Rectifier Filters and Ripple Analysis

About this problem set

Worked problems on smoothing rectifier output: capacitor-input ripple voltage and ripple factor, capacitor sizing to a ripple specification, percentage regulation of a filtered supply, choke-input filters and critical inductance, a head-to-head comparison of the two filter types, and pi-filter ripple attenuation.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

PROBLEM 01

Capacitor-Input Filter Ripple Voltage

Problem Statement

A full-wave bridge rectifier delivers a peak load voltage of \( 20~\text{V} \) (diode drops already subtracted) from a \( 50~\text{Hz} \) supply. A \( 470~\mu\text{F} \) smoothing capacitor is connected across \( R_L = 500~\Omega \).

Find the DC output voltage and the load current.
Find the peak-to-peak ripple voltage and the rms ripple.
Find the ripple factor and compare it with the unfiltered value.

Solution

\( V_m = 20~\text{V} \) (peak at the capacitor), \( R_L = 500~\Omega \), \( C = 470~\mu\text{F} \)
Full wave from \( 50~\text{Hz} \) \( \Rightarrow \) ripple frequency \( f_r = 2f = 100~\text{Hz} \)
Assume the capacitor discharges linearly (ripple \( \ll \) DC), i.e. the sawtooth approximation

Step 1 — The two governing relations

Over one ripple period the capacitor supplies the load almost alone, so \( V_r = I_{dc}/(f_rC) \); and the DC level sits half a ripple below the peak, \( V_{dc} = V_m - V_r/2 \). Solve the pair simultaneously with \( I_{dc} = V_{dc}/R_L \):

\[\begin{aligned} V_{dc} &= V_m - \frac{V_{dc}}{2f_r C R_L}\;\Rightarrow\;V_{dc} = \frac{V_m}{1 + \dfrac{1}{2fCR_L}} \\ \frac{1}{2fCR_L} &= \frac{1}{2(50)(470\times10^{-6})(500)} = 0.04255 \\ V_{dc} &= \frac{20}{1 + 0.04255} = 19.2~\text{V} \\ I_{dc} &= \frac{V_{dc}}{R_L} = 38.4~\text{mA} \end{aligned}\]

(a) \( V_{dc} = 19.2~\text{V} \), \( I_{dc} = 38.4~\text{mA} \)

Step 2 — Ripple voltage

\[\begin{aligned} V_{r(p\text{-}p)} &= \frac{I_{dc}}{f_r C} = \frac{0.038367}{(100)(470\times10^{-6})} = 0.816~\text{V} \\ V_{r(rms)} &= \frac{V_{r(p\text{-}p)}}{2\sqrt3} = 0.236~\text{V} \end{aligned}\]

The \( 2\sqrt3 \) converts the peak-to-peak of a triangular (sawtooth) ripple to rms.

(b) \( V_{r(p\text{-}p)} = 0.816~\text{V} \), \( V_{r(rms)} = 0.236~\text{V} \)

Step 3 — Ripple factor

\[\begin{aligned} r &= \frac{V_{r(rms)}}{V_{dc}} = \frac{1}{4\sqrt3\,fCR_L} \\ &= \frac{1}{4(1.732)(50)(470\times10^{-6})(500)} = 0.01228 = 1.23\% \end{aligned}\]

Direct check: \( 0.236/19.2 = 0.01228 \). Without the capacitor the same rectifier would give \( V_{dc} = 2V_m/\pi = 12.7~\text{V} \) with \( r = 0.483 \).

(c) \( r = 1.23\% \) — the capacitor cuts the ripple by a factor of about 39 and lifts the DC output by \( 6.45~\text{V} \).

Note \( f \) in \( r = 1/(4\sqrt3 fCR_L) \) is the supply frequency: the factor 4 already contains the doubling to \( 100~\text{Hz} \). For a half-wave circuit the same formula carries a 2 instead of a 4.

PROBLEM 02

Sizing the Smoothing Capacitor

Problem Statement

Design. A bridge rectifier with a capacitor-input filter must supply \( 12~\text{V} \) DC at \( 200~\text{mA} \) from the \( 50~\text{Hz} \) mains, with a ripple factor not exceeding \( 1\% \).

