Filters and Ripple

1 Why the Rectifier Output Is Unusable

Chapter 9 ended with a bridge delivering 20.7 V DC accompanied by 10.0 V RMS of alternating content. Put numbers on what that means for a load. The output falls to zero twice in every mains cycle, so a circuit powered from it is completely unpowered for a substantial fraction of every 10 ms. A 5 V logic circuit will not merely misbehave; it will restart. An audio amplifier will reproduce the 100 Hz ripple and its harmonics directly at its output, and since the ear is sensitive at those frequencies, a few millivolts of supply ripple is audible as hum.

The specification a filter must meet is therefore usually written as a ripple factor or as a millivolt figure. Typical requirements are:

Going from 0.483 to 0.01 is an attenuation of about 34 dB, and going to 0.00003 is 84 dB. No single passive component achieves the latter; in practice a filter takes the ripple down to a per cent or so and a regulator — the subject of Chapters 12 and 30 — removes the rest, since a regulator's ripple rejection is typically 60 to 80 dB.

Two properties of the rectifier output determine how hard the filtering is. The first is the ripple factor itself, 1.211 for half-wave and 0.483 for full-wave. The second, and more important, is the ripple frequency. The half-wave output contains a component at the mains frequency of amplitude \(1.11V_{dc}\); the full-wave output contains nothing at all at 50 Hz, and its largest component is at 100 Hz with an RMS value of \(0.471V_{dc}\). Since the reactance of a capacitor halves when the frequency doubles, a given capacitor is twice as effective on the full-wave waveform for that reason alone, on top of the 2.5:1 improvement in ripple factor. The two effects multiply, and a full-wave rectifier with a given capacitor gives about five times less ripple than a half-wave one.

All the filters in this chapter are low-pass networks whose job is to pass DC and block 100 Hz. That framing makes the design goals obvious: the series element should have low resistance and high reactance at 100 Hz — an inductor; the shunt element should have high resistance and low reactance at 100 Hz — a capacitor. Everything that follows is a consequence of arranging those two components in the three possible ways.

2 The Capacitor-Input Filter and the Ripple Derivation

Connect a large capacitor directly across the rectifier output, in parallel with the load. Figure 10.1 shows the circuit and, below it, what happens.

Follow one ripple cycle. As the rectifier output rises towards a peak it exceeds the voltage already on the capacitor, the diodes conduct, and the capacitor charges — rapidly, because the only thing limiting the current is the small series resistance \(r_s\) of the transformer and diodes. At the peak, the rectifier output starts to fall. The capacitor cannot fall as fast, because it can only discharge through \(R_L\); so within a fraction of a degree the diodes become reverse biased and switch off, disconnecting the capacitor from the source entirely. From that moment the capacitor alone supplies the load, decaying with time constant \(R_LC\), until the next peak of the rectified waveform rises above it and the cycle repeats.

The derivation of the ripple follows from that description, using two approximations that are excellent for any practical filter.

Approximation 1: the discharge lasts the whole ripple period. In reality the diodes conduct for a short interval \(\theta\) and the capacitor discharges for the rest; taking the discharge time as the full period \(T_r = 1/f_r\) overestimates it slightly and so overestimates the ripple, which is the safe direction to err.

Approximation 2: the discharge is linear. The capacitor discharges exponentially, \(v = V_m e^{-t/R_LC}\), but if \(R_LC \gg T_r\) — which is the condition for the filter to work at all — only the first part of the exponential is used and it is nearly a straight line. Equivalently, the load draws a nearly constant current \(I_{dc}\).

With a constant current \(I_{dc}\) flowing out of the capacitor for a time \(T_r\), the charge removed is \(Q = I_{dc}T_r\), and the voltage falls by \(Q/C\):

The central result. \(f_r\) is the ripple frequency: 50 Hz for a half-wave rectifier on a 50 Hz mains, 100 Hz for a full-wave one. Equivalently \(V_{r(pp)} = V_{dc}/(f_rCR_L)\).

