By the end of this chapter you should be able to:
- Explain the operation of the centre-tapped and bridge full-wave rectifiers over a complete cycle, identifying which devices conduct when.
- Derive \(V_{dc} = 2V_m/\pi\) and \(V_{rms} = V_m/\sqrt2\) and obtain a ripple factor of 0.483.
- Derive the rectification efficiency \(8/\pi^2\) and explain why it is exactly twice the half-wave value.
- Determine the PIV for each topology and explain why the centre-tapped circuit demands \(2V_m\).
- Compare the transformer utilisation factors of the two circuits and account for the difference.
- Compute the switch-on surge current of a capacitor-input supply and size a series limiting resistor.
- Choose between half-wave, centre-tapped and bridge topologies for a stated set of requirements.
The half-wave rectifier of Chapter 8 discards one half-cycle of every two. Its ripple factor is 1.211, its efficiency 40.5 %, its transformer utilisation 0.287, and the ripple it produces sits at the mains frequency where it is hardest to filter. Every one of those numbers improves — most of them by a factor of two or better — if the negative half-cycle can be turned round and used rather than thrown away.
Two circuits do this, and the choice between them is one of the standard small decisions of power-supply design. The centre-tapped rectifier uses two diodes and a transformer with a tapped secondary, arranging for one half of the winding to drive the load on each half-cycle. The bridge uses four diodes and an ordinary untapped secondary, reversing the connection to the load electronically instead of magnetically. They give identical output waveforms and identical ripple, and they differ in exactly three respects: the number of diodes, the peak inverse voltage each diode must block, and how efficiently the transformer's copper is used.
The chapter also introduces the one failure mode that half-wave analysis never raises. A full-wave supply almost always feeds a large reservoir capacitor, and at the instant the supply is switched on that capacitor is a short circuit. The resulting inrush is limited only by the resistance of the transformer winding and the diodes, and for a modest 24 V supply it reaches 42 A — comfortably beyond the surge rating of the diodes it flows through. Sizing the resistor that prevents this is the last design step of the chapter.
1 The Centre-Tapped Rectifier
Wind the secondary with a connection brought out from its midpoint, and the two halves become two sources of equal amplitude whose instantaneous voltages, measured from the tap, are always of opposite sign. Call each half-winding \(V_m\sin\omega t\) and \(-V_m\sin\omega t\) with respect to the tap. Connect a diode to each end, join their cathodes, and take the load from that junction back to the tap.
On the positive half-cycle the top of the secondary is positive with respect to the tap, so \(D_1\) is forward biased and \(D_2\) is reverse biased. Current flows out of the top of the winding, through \(D_1\), down through \(R_L\) and back to the tap. On the negative half-cycle the bottom of the secondary is positive with respect to the tap, \(D_2\) conducts and \(D_1\) blocks, and current flows out of the bottom of the winding, through \(D_2\), and down through \(R_L\) in the same direction as before. That last clause is the whole trick: the load current reverses in the transformer but not in the load, because the two diodes steer it.
The output is therefore a continuous train of half-sinusoids with no gaps, repeating at twice the supply frequency — 100 Hz from a 50 Hz mains. Each diode conducts for half of every cycle and carries an average current of \(I_{dc}/2\), which is a genuine benefit: for a given load current the diodes are worked half as hard as in a half-wave circuit.
Peak inverse voltage. This is where the topology charges for its simplicity, and the analysis must be done carefully. Consider the instant of the positive peak. \(D_1\) is conducting, so its anode and cathode are at essentially the same potential, and the cathode — which is also the top of the load — sits at \(+V_m\) relative to the tap. Now look at \(D_2\). Its cathode is tied to \(D_1\)'s at \(+V_m\). Its anode is connected to the bottom of the secondary, which at this instant is at \(-V_m\) relative to the tap. The reverse voltage across \(D_2\) is therefore
Each diode blocks the whole secondary, end to end, not just its own half. This is the single most important practical fact about the circuit and the one most often got wrong.
The consequence for the transformer follows immediately. To deliver a peak output of \(V_m\) the tapped secondary must be wound for \(V_m\) on each half, so \(2V_m\) peak end to end — twice the copper and twice the insulation class of a bridge secondary for the same output. In the notation of the catalogue, a supply needing 24 V RMS needs a "24-0-24" transformer here and a plain 24 V transformer in a bridge.
Against that, the centre-tapped circuit has one genuine advantage: only one diode is ever in series with the load, so only one forward drop is lost. In a 5 V supply, saving 0.7 V out of 5 V is worth 14 % of the output, and centre-tapped rectifiers persist in low-voltage high-current supplies for precisely this reason. In a 300 V supply the same saving is 0.2 % and nobody cares.
2 The Bridge Rectifier
The bridge achieves the same reversal with four diodes and no tap. Draw them as a diamond: the transformer connects across one diagonal, the load across the other.