Find the minimum smoothing capacitance and choose a standard value.
With that standard value, find the actual ripple factor and peak-to-peak ripple.
Find the required peak and rms secondary voltage of the transformer.

Solution

\( V_{dc} = 12~\text{V} \), \( I_{dc} = 200~\text{mA} \Rightarrow R_L = V_{dc}/I_{dc} = 60~\Omega \)
\( r \le 0.01 \), \( f = 50~\text{Hz} \), full wave \( \Rightarrow f_r = 100~\text{Hz} \)
Silicon bridge: \( 2V_D = 1.4~\text{V} \) lost in the conduction path

Step 1 — Solve the ripple-factor formula for C

\[\begin{aligned} r &= \frac{1}{4\sqrt3\,fCR_L}\;\Longrightarrow\; C = \frac{1}{4\sqrt3\,f\,r\,R_L} \\ C &= \frac{1}{4(1.732)(50)(0.01)(60)} = 4810~\mu\text{F} \end{aligned}\]

(a) \( C_{min} = 4810~\mu\text{F} \) → choose the next standard size up, \( C = 6800~\mu\text{F} \), \( 25~\text{V} \) electrolytic. (A \( 4700~\mu\text{F} \) part would miss the specification.)

Step 2 — Performance with the chosen capacitor

\[\begin{aligned} r &= \frac{1}{4\sqrt3 (50)(6800\times10^{-6})(60)} = 0.007075 = 0.708\% \\ V_{r(p\text{-}p)} &= \frac{I_{dc}}{2fC} = \frac{0.2}{2(50)(6800\times10^{-6})} = 0.294~\text{V} \end{aligned}\]

The \( 1\% \) specification alone would have allowed \( V_{r(p\text{-}p)} = 2\sqrt3\,rV_{dc} = 0.416~\text{V} \); the standard capacitor beats it.

(b) \( r = 0.708\% \), \( V_{r(p\text{-}p)} = 0.294~\text{V} \)

Step 3 — Transformer secondary

The DC level sits half a ripple below the peak of the capacitor voltage, and the secondary peak must also cover the two diode drops.

\[\begin{aligned} V_{m,\text{load}} &= V_{dc} + \frac{V_{r(p\text{-}p)}}{2} = 12 + 0.147 = 12.1~\text{V} \\ V_{m,\text{sec}} &= 12.1 + 1.4 = 13.5~\text{V} \\ V_{s} &= \frac{13.5}{\sqrt2} = 9.58~\text{V (rms)} \end{aligned}\]

Standard secondaries bracket this value, so check both:

Secondary (rms)	\( V_m \)	Resulting \( V_{dc} \)	Verdict
\( 9~\text{V} \)	\( 12.7~\text{V} \)	\( 11.2~\text{V} \)	too low
\( 10~\text{V} \)	\( 14.1~\text{V} \)	\( 12.6~\text{V} \)	usable margin
\( 12~\text{V} \)	\( 17~\text{V} \)	\( 15.4~\text{V} \)	\( 28.5\% \) high

(c) Calculated secondary \( = 9.58~\text{V} \) rms → specify a \( 10~\text{V} \), \( 500~\text{mA} \) secondary, which gives \( 12.6~\text{V} \) — a small margin to absorb the winding drop at full load.

Discharge time constant \( \tau = R_LC = 60 \times 6800~\mu\text{F} = 408~\text{ms} \), which is 40.8 ripple periods — the linear-discharge assumption used throughout is well satisfied.

PROBLEM 03

Percentage Regulation of a Filtered Supply

Problem Statement

A capacitor-filtered bridge supply measures \( 18.5~\text{V} \) on open circuit and \( 16.2~\text{V} \) when delivering \( 300~\text{mA} \). The smoothing capacitor is \( 4700~\mu\text{F} \) and the mains frequency is \( 50~\text{Hz} \).

