By the end of this chapter you should be able to:
- Describe the operation of the half-wave rectifier over a complete cycle and sketch the load and diode waveforms.
- Derive \(V_{dc} = V_m/\pi\) and \(V_{rms} = V_m/2\) from the defining integrals, and obtain the corresponding currents.
- Define the ripple factor from first principles and evaluate it as 1.21 for the half-wave circuit.
- Define and compute the form factor (1.571) and peak factor (2), and relate them to the ripple factor.
- Derive the rectification efficiency \(4/\pi^2\) and explain what the missing 59 % becomes.
- Determine the peak inverse voltage and the transformer utilisation factor, and use them to specify components.
- Quantify the effect of the diode's forward drop and bulk resistance, and carry out a complete supply design.
Almost every piece of electronic equipment needs a steady direct voltage, and almost every source of energy that reaches a building is alternating. Something must convert one into the other, and the first stage of that conversion is rectification: forcing current to flow in only one direction. Chapter 6 showed that a p-n junction does exactly that, passing milliamperes forward at 0.7 V and nanoamperes backward at 20 V. This chapter takes one diode, one transformer and one load resistor, and works out precisely what comes out.
The circuit is trivial and the analysis is not, because the output is neither DC nor AC but a train of half-sinusoids — a waveform with a non-zero average and a large alternating component on top of it. Describing such a waveform takes four separate numbers, and the whole of this chapter is about deriving and interpreting them: the average or DC value, the RMS value, the ripple factor and the efficiency with which mains power is converted into useful DC. Two further numbers decide whether the circuit will survive: the peak inverse voltage the diode must block, and the transformer rating the circuit demands.
The half-wave rectifier is, by every one of those measures, a poor circuit. Its ripple factor is 1.21, meaning the alternating component is larger than the DC it is riding on; its theoretical efficiency is 40.5 %; and it uses a transformer at 28.7 % of the rating it forces you to buy. Chapter 9 does better on all three counts. It is worth studying anyway, partly because it is still used wherever the current is small and the cost matters more than the ripple, and mainly because every definition in this chapter is used unchanged for the better circuits that follow.
1 The Circuit and What It Does
Figure 8.1 shows the whole circuit: a transformer, one diode and a load. The transformer is not incidental. It performs two distinct jobs, and a design that omits it must find another way to do both. First, it scales the mains voltage to whatever the load needs, through the turns ratio \(N_1:N_2\); a 230 V mains and a 12 V supply need a ratio of about 19:1. Second, and more importantly, it isolates the output from the mains, so that no part of the low-voltage circuit is connected to the live conductor. A transformerless rectifier connected straight to the mains produces an output whose "ground" may be 325 V from earth, which is why such circuits appear only inside sealed appliances.
Trace one cycle of the secondary voltage \(v_s = V_m\sin\omega t\).
Positive half-cycle (\(0 < \omega t < \pi\)). The anode is positive with respect to the cathode, the diode conducts, and the circuit is a series loop of \(v_s\), the diode and \(R_L\). Neglecting the drop across the diode for the moment, the whole of \(v_s\) appears across the load: \(v_o = V_m\sin\omega t\) and \(i_o = (V_m/R_L)\sin\omega t\).
Negative half-cycle (\(\pi < \omega t < 2\pi\)). The anode is now negative, the diode blocks, and only its reverse saturation current — nanoamperes — flows. With no current there is no drop across \(R_L\), so \(v_o = 0\) and the entire secondary voltage appears across the diode.
The middle trace of Figure 8.2 is the output: a train of positive half-sinusoids separated by flat gaps, repeating at the supply frequency of 50 Hz. This is the key structural fact about the half-wave circuit and the reason it filters so badly — the ripple appears at the mains frequency, and there is a full 10 ms interval in every cycle during which the source delivers nothing at all. Chapter 9's circuits ripple at 100 Hz and never leave the load unsupplied for more than 10 ms; Chapter 10 shows what that does to the capacitor you need.