On the positive half-cycle, with the left-hand secondary terminal positive, current leaves that terminal, finds \(D_1\) forward biased into the top node, passes down through the load, and returns through \(D_3\) to the right-hand terminal. \(D_2\) and \(D_4\) are reverse biased and carry nothing. On the negative half-cycle the right-hand terminal is positive, so \(D_2\) conducts into the top node, the current again passes down through the load, and \(D_4\) returns it to the left-hand terminal. The two conducting diodes swap over; the direction of load current does not.
Three consequences distinguish the bridge from the centre-tapped circuit.
Two diode drops, not one. Two diodes are in series with the load at every instant, so the peak output is \(V_m - 2V_F\), typically \(V_m - 1.4\) V. This is the bridge's only real disadvantage.
PIV is \(V_m\), not \(2V_m\). At the positive peak, \(D_1\) and \(D_3\) conduct and hold the top node at \(V_m\) (ignoring drops) and the bottom node at zero. \(D_2\) has its cathode at the top node, \(V_m\), and its anode at the right-hand secondary terminal, which is at 0 V; so it blocks \(V_m\). The same argument applies to \(D_4\). Each diode blocks half of what a centre-tapped diode would, which halves the voltage rating you must buy.
No tap and no DC in the winding. The secondary carries a symmetrical alternating current with no DC component, so the core is not magnetically biased, and the whole winding is used on both half-cycles rather than half of it at a time. That is why the transformer utilisation is so much better, as the next section quantifies.
In a bridge, neither output terminal is connected to either secondary terminal, so the output floats with respect to the transformer. That is normally an advantage — you can earth whichever output rail you please. But it means you cannot share one secondary winding between two bridges to make a split supply, and it means that a single transformer feeding two bridges with a common earth will short two of the diodes. Split \(\pm\) supplies use a centre-tapped secondary with either two bridges on separate windings, or one bridge with the tap as ground — a configuration that is a pair of centre-tapped rectifiers, not a bridge.
A practical note: bridges are almost always bought as a single moulded package with four leads marked \(\sim\), \(\sim\), \(+\) and \(-\). The four diodes inside are matched, thermally coupled and cheaper than four discrete devices, and the package can be bolted to a heat sink. Buying four separate diodes and wiring a diamond is an examination exercise, not a design.
3 Average, RMS, Ripple and Efficiency
Both topologies give the same output waveform, \(|V_m\sin\omega t|\), so one analysis serves for both. The period is now \(\pi\) rather than \(2\pi\), which is the whole source of the improvement.
Exactly twice the half-wave value, because the same area is now averaged over half the interval. \(I_{dc} = 2I_m/\pi\).
The same RMS as the original sinusoid, which makes sense: squaring destroys the sign, so a rectified sinusoid and a sinusoid have identical mean squares.
The form factor and ripple factor follow at once:
Most texts quote 0.482, which comes from rounding the form factor to 1.11 before squaring. The exact value is 0.4834; the difference is immaterial in practice but it is worth knowing which of the two you have written down and why.
The peak factor is \(V_m/(V_m/\sqrt2) = \sqrt2 = 1.414\), the same as a sinusoid and much gentler than the half-wave circuit's 2.
Efficiency. Following Chapter 8's definition exactly, with \(r_f\) the total series resistance in the conducting path:
Quoted in most textbooks as 81.2 %, again from a rounded intermediate. Exactly twice the half-wave figure of 40.5 %, and for exactly the reason you would guess: the same waveform is being delivered twice as often.
Both changes help the filter, and the second helps more than the first. The Fourier series of the full-wave output is \(v_o = (2V_m/\pi)\left[1 - \tfrac{2}{3}\cos2\omega t - \tfrac{2}{15}\cos4\omega t - \cdots\right]\), which has no component at the mains frequency. The largest ripple term is at 100 Hz with an RMS of \(0.471V_{dc}\). A given capacitor has half the reactance at 100 Hz that it has at 50 Hz, so the same component filters roughly five times better once the topology is changed — a factor of 2.5 from the ripple factor and a further 2 from the frequency.
4 Conduction Intervals and Diode Currents
The output voltage of the two topologies is identical, but the currents inside them are not, and it is the currents that decide which components you must buy. Working them out is a matter of asking, for each device, how long it conducts and what it carries while it does.
In both circuits each diode conducts for exactly half of every mains cycle — \(\pi\) radians out of \(2\pi\) — and carries a half-sinusoid of peak \(I_m\) during that interval. So for each diode
Halving the average current per diode relative to a half-wave circuit is a real benefit: it halves the junction heating in each device and lets you use a smaller part, or run a given part cooler.