Find the percentage voltage regulation and the equivalent output resistance.
How much of that output resistance is contributed by the capacitor itself?
Predict the full-load voltage and regulation if \( C \) is raised to \( 10\,000~\mu\text{F} \).

Solution

\( V_{NL} = 18.5~\text{V} \), \( V_{FL} = 16.2~\text{V} \) at \( I_{FL} = 300~\text{mA} \)
\( C = 4700~\mu\text{F} \), \( f = 50~\text{Hz} \Rightarrow f_r = 100~\text{Hz} \)

Step 1 — Regulation and output resistance

\[\begin{aligned} \%\text{Reg} &= \frac{V_{NL} - V_{FL}}{V_{FL}}\times100 = \frac{18.5 - 16.2}{16.2}\times100 = 14.2\% \\ R_{out} &= \frac{\Delta V}{\Delta I} = \frac{2.3}{0.3} = 7.67~\Omega \end{aligned}\]

(a) \( \%\text{Reg} = 14.2\% \), \( R_{out} = 7.67~\Omega \)

Step 2 — The capacitor's share

From \( V_{dc} = V_m - I_{dc}/(2fC) \), the filter droops linearly with load current, so it behaves as a resistance \( 1/(2fC) \) in series with the ideal peak source.

\[\begin{aligned} R_{C} &= \frac{1}{2fC} = \frac{1}{2(50)(4700\times10^{-6})} = 2.13~\Omega \\ R_{\text{winding + diodes}} &= R_{out} - R_C = 7.67 - 2.13 = 5.54~\Omega \end{aligned}\]

(b) The capacitor accounts for \( 2.13~\Omega \), i.e. \( 27.8\% \) of the droop; the rest is transformer and diode resistance.

Step 3 — Effect of a larger capacitor

\[\begin{aligned} R_C' &= \frac{1}{2(50)(10000\times10^{-6})} = 1~\Omega \\ R_{out}' &= 5.54 + 1 = 6.54~\Omega \\ V_{FL}' &= 18.5 - (6.54)(0.3) = 16.5~\text{V} \\ \%\text{Reg}' &= \frac{18.5 - 16.5}{16.5}\times100 = 11.9\% \end{aligned}\]

More capacitance improves regulation only up to the floor set by the winding and diode resistance — here regulation cannot fall below \( 9.87\% \) no matter how large \( C \) is made. Beyond that point a series regulator, not a bigger capacitor, is the answer.

PROBLEM 04

Choke-Input Filter and Critical Inductance

Problem Statement

A full-wave rectifier with a peak output of \( 25~\text{V} \) feeds an L-section (choke-input) filter with \( L = 5~\text{H} \) and \( C = 100~\mu\text{F} \) into \( R_L = 200~\Omega \). The supply is \( 50~\text{Hz} \).

Find the critical inductance and check that the choke is large enough.
Find the DC output voltage and the ripple factor.
What is the largest \( R_L \) (lightest load) for which this choke still maintains continuous current?

Solution

\( V_m = 25~\text{V} \), \( R_L = 200~\Omega \), \( L = 5~\text{H} \), \( C = 100~\mu\text{F} \)
\( f = 50~\text{Hz} \Rightarrow \omega = 2\pi(50) = 314.16~\text{rad/s} \); ripple component at \( 2\omega \)
Choke resistance neglected; only the dominant second-harmonic ripple is retained

Step 1 — Critical inductance

Current through the choke stays continuous provided the peak ripple current never exceeds the DC current, which gives the standard condition:

\[\begin{aligned} L_c &= \frac{R_L}{3\omega} = \frac{200}{3(314.16)} = 0.2122~\text{H} = 212~\text{mH} \end{aligned}\]

(a) \( L_c = 0.212~\text{H} \); the \( 5~\text{H} \) choke is 23.6 times larger, so conduction is comfortably continuous.