The bottom trace is the diode voltage, and it repays study because it decides the component you must buy. While the diode conducts, the voltage across it is its forward drop, essentially zero on this scale. While it blocks, no current flows, so no voltage is dropped across \(R_L\), and Kirchhoff's voltage law around the loop puts the whole secondary voltage across the diode. The most negative value it reaches is \(-V_m\), and that is the peak inverse voltage, or PIV:
For an unfiltered load. Add the reservoir capacitor of Chapter 10 and the picture changes completely: the capacitor holds the cathode at \(+V_m\) while the secondary swings to \(-V_m\), so the diode sees \(2V_m\). This is one of the most commonly missed points in supply design, and it is worth writing on the schematic.
2 Average and RMS Values
The output is periodic with period \(T = 2\pi/\omega\), so its DC component is its average over one full period. Only the first half of the period contributes:
This is what a moving-coil meter, which responds to the average, would read. Note the factor \(1/2\pi\) and not \(1/\pi\): the average is taken over the whole cycle, including the half in which nothing happens.
The corresponding DC current follows immediately, since the load is resistive:
The RMS value is the square root of the mean of the square, and it is the quantity that determines heating in a resistor, the rating of a fuse and the loss in a transformer winding:
The \(\cos2\omega t\) term integrates to zero over the half-cycle, which is why the algebra is so short. Compare this with the RMS of a full sinusoid, \(V_m/\sqrt2 = 0.707V_m\): removing half the waveform reduces the RMS by \(\sqrt2\), because it removes half the squared area.
It is worth recording the third descriptor as well. The peak value is \(V_m\) itself, reached once per cycle for an instant. So the same waveform is described by three quite different numbers — peak \(V_m\), RMS \(0.500V_m\), average \(0.318V_m\) — and every specification in a rectifier datasheet names which one it means. A diode's \(I_{F(AV)}\) rating is an average, because it is set by junction heating over many cycles; its \(I_{FSM}\) rating is a peak, because it is set by the fusing of the bond wires; and a transformer's current rating is an RMS, because it is set by copper loss. Quoting the wrong one of the three is the commonest way to destroy a rectifier that appeared to be correctly specified.
These two results are worth holding side by side, because their ratio is where all the remaining figures of merit come from. \(V_{rms}/V_{dc} = (V_m/2)/(V_m/\pi) = \pi/2 = 1.571\). A pure DC waveform would give a ratio of exactly 1, since its RMS and its average coincide. Anything above 1 is a measure of how much alternating content is present, and 1.571 is a great deal.
Put a moving-coil (analogue) multimeter on DC across a half-wave rectifier's output and it reads \(V_m/\pi\), because it averages. Put a true-RMS meter on AC and it reads the alternating part alone, \(\sqrt{V_{rms}^2 - V_{dc}^2}\). Put a cheap averaging AC multimeter on it and it reads nonsense, because such meters assume a sinusoid and apply a fixed 1.11 scaling factor. Half of all confusing bench measurements of rectifier outputs come from this last case.
3 Ripple Factor, Form Factor and Peak Factor
The output can be regarded as a DC component with an unwanted AC component superimposed. Write \(v_o = V_{dc} + v_{ac}\). By the definition of the average, \(v_{ac}\) has zero mean, so the mean square of the total splits cleanly:
The cross term \(2V_{dc}\overline{v_{ac}}\) vanishes because \(\overline{v_{ac}} = 0\). This is the same decomposition used for noise power, and it is the reason RMS values of independent components add in quadrature.
The ripple factor \(\gamma\) is defined as the ratio of the unwanted to the wanted:
Everything now follows from the ratio computed at the end of the last section. Substituting \(V_{rms}/V_{dc} = \pi/2\):
An unfiltered half-wave rectifier delivering 10 V DC also delivers 12.1 V RMS of alternating rubbish. Powering anything sensitive from it directly is out of the question, and the filter of Chapter 10 has to remove a component larger than the signal it is protecting. The full-wave circuits of Chapter 9 reduce this to 0.483, a factor of 2.5 better before any filtering at all.