To convert this into a ripple factor, note that the waveform is very nearly a sawtooth, and the RMS value of a triangular wave is its peak-to-peak value divided by \(2\sqrt3\):

For a full-wave rectifier on 50 Hz mains, \(2\sqrt3\times100 = 346.4\), so with \(C\) in microfarads and \(R_L\) in ohms, \(\gamma = 2887/(CR_L)\). For half-wave the constant doubles to 5774; for a 60 Hz mains it becomes 2406, which is the familiar "2410" of American textbooks.

The DC output is the mean of a waveform running from \(V_m\) down to \(V_m - V_{r(pp)}\), so

Note what this says: with a capacitor fitted, the DC output climbs from \(0.637V_m\) to nearly \(V_m\) — a 57 % increase for free. That, quite as much as the ripple reduction, is why capacitor-input filters are universal.

3 Exact and Approximate Treatments

It is worth knowing how good the two approximations are, because the answer changes with the design and because the approximate formula is used to choose components that the exact behaviour then has to justify.

The exact discharge is \(v(t) = V_{m}e^{-t/R_LC}\) over the non-conducting interval. Expanding the exponential,

The linear approximation keeps only the first term, so it overestimates the ripple by about \(T_d/2R_LC\), which is half the fractional ripple. For a 1 % design the error from this source is 0.5 %, entirely negligible.

The second approximation is the more significant one. Taking the discharge time \(T_d\) as the whole ripple period ignores the conduction interval, and the error is exactly the conduction duty cycle. Compare the formula with a numerical solution of the full circuit for the design used later in this chapter — a bridge from a 17 V RMS secondary with \(r_s = 0.5\ \Omega\), \(C = 10\,000\) µF and \(R_L = 40\ \Omega\):

The ripple formula is 23 % pessimistic, which is exactly the conduction duty cycle it neglected — and being pessimistic, it is safe to design with. The \(V_{dc}\) formula is 6.8 % optimistic, and that error is not safe: it comes from ignoring \(r_s\) entirely. Because the charging current flows in short tall pulses, its RMS value is 1.18 A for a load of only 0.52 A, and \(I_{rms}^2r_s\) is a real loss. The rule that follows is worth stating plainly: use the approximate formula to choose \(C\), and never to predict the output voltage. Predict the output voltage from the transformer's regulation figure and a simulation or a measurement, and always leave headroom.

4 Conduction Angle and the Diode Current Spike

The diodes conduct only while the rectified sinusoid is above the capacitor voltage. Taking the peak as the origin, the rectifier output during the relevant interval is \(V_{pk}\cos\theta\), and conduction begins when this rises through the minimum capacitor voltage \(V_{pk} - V_{r(pp)}\):

The conduction interval runs from \(-\theta_c\) to 0 in each ripple half-cycle, so the fraction of the period spent conducting is \(\theta_c/\pi\). For \(V_{pk} = 22.64\) V and \(V_{r(pp)} = 0.5\) V this gives \(\theta_c = 0.2105\) rad \(= 12.06^\circ\), a conduction fraction of 6.7 %.

Now apply charge conservation. Over one ripple period the load removes \(Q = I_{dc}T_r\) coulombs, and every one of them must be replaced during the conduction interval. If the interval is a fraction \(k\) of the period, the average current during conduction must be \(I_{dc}/k\). For the numbers above, \(k = 0.067\) and \(I_{dc} = 0.5\) A, giving an average of 7.5 A during conduction — and since the pulse is roughly triangular, a peak of nearly twice that.

The formal result comes from differentiating the capacitor voltage. During conduction the capacitor is following the source, \(v = V_{pk}\cos\omega t\), so the capacitor current is \(i_C = C\,dv/dt = -\omega CV_{pk}\sin\omega t\), largest in magnitude at the start of conduction:

For the design above, \(\omega C V_{pk}\sin\theta_c = 314.2\times0.01\times22.64\times\sin(0.2105) = 14.87\) A, giving \(I_{D(pk)} = 15.37\) A — thirty times the DC load current.