Step 2 — Output voltage and ripple

With continuous conduction the choke holds the current constant, so the DC output is simply the average of the rectified wave — the capacitor cannot charge to the peak as it does in a capacitor-input filter.

\[\begin{aligned} V_{dc} &= \frac{2V_m}{\pi} = \frac{2(25)}{\pi} = 15.9~\text{V} \end{aligned}\]

The ripple divider is \( X_C \) against \( X_L \) at the ripple frequency \( 2\omega \); the second-harmonic term of a full-wave wave has rms value \( (\sqrt2/3)V_{dc} \):

\[\begin{aligned} X_L &= 2\omega L = 2(314.16)(5) = 3140~\Omega \\ X_C &= \frac{1}{2\omega C} = \frac{1}{2(314.16)(100\times10^{-6})} = 15.92~\Omega \\ r &= \frac{\sqrt2}{3}\cdot\frac{X_C}{X_L} = \frac{\sqrt2}{12\,\omega^2LC} \\ &= \frac{\sqrt2}{12(314.16)^2(5)(100\times10^{-6})} = 0.002388 = 0.239\% \end{aligned}\]

(b) \( V_{dc} = 15.9~\text{V} \), \( r = 0.239\% \) i.e. \( V_{r(rms)} = 38~\text{mV} \)

Step 3 — Lightest permissible load

Rearranging \( L \ge R_L/3\omega \) for the load:

\[\begin{aligned} R_{L(max)} &= 3\omega L = 3(314.16)(5) = 4712~\Omega \end{aligned}\]

(c) \( R_{L(max)} = 4.712~\text{k}\Omega \) — above this the output jumps towards \( V_m \) and the ripple worsens sharply.

This is why choke-input supplies carry a bleeder resistor: it guarantees a minimum load so the choke never falls below its critical inductance.

PROBLEM 05

Capacitor-Input Versus Choke-Input Filter

Problem Statement

A full-wave rectifier producing a \( 25~\text{V} \) peak at \( 50~\text{Hz} \) is to feed \( R_L = 200~\Omega \). Two filters are considered:

Filter A — a single shunt capacitor \( C = 470~\mu\text{F} \).
Filter B — an L-section choke input, \( L = 5~\text{H} \) followed by \( C = 100~\mu\text{F} \).

Compute \( V_{dc} \) and the ripple factor for each.
The load is now increased to draw four times the current (\( R_L = 50~\Omega \)). What happens to each filter?
Which filter would you specify, and why?

Solution

\( V_m = 25~\text{V} \), \( f = 50~\text{Hz} \), \( \omega = 314.16~\text{rad/s} \)
A: \( C = 470~\mu\text{F} \). B: \( L = 5~\text{H} \), \( C = 100~\mu\text{F} \)

Step 1 — Filter A (capacitor input)

\[\begin{aligned} V_{dc} &= \frac{V_m}{1 + 1/(2fCR_L)} = \frac{25}{1 + 0.1064} = 22.6~\text{V} \\ I_{dc} &= 113~\text{mA},\qquad V_{r(p\text{-}p)} = \frac{I_{dc}}{2fC} = 2.4~\text{V} \\ r &= \frac{1}{4\sqrt3 fCR_L} = 0.03071 = 3.07\% \end{aligned}\]

Step 2 — Filter B (choke input)

\[\begin{aligned} V_{dc} &= \frac{2V_m}{\pi} = 15.9~\text{V} \quad(\text{load independent}) \\ I_{dc} &= 79.6~\text{mA} \\ r &= \frac{\sqrt2}{12\omega^2LC} = 0.002388 = 0.239\% \end{aligned}\]

	Filter A (C-input)	Filter B (L-section)
\( V_{dc} \)	\( 22.6~\text{V} \)	\( 15.9~\text{V} \)
Ripple factor	\( 3.07\% \)	\( 0.239\% \)
Output vs load	falls with load	essentially constant
Diode surge current	large	small (choke limits di/dt)
Size / cost	small, cheap	bulky, expensive

(a) A: \( 22.6~\text{V} \), \( r = 3.07\% \). B: \( 15.9~\text{V} \), \( r = 0.239\% \).

