideal waveforms,
\[r_{\rm half}=\sqrt{\frac{\pi^2}{4}-1}=1.211, \qquad r_{\rm full}=\sqrt{\frac{\pi^2}{8}-1}=0.483.\]Rectification efficiency is the dc load power divided by the ac power entering the rectifying path:
\[\eta_r=\frac{P_{dc}}{P_{ac}}.\]If diode resistance and transformer losses are neglected,
\[\boxed{\eta_{r,\rm half}=\frac4{\pi^2}=0.405}, \qquad \boxed{\eta_{r,\rm full}=\frac8{\pi^2}=0.811}.\]These are limiting values, not percentages of voltage converted.
Voltage regulation
Rectifier output falls when load current increases because of transformer, diode, and source resistances. A standard measure is
\[\boxed{\%\text{ regulation} =\frac{V_{NL}-V_{FL}}{V_{FL}}\times100},\]where $V_{NL}$ and $V_{FL}$ are the no-load and specified full-load dc voltages. Smaller positive regulation is better. This quantity describes load dependence; it is distinct from the operation of a voltage-regulator circuit.
Filters
A shunt capacitor charges near each rectified peak and discharges through $R_L$ between peaks. If the ripple is small, load current is approximately constant, and diode conduction time is short,
\[V_{r,pp}\simeq\frac{I_L}{f_rC}, \qquad V_{r,rms}\simeq\frac{V_{r,pp}}{2\sqrt3},\]where $f_r=f$ for half-wave and $f_r=2f$ for full-wave rectification. With $I_L\simeq V_{dc}/R_L$,
\[\boxed{r\simeq\frac1{2\sqrt3\,f_rR_LC}}.\]$R_LC$ has units seconds, so the expression is dimensionless. The approximation fails for large ripple, rapidly changing load, or significant source resistance.
A series inductor has reactance $X_L=2\pi f_rL$: it passes dc but opposes ripple current. A shunt capacitor has $X_C=1/(2\pi f_rC)$ and bypasses ripple across the load. LC and $\pi$ filters combine these actions; their useful smoothing requires $X_L$ large and $X_C$ small at the ripple frequency while dc voltage and component current ratings remain adequate.
Solved Problems
1. Complete ideal full-wave output calculation
An ideal full-wave rectifier supplies $R_L=500\,\Omega$ from a sinusoidal load peak $V_m=18.0\,\mathrm V$.
\[V_{dc}=\frac{2V_m}{\pi}=\frac{36.0}{\pi}=11.46\,\mathrm V, \qquad V_{rms}=\frac{18.0}{\sqrt2}=12.73\,\mathrm V.\]Therefore
\[I_{dc}=22.92\,\mathrm{mA}, \qquad I_{rms}=25.46\,\mathrm{mA}.\]The output powers are
\[P_{dc}=\frac{V_{dc}^2}{R_L}=0.2626\,\mathrm W, \qquad P_{ac}=\frac{V_{rms}^2}{R_L}=0.3240\,\mathrm W.\]Hence $\eta_r=P_{dc}/P_{ac}=8/\pi^2=0.8106$ exactly within the stated ideal model. Units and the inequality $P_{dc}<P_{ac}$ provide useful checks.
2. Capacitor-filter ripple
A full-wave rectifier operates from $50.0\,\mathrm{Hz}$ mains, so $f_r=100\,\mathrm{Hz}$. With $I_L=40.0\,\mathrm{mA}$ and $C=1000\,\mu\mathrm F$,
\[V_{r,pp}\simeq\frac{0.0400}{(100)(1000\times10^{-6})}=0.400\,\mathrm V,\] \[V_{r,rms}\simeq\frac{0.400}{2\sqrt3}=0.1155\,\mathrm V.\]If $V_{dc}=12.0\,\mathrm V$, then $r=0.1155/12.0=0.00962$, or $0.962\%$. The small ratio supports the approximation used.
3. Load regulation
A supply gives $V_{NL}=13.2\,\mathrm V$ with no load and $V_{FL}=12.0\,\mathrm V$ at rated current. Thus
\[\%\text{ regulation} =\frac{13.2-12.0}{12.0}\times100=10.0\%.\]The positive sign means the terminal voltage falls as load is applied; a smaller value would indicate a stiffer source.
Descriptive Questions
- Derive the average and RMS output values of ideal half-wave and full-wave rectifiers.
- Derive ripple factor from the separation of dc and ac RMS components.
- Define rectification efficiency and voltage regulation, explaining why they measure different properties.
- Explain the operation and design conditions of capacitor, inductor, LC, and $\pi$ filters.
Numerical Problems
-
A rectifier supply reads $15.0\,\mathrm V$ at no load and $13.5\,\mathrm V$ while delivering $0.300\,\mathrm A$. Model its droop by a constant series resistance. Infer that resistance, predict the output at $0.200\,\mathrm A$, and find the load regulation at the new operating point.
Final answer: $R_{int}=(15.0-13.5)/0.300=5.00\,\Omega$. At $0.200\,\mathrm A$, $V_o=15.0-(0.200)(5.00)=14.0\,\mathrm V$, giving regulation $[(15.0-14.0)/14.0]\times100=7.143\%$.
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A rectifier output has dc component $V_{dc}=12.0\,\mathrm V$ and ripple factor $r=4.00\%$ across $R_L=600\,\Omega$. Without assuming a half-wave or full-wave shape, find the RMS ripple voltage, total RMS output voltage, dc load power, and ac ripple power.
Final answer: $V_{ac,rms}=rV_{dc}=0.480\,\mathrm V$, $V_{rms}=\sqrt{V_{dc}^2+V_{ac,rms}^2}=12.010\,\mathrm V$, $P_{dc}=V_{dc}^2/R_L=0.240\,\mathrm W$, and $P_{ac}=V_{ac,rms}^2/R_L=0.384\,\mathrm{mW}$.
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A capacitor-input supply operates from $50.0\,\mathrm{Hz}$ mains and gives $V_{dc}=12.0\,\mathrm V$ across $R_L=600\,\Omega$. With $C=1000\,\mu\mathrm F$, the measured peak-to-peak ripple is $0.200\,\mathrm V$. Use the small-ripple model to infer the ripple frequency, identify whether the rectification is half-wave or full-wave, and find the ripple factor.
Final answer: $I_L=V_{dc}/R_L=20.0\,\mathrm{mA}$ and $f_r=I_L/(CV_{r,pp})=100\,\mathrm{Hz}$, so the circuit is full-wave. Also $V_{r,rms}=0.200/(2\sqrt3)=57.7\,\mathrm{mV}$ and $r=V_{r,rms}/V_{dc}=0.00481$, or $0.481\%$.
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At $f_r=100\,\mathrm{Hz}$, find $X_L$ for $L=2.00\,\mathrm H$ and $X_C$ for $C=47.0\,\mu\mathrm F$.
Final answer: $X_L=2\pi f_rL=1.257\,\mathrm{k\Omega}$ and $X_C=1/(2\pi f_rC)=33.86\,\Omega$, giving large series and small shunt ripple impedances.
The waveform integrals, ripple and efficiency identities, filter estimates, and all numerical values are reproducible in rectifier-output-filter-check.mac; every printed residual is zero.
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