That figure is the answer for an ideal source with no series resistance. A real supply has \(r_s\), and it changes the picture completely, because the resistance limits the charging current and thereby forces the conduction interval to widen until enough charge has been transferred. Numerically solving the same circuit with \(r_s = 0.5\ \Omega\):

Series resistance is therefore not simply a loss: it is what keeps a capacitor-input supply's diode currents finite. A perfect transformer would destroy the rectifier. This is the same insight as the surge limiter of Chapter 9, applied to the steady state rather than to switch-on, and it is why designers of high-current supplies sometimes deliberately add a fraction of an ohm.

The practical consequences are three. The diode must be selected on its repetitive peak and RMS ratings, not merely on \(I_{F(AV)}\); for the design above the ratios are \(I_{pk}/I_{dc} = 6.3\) and \(I_{rms}/I_{dc} = 2.2\), and a "1 A" rectifier is not adequate for 0.52 A of DC. The capacitor must be rated for that RMS ripple current, which heats it internally through its equivalent series resistance and is the dominant cause of electrolytic failure. And the transformer must be specified for an RMS secondary current more than twice the DC output, which is why a transformer that comfortably drives a resistive load may overheat driving a capacitor-input supply of the same DC rating.

5 The Choke-Input Filter and Critical Inductance

Put an inductor in series before the capacitor and the behaviour changes fundamentally. An inductor opposes changes of current, so if it is large enough the current through it never falls to zero: the rectifier conducts continuously, one diode pair handing over to the other at each crossing, and the tall charging pulses disappear.

Because the input current is continuous, the inductor sees the rectifier's average voltage rather than its peak, and the output is

Against \(V_{dc}\approx V_m\) for a capacitor input. From the same 17 V secondary the choke-input filter gives 14.4 V where the capacitor gives 21 V — a 32 % penalty in output voltage, paid for a large improvement in regulation and in diode stress.

Critical inductance. Continuous conduction is not automatic. The rectifier output is \(V_{dc}\) plus a ripple whose dominant component is the second harmonic, of amplitude \(4V_m/3\pi = \tfrac{2}{3}V_{dc}\). That ripple voltage drives an alternating current through the inductor of peak value \(\tfrac{2}{3}V_{dc}/(2\omega L)\), and the current stays positive only if that peak does not exceed the DC current \(V_{dc}/R_L\):

The critical inductance. With \(\omega = 2\pi\times50 = 314.16\) rad/s, \(3\omega = 942.5\), so \(L_c = R_L/942.5\) at 50 Hz. (On a 60 Hz mains the divisor is 1131, which is the familiar \(R/1130\) of American texts.) For \(R_L = 40\ \Omega\), \(L_c = 42.4\) mH.

The condition depends on \(R_L\), and \(R_L\) rises without limit as the load is removed. At no load, \(L_c\) is infinite and no practical inductor satisfies it: conduction becomes discontinuous, the filter reverts to capacitor-input behaviour, and the output rises from \(0.637V_m\) towards \(V_m\). Figure 10.3 shows this as the steep rise at the left of the choke-input curve. Two remedies exist, and both are used:

• A bleeder resistor permanently across the output, drawing enough current to keep the effective \(R_L\) below \(3\omega L\). With \(L = 100\) mH, \(3\omega L = 94.2\ \Omega\), so a bleeder of 94 \(\Omega\) or less holds the supply in continuous conduction at any load. The bleeder also discharges the reservoir capacitor when the equipment is switched off, which is a safety requirement in its own right in high-voltage supplies.
• A swinging choke, deliberately designed so that its core saturates partially at high current. Its inductance is large at light load, where \(L_c\) is large, and falls at heavy load, where a large inductance is not needed. It is a neat solution to a condition that scales the wrong way, and it was standard in valve equipment.

Ripple factor. Above the critical inductance the filter is a straightforward reactive divider at the ripple frequency \(2\omega\). The inductor's reactance \(X_L = 2\omega L\) is placed in series and the capacitor's \(X_C = 1/2\omega C\) in shunt, and since the capacitor's reactance is far smaller than \(R_L\) the load hardly loads the divider:

The factor 0.4714 is the ripple presented by the rectifier, \(\tfrac{2}{3}\) of \(V_{dc}\) as an amplitude and hence \(\tfrac{2}{3}/\sqrt2\) as an RMS fraction. At 50 Hz, \(4\omega^2 = 3.948\times10^5\), so with \(L\) in henries and \(C\) in microfarads, \(\gamma = 1.19/(LC)\). At 60 Hz the constant is 0.83, which is the value most textbooks print — check which mains frequency a quoted formula assumes.