Step 3 — Four times the load current

\[\begin{aligned} \text{A: } V_{dc} &= \frac{25}{1 + 1/(2(50)(470\mu)(50))} = 17.5~\text{V},\quad r = 12.3\% \\ \text{B: } V_{dc} &= \frac{2V_m}{\pi} = 15.9~\text{V},\quad r = 0.239\%\;\text{(unchanged)} \end{aligned}\]

Filter B is unaffected because \( R_L = 50~\Omega \) is still far below \( 3\omega L = 4.712~\text{k}\Omega \), so conduction stays continuous.

(b) Filter A droops by \( 22.4\% \) and its ripple worsens to \( 12.3\% \); Filter B holds \( 15.9~\text{V} \) and \( 0.239\% \) unchanged.

(c) For a light, fixed load choose A — it gives \( 6.68~\text{V} \) more output from a cheap component. For a heavy or varying load choose B, whose regulation and ripple are almost independent of load current.

The trade is fundamental: a capacitor-input filter charges to the peak but only while the load is light, whereas a choke-input filter fixes the output at the average and holds it there.

PROBLEM 06

Ripple Attenuation of a Pi Filter

Problem Statement

A \( \pi \) (CLC) filter follows a full-wave rectifier whose peak output is \( 30~\text{V} \) at \( 50~\text{Hz} \). The filter uses \( C_1 = C_2 = 100~\mu\text{F} \) and \( L = 10~\text{H} \), feeding \( R_L = 500~\Omega \).

Find the reactances at the ripple frequency.
Find the ripple factor of the complete filter.
By what factor has the \( L \)-\( C_2 \) section improved on \( C_1 \) alone?

Solution

\( C_1 = C_2 = 100~\mu\text{F} \), \( L = 10~\text{H} \), \( R_L = 500~\Omega \)
\( f = 50~\text{Hz} \Rightarrow \omega = 314.16~\text{rad/s} \), ripple at \( 2\omega \)
Assume \( X_{C_2} \ll R_L \) so the output section is a pure reactive divider

Step 1 — Reactances at 100 Hz

\[\begin{aligned} X_{C_1} = X_{C_2} &= \frac{1}{2\omega C} = \frac{1}{2(314.16)(100\times10^{-6})} = 15.92~\Omega \\ X_L &= 2\omega L = 2(314.16)(10) = 6.283~\text{k}\Omega \end{aligned}\]

Check the assumption: \( X_{C_2} = 15.9~\Omega \ll R_L = 500~\Omega \) ✓, and \( X_L = 6.28~\text{k}\Omega \gg X_{C_2} \) ✓.

(a) \( X_{C_1} = X_{C_2} = 15.9~\Omega \), \( X_L = 6.28~\text{k}\Omega \)

Step 2 — Ripple factor

\( C_1 \) sets the ripple entering the filter (a factor \( \sqrt2 X_{C_1}/R_L \)); the \( L \)-\( C_2 \) section then divides it again by \( X_{C_2}/X_L \):

\[\begin{aligned} r &= \sqrt2\;\frac{X_{C_1}}{R_L}\cdot\frac{X_{C_2}}{X_L} = \frac{\sqrt2\,X_{C_1}X_{C_2}}{R_L X_L} = \frac{\sqrt2}{8\omega^3C_1C_2LR_L} \\ &= \frac{\sqrt2 (15.92)(15.92)}{(500)(6283.2)} = 0.000114 = 0.0114\% \end{aligned}\]

With \( V_{dc} \approx 25~\text{V} \) at the input capacitor this is only \( V_{r(rms)} = 2.85~\text{mV} \) of ripple.

(b) \( r = 0.000114 \), i.e. \( 0.0114\% \)

Step 3 — Improvement over a single capacitor

\[\begin{aligned} r_{C_1\text{ alone}} &= \frac{1}{4\sqrt3 fC_1R_L} = 0.05774 = 5.77\% \\ \text{improvement} &= \frac{0.05774}{0.000114} = 506\times \end{aligned}\]

The \( \pi \) filter keeps the capacitor-input filter's high DC output (\( \approx V_m \)) while buying choke-like smoothing — but it inherits the poor regulation of \( C_1 \), because the choke is no longer facing the rectifier.