Two related ratios are quoted alongside it, and the definitions are worth keeping straight because examinations ask for them by name.
The form factor is the ratio of RMS to average:
and the relation \(\gamma = \sqrt{\text{FF}^2-1}\) means it carries exactly the same information. For reference, a pure sinusoid has FF \(= 1.111\), a square wave has FF \(= 1\), and the closer to 1 the better.
The peak factor (or crest factor) is the ratio of peak to RMS:
against \(\sqrt2 = 1.414\) for a sinusoid. A high peak factor is a warning about stress rather than about quality: it says that the components must survive a peak much larger than the heating current suggests. The capacitor-input filter of Chapter 10 pushes the peak factor of the diode current to six or more, and that is what destroys rectifiers.
A half-wave rectifier is fed from a 24 V RMS secondary and feeds a 100 \(\Omega\) load. Neglecting the diode drop, \(V_m = 24\sqrt2 = 33.94\) V. Then \(V_{dc} = 33.94/\pi = 10.80\) V and \(I_{dc} = 108.0\) mA; \(V_{rms} = 33.94/2 = 16.97\) V and \(I_{rms} = 169.7\) mA. The alternating content is \(V_{ac,rms} = \sqrt{16.97^2-10.80^2} = 13.08\) V, so \(\gamma = 13.08/10.80 = 1.211\) as derived. The load dissipates \(I_{rms}^2R_L = (0.1697)^2\times100 = 2.88\) W, of which only \(I_{dc}^2R_L = (0.108)^2\times100 = 1.17\) W is DC power. The remaining 1.71 W is ripple heating, useless and unavoidable in this topology.
4 What the Ripple Is Made Of
The ripple factor says how much unwanted alternating content there is, but not what frequencies it occupies, and a filter designer needs the second answer as much as the first. Expanding the half-wave output as a Fourier series answers it exactly. The waveform is even about \(\omega t = \pi/2\) within its conducting half and zero elsewhere, and the standard result is
The general term of the bracket is \(\cos n\omega t/(n^2-1)\) for even \(n\). Only even harmonics appear beyond the fundamental, because the waveform repeats every full cycle but its two halves are not related by a simple sign change.
Reading the coefficients off:
| Component | Frequency | Amplitude | RMS | As a fraction of \(V_{dc}\) |
|---|---|---|---|---|
| DC | 0 | \(V_m/\pi\) | 0.3183 \(V_m\) | 1 |
| Fundamental | 50 Hz | \(V_m/2\) | 0.3536 \(V_m\) | 1.111 |
| 2nd harmonic | 100 Hz | \(2V_m/3\pi\) | 0.1500 \(V_m\) | 0.471 |
| 4th harmonic | 200 Hz | \(2V_m/15\pi\) | 0.0300 \(V_m\) | 0.094 |
| 6th harmonic | 300 Hz | \(2V_m/35\pi\) | 0.0129 \(V_m\) | 0.040 |
Two checks confirm the series. Summing the squares of the RMS entries, \(0.3183^2+0.3536^2+0.1500^2+0.0300^2+\cdots = 0.2500\), whose square root is \(0.500V_m\) — the RMS value derived by integration in Section 2. And discarding the DC term leaves \(\sqrt{0.2500-0.1013} = 0.3856V_m\) of alternating content, so \(\gamma = 0.3856/0.3183 = 1.211\), as before. The Fourier picture and the integral picture agree, as they must.
What the table adds is the bad news about filtering. The largest single ripple component sits at the mains frequency itself, with an amplitude 1.11 times the DC output. A low-pass filter must attenuate 50 Hz by whatever factor the specification demands while passing DC untouched, and that requires a corner frequency far below 50 Hz — which for a capacitor-input filter means a large capacitance and for a choke-input filter a large inductance.
The full-wave output of Chapter 9 has no component at 50 Hz at all. Its series is \(v_o = (2V_m/\pi)[1 - \tfrac{2}{3}\cos2\omega t - \tfrac{2}{15}\cos4\omega t - \cdots]\), so the lowest ripple frequency is 100 Hz with an RMS of \(0.471V_{dc}\) — the same fractional size as the half-wave second harmonic, but now the largest term instead of the third-largest. A filter attenuating as \(1/f\) or \(1/f^2\) therefore does between two and four times better on the same components, on top of the ripple factor already having fallen from 1.211 to 0.483.