Take the 24 V RMS, 100 \(\Omega\) example again. For the bridge, \(I_m = 32.54/100 = 325.4\) mA, so each diode carries 103.6 mA average, 162.7 mA RMS and 325.4 mA peak, and dissipates \(V_FI_{D(\text{av})} = 0.7\times0.1036 = 72.5\) mW. Four diodes give 290 mW in total, which is the same as \(2V_F I_{dc} = 1.4\times0.2072\) W — a useful cross-check, since the two conducting drops are in series with the whole load current at all times. For the centre-tapped circuit, \(I_m = 332.4\) mA and only two diodes are involved, so the total diode dissipation is 148 mW, half as much.
Where the two circuits genuinely differ is in the transformer current, and this is what the TUF of the next section is measuring. In a bridge the secondary carries a complete sinusoid: current flows out of one end on one half-cycle and out of the other on the next, so the winding sees \(I_{s(\text{rms})} = I_m/\sqrt2 = 230.1\) mA. In a centre-tapped circuit each half-winding carries a half-sinusoid, \(I_m/2 = 166.2\) mA RMS, and rests for the other half-cycle. The bridge's single winding is therefore worked harder but there is only one of it, while the centre-tapped circuit has two windings each working at 72 % of that current — more copper, less used.
- Average \(I_{F(\text{AV})}\) — sets the junction temperature through \(V_FI_{av}\), and is the rating printed on the front of a rectifier datasheet. A "1 A" diode is a 1 A average diode.
- RMS \(I_{F(\text{RMS})}\) — sets the heating in the bulk resistance and in every wire and winding in the path. This is the number a transformer and a fuse are rated by.
- Peak \(I_{FSM}\) and \(I_{FRM}\) — set by the fusing of the bond wires, and attacked by the switch-on surge and by the repetitive spikes of a capacitor-input filter.
With a resistive load these three stand in the ratio \(1 : \pi/2 : \pi\), or \(1:1.57:3.14\). Chapter 10 shows that a reservoir capacitor pushes the same ratio out to something like \(1:2.4:6.6\), which is why a rectifier that is comfortable driving a resistor may fail driving a capacitor at the same DC current.
One further conduction detail matters for the centre-tapped circuit. The two diodes must never conduct simultaneously, and they cannot, since one half-winding is always negative when the other is positive. But the transition is not instantaneous in a real device: the reverse recovery of Chapter 6 keeps the outgoing diode conducting for a few microseconds after the crossing, briefly short-circuiting the whole secondary through both diodes. At 50 Hz with a 30 µs recovery time this costs a fraction of a per cent of a cycle and merely generates a little conducted interference; in a 100 kHz switching supply the same event would be catastrophic, which is why fast or Schottky rectifiers are mandatory there.
5 Transformer Utilisation Compared
The TUF was defined in Chapter 8 as the ratio of DC output power to the VA rating the transformer must be built to. It penalises circuits that force the transformer to carry more RMS current, or to be wound for more voltage, than the DC output justifies. The three topologies differ sharply.
Bridge. The secondary carries a full sinusoid of current, RMS \(I_m/\sqrt2\), at an RMS voltage \(V_m/\sqrt2\). So
The best of the three, and numerically equal to the efficiency — a coincidence of this particular circuit, not a general rule.
Centre-tapped. Each half-winding conducts for only half of each cycle, so its current is a half-sinusoid with RMS \(I_m/2\), and its voltage is \(V_m/\sqrt2\) RMS. The secondary VA is the sum over both halves, \(2\times(V_m/\sqrt2)(I_m/2) = V_mI_m/\sqrt2\):
The primary, however, carries a full sinusoid like the bridge's, so \(\text{TUF}_{\text{pri}} = 0.8106\). Since the transformer must be built for both, the figure quoted is the mean:
This averaging convention is where the familiar 0.693 comes from, and it is worth knowing that it is a convention: the secondary alone is only 0.573.
Setting the three side by side:
| Topology | TUF | Transformer VA per watt of DC |
|---|---|---|
| Half-wave | 0.2866 | 3.49 VA/W, plus a gapped core for the DC magnetisation |
| Centre-tapped | 0.6919 | 1.45 VA/W, but the secondary is wound for \(2V_m\) |
| Bridge | 0.8106 | 1.23 VA/W |
Two things about the half-wave entry deserve emphasis, because the raw TUF understates how bad it is. First, the transformer must be built to tolerate a DC magnetising component of \(I_m/\pi\) in the secondary, which no full-wave circuit produces; that means a gapped or oversized core, and gapping reduces the inductance and raises the magnetising current, so the penalty compounds. Second, a transformer running at a TUF of 0.287 is dissipating copper loss proportional to \(I_{rms}^2\) while delivering DC proportional to \(I_{dc}^2\), and the ratio of those is the square of the form factor, 2.47. The winding runs hot for the output it gives.