The most attractive property of the choke-input filter is that \(\gamma\) does not contain \(R_L\) at all. Its ripple is independent of load current, whereas a capacitor-input filter's ripple is directly proportional to it. Combined with the flat \(0.637V_m\) output, that gives the excellent regulation shown in Figure 10.3: from 153 mA to 750 mA the output falls only from 14.36 V to 14.19 V, a regulation of 1.2 %, against 10 % for the capacitor.

6 LC, π and the Choice Between Them

The three arrangements of one inductor and one or two capacitors give three filters with very different characters.

The \(\pi\)-section filter puts a capacitor first, then an inductor, then a second capacitor. It is a capacitor-input filter followed by an L-section, and it inherits the properties of both: the output voltage is close to \(V_m\) as in a capacitor filter, the diode currents are the tall spikes of a capacitor filter, and the regulation is as poor — but the residual ripple is far smaller, because the L-section attenuates what the first capacitor leaves.

Its ripple factor is the capacitor-input result multiplied by the reactive divider ratio:

At 50 Hz with \(L\) in henries, capacitances in microfarads and \(R_L\) in ohms, \(\gamma_\pi = 5700/(LC_1C_2R_L)\). On a 60 Hz mains the constant is 3300. The frequently quoted "5700" is therefore the 50 Hz constant, while the equally frequently quoted "0.83/LC" for the L-section is the 60 Hz one — a pair of inconsistencies that has confused generations of students.

The \(\pi\) filter's advantage is dramatic — a factor of 50 better than the capacitor alone in that example — and it was universal in valve equipment, where a 10 H choke and a 40 µF capacitor feeding a 10 k\(\Omega\) load were ordinary. In modern low-voltage supplies chokes have almost vanished, for three good reasons: an inductor able to carry an ampere without saturating is large, heavy and expensive; its winding resistance introduces the resistive droop the choke was supposed to avoid; and a three-terminal regulator costing a fraction as much achieves 60–80 dB of ripple rejection with no magnetics at all. The choke survives where regulators cannot follow: at high voltage, at high current where a regulator's dissipation would be prohibitive, and in switch-mode supplies, where the inductor operates at 100 kHz and is consequently a thousand times smaller.

One cheap substitute is worth knowing. Replacing the inductor of a \(\pi\)-section with a resistor gives the RC \(\pi\) filter, whose ripple factor is \(\gamma_{RC} = \gamma_C\times X_{C2}/R\). It costs DC output voltage in proportion to the load current, so it is used only where the current is small and constant — a preamplifier stage drawing a few milliamperes decoupled from a shared rail, for instance, where 100 \(\Omega\) and 100 µF give 15 dB of extra rejection for pennies.

7 Regulation and a Worked Design

Percentage regulation measures how much the output sags between no load and full load:

A perfect supply has zero regulation. Note the denominator is the full-load value, not the no-load one; both conventions exist and they differ by a few per cent, so state which you are using.

For a capacitor-input filter the sag has two causes. The ripple term \(I_{dc}/2f_rC\) grows in proportion to the load current, and the resistive term grows faster still, because the RMS charging current rises with load and the loss is \(I_{rms}^2r_s\). For a choke-input filter the output is pinned at \(0.637V_m\) and the only sag is resistive, so regulation is far better — provided the load never falls below the critical current.

Two closing observations. First, the design met its ripple specification with a component chosen from a formula that is 23 % pessimistic, so the delivered ripple is 0.59 % rather than 0.72 % — the conservatism is deliberate and worth keeping. Second, if the 7.9 % regulation is unacceptable there are two routes: increase \(C\), which improves the ripple term but not the resistive one and makes the diode currents worse; or add a regulator, which fixes both and is almost always the right answer. Chapter 12 builds the simplest such regulator from a Zener diode, and Chapter 30 develops the series and integrated-circuit regulators that finish the job.

8 Summary and Key Results

9 Common Mistakes

10 Chapter Review