The 0.471 figure is worth remembering: it is the fractional ripple presented to any filter fed from a full-wave rectifier, and it is the starting point of every derivation in Chapter 10.
5 Rectification Efficiency and Transformer Utilisation
Rectification efficiency asks what fraction of the power delivered by the source ends up as useful DC power in the load. Both quantities are computed for the same circuit; the numerator uses the average current, the denominator the RMS current, and the denominator must include the power lost in the diode's forward resistance \(r_f\) and the transformer winding resistance, lumped together:
Most textbooks round this to 40.6 %. The exact value of \(4/\pi^2\) is 0.40528, so 40.5 % is the honest figure; the traditional 40.6 % comes from carrying too few digits through an intermediate step. Either way the message is the same.
It is worth being clear about what the missing 59.5 % is, because it is not lost as heat in the diode. Set \(r_f = 0\) and the diode dissipates nothing at all, yet the efficiency is still only 40.5 %. The shortfall is definitional: \(P_{ac}\) is the total power delivered to the load, and only the fraction \((I_{dc}/I_{rms})^2\) of it is DC. The rest is ripple power, which heats the load just as effectively but carries no information and can do no work in a DC circuit. Half the loss comes from discarding one half-cycle and the rest from the poor shape of the half-cycle that survives.
The effect of \(r_f\) is easy to quantify. With \(R_L = 100\ \Omega\) and \(r_f = 20\ \Omega\), \(\eta = 40.53/1.2 = 33.8\) %, and the DC output falls from 6.37 V to 5.31 V for a 20 V peak. A rule of thumb: keep \(r_f\) below 5 % of \(R_L\) and the efficiency penalty stays under 5 %.
Transformer utilisation factor. The transformer must be specified in volt-amperes, and volt-amperes are computed from RMS quantities. But the useful output is DC power. The TUF is the ratio:
For the half-wave circuit, \(V_{s,rms} = V_m/\sqrt2\) and the secondary current is the load current, with RMS value \(I_m/2\):
To obtain 1 W of DC you must buy a transformer rated at \(1/0.287 = 3.5\) VA. The bridge rectifier of Chapter 9 achieves 0.812, needing only 1.23 VA for the same watt.
There is a further penalty the TUF conceals. The secondary current in a half-wave rectifier flows in one direction only, so it has a DC component of \(I_m/\pi\) which magnetises the core in one direction and drives it towards saturation. The transformer must be built with an air gap or an oversized core to tolerate this, which raises its cost above what the VA figure alone suggests. No full-wave topology has this problem, because their secondary currents are symmetric. This, more than the ripple, is why half-wave rectification is confined to low-power applications.
6 What a Real Diode Costs
All the results so far assume an ideal diode. Reinstating the constant-drop and piecewise-linear models of Chapter 6 changes the picture in two ways, and the change is negligible at high voltage and severe at low.
With a forward drop \(V_F\) and a total series resistance \(r_f\) (diode bulk plus transformer winding), the diode does not conduct until \(v_s\) exceeds \(V_F\). Conduction therefore starts at a cut-in angle
and ends symmetrically at \(\pi - \theta_1\). During conduction the output is \((v_s - V_F)R_L/(R_L+r_f)\), so the peak is reduced and the base of the half-sinusoid is cut away.
Figure 8.3 draws the case that matters: \(V_m = 5\) V into \(R_L = 120\ \Omega\), with \(V_F = 0.7\) V and \(r_f = 1.5\ \Omega\). The cut-in angle is \(\arcsin(0.7/5) = 0.1405\) rad \(= 8.05^\circ\), the peak output falls from 5 V to 4.247 V, and the average falls from \(5/\pi = 1.592\) V to 1.242 V — a loss of 22.0 %.
Now scale the same diode up to larger supplies, keeping the load at 120 \(\Omega\):
| \(V_m\) | Ideal \(V_{dc}\) | Real \(V_{dc}\) | Loss | Cut-in angle |
|---|---|---|---|---|
| 5 V | 1.592 V | 1.242 V | 22.0 % | 8.05° |
| 10 V | 3.183 V | 2.806 V | 11.9 % | 4.01° |
| 20 V | 6.366 V | 5.946 V | 6.6 % | 2.01° |
| 50 V | 15.92 V | 15.37 V | 3.4 % | 0.80° |
The loss is roughly \(V_F/V_m\) plus the resistive term, exactly as the model-error analysis of Chapter 7 predicts, and it is the reason a 5 V supply must never be built by rectifying a 5 V peak. It is also the reason Schottky diodes, with their 0.3 V drop, dominate low-voltage rectification: replacing 0.7 V with 0.3 V in the \(V_m = 5\) V row recovers about 8 % of the output.
There is a second, quieter consequence of the forward drop that the table does not show: the shape of the current pulse changes. The conduction interval shrinks from \(\pi\) to \(\pi - 2\theta_1\), so at \(V_m = 5\) V the diode conducts for \(163.9^\circ\) rather than \(180^\circ\), and the current within that interval is the difference between a sinusoid and a constant rather than a pure sinusoid. The ripple factor consequently rises slightly — the worked design at the end of this chapter comes out at 1.222 instead of 1.211 — because chopping the base off each half-sinusoid removes proportionally more average than RMS. The effect is small here and becomes large in Chapter 10, where the reservoir capacitor cuts the conduction angle to a few degrees.
The dissipation in the diode is worth estimating at the design stage, and the piecewise-linear model makes it easy. The forward drop contributes \(V_FI_{dc}\), because the drop is nearly constant while current flows and the time-average of the current is \(I_{dc}\); the bulk resistance contributes \(r_fI_{rms}^2\), because resistive loss follows the mean square. For the design of Section 6 those are 70 mW and 37 mW, a total of 107 mW — comfortable for a 1N400x in free air, whose thermal resistance of about 50 K/W puts the junction 5 K above the lead temperature.
Two further real-device effects belong here. The reverse leakage flows during the blocking half-cycle and subtracts a small negative average, but at nanoamperes into 120 \(\Omega\) it is a fraction of a microvolt and is ignored below about 100 °C. And the reverse recovery of Chapter 6 does nothing at 50 Hz — a 1N4007 with \(t_{rr} = 30\ \mu\text{s}\) is in recovery for 0.3 % of each cycle — but it is decisive in the 100 kHz switching supplies that have replaced mains-frequency rectifiers in most modern equipment.
7 A Complete Half-Wave Supply Design
The pieces now assemble into a design procedure. The specification: an unfiltered half-wave supply delivering \(V_{dc} = 12\) V into a 120 \(\Omega\) load, that is \(I_{dc} = 100\) mA, from 230 V 50 Hz mains, using a silicon diode with \(V_F = 0.7\) V and a total series resistance of 1.5 \(\Omega\).
Step 1: first estimate of the peak. Ignoring the diode, \(V_m = \pi V_{dc} = \pi\times12 = 37.70\) V, so a secondary of \(37.70/\sqrt2 = 26.66\) V RMS.
Step 2: correct for the diode. Feeding 37.70 V peak through the real diode gives only 11.51 V, 4.1 % short. Solving numerically for the peak that restores 12.00 V gives \(V_m = 39.26\) V, so the secondary must be \(39.26/\sqrt2 = 27.76\) V RMS. Specify a 28 V transformer; the turns ratio from 230 V is \(230/27.76 = 8.28:1\).
Step 3: the four output numbers. With \(V_m = 39.26\) V, integration of the real waveform gives \(V_{dc} = 12.00\) V, \(V_{rms} = 18.95\) V, \(I_{dc} = 100.0\) mA and \(I_{rms} = 157.9\) mA. The ripple factor is \(\sqrt{(18.95/12.00)^2-1} = 1.222\), slightly worse than the ideal 1.211 because the cut-in has chopped the base off each half-sinusoid.
Step 4: efficiency. \(P_{dc} = I_{dc}^2R_L = (0.100)^2\times120 = 1.200\) W. \(P_{ac} = I_{rms}^2(R_L+r_f) = (0.1579)^2\times121.5 = 3.030\) W. So \(\eta = 39.6\) %, close to the theoretical 40.5 % because \(r_f/R_L\) is only 1.25 %.
Step 5: the diode. Peak current \(I_m = (39.26-0.7)/121.5 = 317\) mA. Average current 100 mA. Dissipation \(\approx V_FI_{dc} + r_fI_{rms}^2 = 0.7\times0.100 + 1.5\times(0.1579)^2 = 0.070 + 0.037 = 0.107\) W. PIV \(= V_m = 39.3\) V. A 1N4001 (1 A, 50 V) would work but leaves only 27 % voltage margin; specify a 1N4004 (1 A, 400 V), which costs the same and tolerates mains transients.
Step 6: the transformer. Secondary VA \(= V_{s,rms}I_{rms} = 27.76\times0.1579 = 4.38\) VA for 1.20 W of DC, a TUF of 0.274 — slightly below the ideal 0.287 because of the diode loss. Specify at least a 5 VA transformer, and one rated to tolerate the 100 mA DC magnetising component.
Step 3 is the one to keep in mind. The design meets its DC specification exactly and is still useless as a power supply, because 12 V DC arrives with 14.7 V RMS of ripple on it. Everything in this chapter has been about the DC quantities; nothing in it has produced a usable output. That takes a reservoir capacitor, and adding one changes the analysis so completely — the conduction angle collapses to a few degrees, the diode current becomes a narrow spike many times \(I_{dc}\), and the PIV doubles to \(2V_m\) — that Chapter 10 has to start again from the beginning.
Before that, Chapter 9 fixes the topology. Using both half-cycles doubles the DC output for the same peak, cuts the ripple factor from 1.211 to 0.483, doubles the ripple frequency to 100 Hz so that any given capacitor works twice as well, doubles the efficiency to 81 %, and removes the transformer's DC magnetisation entirely. It costs one extra diode, or three.
8 Summary and Key Results
| Quantity | Expression | Value |
|---|---|---|
| DC (average) output | \(V_{dc} = V_m/\pi\) | 0.318 \(V_m\); \(I_{dc} = I_m/\pi\) |
| RMS output | \(V_{rms} = V_m/2\) | 0.500 \(V_m\); \(I_{rms} = I_m/2\) |
| AC (ripple) component | \(V_{ac} = \sqrt{V_{rms}^2-V_{dc}^2}\) | 0.386 \(V_m\) |
| Ripple factor | \(\gamma = \sqrt{(V_{rms}/V_{dc})^2-1}\) | 1.2114 — ripple exceeds the DC |
| Form factor | \(\text{FF} = V_{rms}/V_{dc} = \pi/2\) | 1.5708 (sinusoid 1.111, square wave 1) |
| Peak factor | \(\text{PF} = V_m/V_{rms}\) | 2.000 (sinusoid 1.414) |
| Rectification efficiency | \(\eta = (4/\pi^2)/(1+r_f/R_L)\) | 40.5 % maximum; 33.8 % if \(r_f = 0.2R_L\) |
| Peak inverse voltage | PIV | \(V_m\) unfiltered; \(2V_m\) with a reservoir capacitor |
| Transformer utilisation | \(\text{TUF} = 2\sqrt2/\pi^2\) | 0.2866 — 3.5 VA of transformer per watt of DC |
| Ripple frequency | \(f_r = f_{\text{line}}\) | 50 Hz — the load is unsupplied for 10 ms in every cycle |
| Cut-in angle | \(\theta_1 = \arcsin(V_F/V_m)\) | 8.05° at \(V_m\) = 5 V; 0.80° at \(V_m\) = 50 V |
| Real-diode loss in \(V_{dc}\) | \(\approx V_F/V_m + r_f/R_L\) | 22.0 % at \(V_m\) = 5 V; 3.4 % at \(V_m\) = 50 V |
| Worked design | 12 V, 100 mA into 120 \(\Omega\) | 28 V secondary, \(V_m\) = 39.3 V, \(\eta\) = 39.6 %, PIV 39.3 V, 4.4 VA |
9 Common Mistakes
The half-sinusoid has an average of \(2V_m/\pi = 0.637V_m\) over its own half-cycle, and that value is correct for the full-wave circuit of Chapter 9. For the half-wave circuit the average must be taken over the whole period, including the silent half, which halves it to \(V_m/\pi = 0.318V_m\). Writing \(\frac{1}{\pi}\int_0^\pi\) instead of \(\frac{1}{2\pi}\int_0^\pi\) doubles every DC quantity in the chapter and makes the efficiency come out at 81 % for a half-wave rectifier.
With a resistive load the diode blocks a peak of \(V_m\). Add a reservoir capacitor and the cathode is held near \(+V_m\) by the charged capacitor while the anode swings to \(-V_m\), so the diode blocks \(2V_m\). A supply built with 50 V diodes on a 24 V secondary (\(V_m = 33.9\) V) is fine unfiltered and fails immediately once the capacitor is fitted, because the diodes then see 67.9 V. Always ask what holds the cathode up.
The 40.5 % figure does not mean that 59.5 % of the input power is dissipated somewhere. With an ideal diode nothing is dissipated except in the load, and the load receives every watt the transformer delivers. The 40.5 % is the fraction of the load power that is DC; the rest is ripple heating the same resistor. A supply feeding a filter recovers much of it, which is why a filtered half-wave supply is far better than 40.5 % efficient in the everyday sense.
10 Chapter Review
1. A half-wave rectifier is fed from a 15 V RMS secondary into a 220 \(\Omega\) load. Taking the diode as ideal, find \(V_m\), \(V_{dc}\), \(I_{dc}\), \(V_{rms}\), the ripple factor and the DC power delivered.
\(V_m = 15\sqrt2 = 21.21\) V. Then \(V_{dc} = V_m/\pi = 21.21/3.1416 = 6.752\) V and \(I_{dc} = 6.752/220 = 30.69\) mA. The RMS output is \(V_m/2 = 10.61\) V, giving \(I_{rms} = 48.22\) mA. The ripple factor is \(\sqrt{(10.61/6.752)^2-1} = \sqrt{2.4674-1} = 1.211\), which is of course the universal half-wave value and did not need the numbers at all. The DC power is \(I_{dc}^2R_L = (0.03069)^2\times220 = 0.2072\) W, while the total power in the load is \(I_{rms}^2R_L = (0.04822)^2\times220 = 0.5116\) W. The ratio 0.2072/0.5116 = 0.405 is the rectification efficiency, confirming \(4/\pi^2\). Note that 0.30 W of the half-watt reaching the load is ripple.
2. Derive the ripple factor of a half-wave rectifier from first principles and explain why it exceeds unity.
Split the output into its average and the remainder: \(v_o = V_{dc}+v_{ac}\), where \(v_{ac}\) has zero mean by construction. Squaring and averaging, the cross term \(2V_{dc}\overline{v_{ac}}\) vanishes, so \(V_{rms}^2 = V_{dc}^2+V_{ac,rms}^2\) and \(V_{ac,rms} = \sqrt{V_{rms}^2-V_{dc}^2}\). The ripple factor is defined as \(\gamma = V_{ac,rms}/V_{dc} = \sqrt{(V_{rms}/V_{dc})^2-1}\). For the half-wave circuit \(V_{rms} = V_m/2\) and \(V_{dc} = V_m/\pi\), so the ratio is \(\pi/2 = 1.5708\) and \(\gamma = \sqrt{1.5708^2-1} = \sqrt{1.4674} = 1.2114\). It exceeds unity because the waveform spends half of every cycle at zero: a signal that is at its peak for a short time and at zero for a long time has a mean square much larger than the square of its mean. Quantitatively, \(V_{ac,rms} = 0.386V_m\) against \(V_{dc} = 0.318V_m\). Any waveform whose form factor exceeds \(\sqrt2\) has \(\gamma > 1\).
3. Why is the transformer utilisation factor of a half-wave rectifier as poor as 0.287, and what additional problem does the DC secondary current create?
TUF is \(P_{dc}\) divided by the VA rating the transformer must be built to, and the two are computed from different averages. \(P_{dc} = V_{dc}I_{dc} = (V_m/\pi)(I_m/\pi) = V_mI_m/\pi^2\). The transformer must be rated for the RMS values it actually carries, \(V_{s,rms}I_{s,rms} = (V_m/\sqrt2)(I_m/2) = V_mI_m/(2\sqrt2)\). The ratio is \(2\sqrt2/\pi^2 = 0.2866\). Physically, the winding is heated by an RMS current that flows for only half the time but is \(\pi/2\) times larger than the average, so the copper is worked hard for little DC output. The extra problem is that the secondary current is unidirectional, with a DC component of \(I_m/\pi\). That component produces a steady magnetomotive force in the core which is not cancelled by anything in the primary, so the core sits biased towards saturation, its incremental permeability falls, and the magnetising current and losses rise. The remedy is a gapped or oversized core, which raises the cost still further. Full-wave topologies have symmetric secondary currents and no such bias.
4. A half-wave rectifier must deliver 5 V DC into a 50 \(\Omega\) load using a silicon diode with \(V_F = 0.7\) V and \(r_f = 2\ \Omega\). Estimate the required secondary voltage and comment on the design.
Ignoring the diode entirely, \(V_m = \pi\times5 = 15.71\) V. The diode's forward drop removes roughly \(V_F/\pi\) from the average, that is \(0.7/3.1416 = 0.223\) V, and the series resistance costs a further factor \(R_L/(R_L+r_f) = 50/52 = 0.9615\), so a first correction gives \(V_{dc}\approx(15.71/\pi - 0.223)\times0.9615 = 4.59\) V. Scaling up, \(V_m\) must be about \(15.71\times5/4.59 = 17.1\) V, so the secondary is \(17.1/\sqrt2 = 12.1\) V RMS; specify 12 V and accept a small shortfall, or 15 V with a slightly high output. The comment: with \(I_{dc} = 100\) mA the peak current is \((17.1-0.7)/52 = 315\) mA, the diode dissipates about \(0.7\times0.1+2\times0.157^2 = 0.12\) W, and the PIV is 17.1 V. The design is workable but poor — the total real-diode penalty is about 9 %, and a Schottky with \(V_F = 0.3\) V would recover most of it. More importantly, an unfiltered 5 V rail with \(\gamma = 1.21\) is not a 5 V rail in any useful sense; the specification is incomplete without a ripple requirement.
5. The output of a half-wave rectifier is measured with three meters: an analogue DC voltmeter, a true-RMS AC voltmeter and an oscilloscope. If \(V_m = 20\) V, what does each show?
The analogue DC voltmeter is a moving-coil movement whose deflection follows the average of the applied waveform, so it reads \(V_{dc} = 20/\pi = 6.37\) V. The true-RMS AC voltmeter is AC-coupled, so it blocks the DC component and measures the RMS of what remains: \(V_{ac,rms} = \sqrt{V_{rms}^2-V_{dc}^2} = \sqrt{10.00^2-6.37^2} = 7.71\) V. Note this is not the 10.00 V total RMS — a DC-coupled true-RMS meter would read 10.00 V, and the difference between the two readings catches people out routinely. The oscilloscope shows the actual waveform: half-sinusoids of 20 V peak occupying 10 ms, separated by 10 ms of zero, repeating at 50 Hz. Measured on the scope, the peak-to-peak is 20 V, the peak is 20 V, and the mean cursor would read 6.37 V. The three instruments disagree not because any of them is wrong but because "the voltage" of a non-sinusoidal waveform is not a single number.