The economic argument now assembles itself. For 100 W of DC, a bridge needs a 123 VA transformer, a centre-tapped circuit a 145 VA one wound for twice the voltage, and a half-wave circuit a 349 VA one with a special core. Two extra diodes cost a fraction of that difference at any power above a few watts, and in a mains supply the diodes are not the expensive part — the transformer, the capacitor and the heat sink are.
6 Switch-On Surge and the Limiting Resistor
Every practical full-wave supply has a reservoir capacitor across its output. Chapter 10 analyses what that capacitor does in steady state; here we need only the first cycle after the switch is closed, because that is when the diodes are in danger.
At the instant of switch-on the capacitor is discharged, so it holds the output at 0 V. If the switch happens to close at the peak of the mains cycle — and over many switch-ons it eventually will — the full secondary peak appears across nothing but the series resistance of the path: the transformer's secondary and reflected primary resistance, plus the bulk resistance of the two conducting diodes. Call the total \(R_s\). Then
The load resistance does not appear: the capacitor is a short circuit at \(t = 0\), and all of the current goes into charging it. The surge decays with time constant \(R_sC\), which is milliseconds.
A bridge supply runs from a 24 V RMS secondary into a 2200 µF reservoir and a 100 \(\Omega\) load. The transformer and diodes together contribute \(R_s = 0.8\ \Omega\). The rectifiers are 1N4007, rated \(I_{F(AV)} = 1\) A and \(I_{FSM} = 30\) A for one half-cycle.
Unprotected surge. \(V_m = 24\sqrt2 = 33.94\) V, so \(I_{\text{surge}} = 33.94/0.8 = 42.4\) A. That is 141 % of the diodes' non-repetitive surge rating, and it recurs at every switch-on. The design will fail, intermittently and unpredictably — the worst kind of failure, because switching on at a zero crossing produces no surge at all and the fault is not reproducible on the bench.
Sizing the resistor. To hold the surge to the 30 A rating, the total path resistance must be at least \(33.94/30 = 1.131\ \Omega\), so add \(R_{\text{lim}} = 1.131 - 0.8 = 0.33\ \Omega\). A standard 0.33 \(\Omega\) resistor gives a surge of \(33.94/1.13 = 30.0\) A, exactly on the limit; 0.47 \(\Omega\) gives 26.7 A and a sensible margin.
What it costs in steady state. Simulating the completed circuit, the RMS current in the rectifier branch is 0.688 A, so the 0.33 \(\Omega\) resistor dissipates \(I_{rms}^2R = 0.688^2\times0.33 = 0.156\) W — a 0.5 W part is ample. The DC output falls from 30.69 V to 30.27 V, a loss of 0.42 V or 1.4 %. That is the price of the protection, and it is cheap.
How long the surge lasts. The charging time constant is \(R_sC = 1.13\times2200\times10^{-6} = 2.5\) ms, so the capacitor is essentially charged within about three time constants, or 7.5 ms — less than one mains cycle. The diode's \(I_{FSM}\) rating is specified for 10 ms, which is exactly the right comparison to make.
Three practical variations are worth knowing. A negative-temperature-coefficient thermistor replaces the fixed resistor with one that is 5 \(\Omega\) when cold and 0.1 \(\Omega\) once warmed by the load current; the surge above would fall to \(33.94/5.8 = 5.9\) A, and the steady-state loss almost vanishes. Its weakness is that it does not protect against a switch-off and immediate switch-on, because it has not had time to cool. A relay-bypassed resistor uses a larger resistor shorted out by a relay after a few hundred milliseconds, and is standard in equipment above a few hundred watts. And in very large supplies the limiting is done on the primary side, where the currents are smaller.
Do not confuse this surge with the repetitive peak current of the steady state, which Chapter 10 shows to be six or more times \(I_{dc}\) in a capacitor-input filter. Both stress the same diode, but they are checked against different ratings — \(I_{FSM}\) for the once-only inrush, \(I_{FRM}\) and \(I_{F(AV)}\) for the repetitive spikes.
7 Choosing Between the Three Topologies
All of the results of Chapters 8 and 9 can now be collected. The table assumes ideal diodes, a resistive load and a peak secondary voltage \(V_m\) measured at the rectifier input.
| Quantity | Half-wave | Centre-tapped | Bridge |
|---|---|---|---|
| Diodes | 1 | 2 | 4 |
| Transformer | Simple, but DC-biased core | Centre-tapped, wound for \(2V_m\) | Simple, wound for \(V_m\) |
| \(V_{dc}\) | \(V_m/\pi = 0.318V_m\) | \(2V_m/\pi = 0.637V_m\) | \(2V_m/\pi = 0.637V_m\) |
| \(V_{rms}\) | \(V_m/2\) | \(V_m/\sqrt2\) | \(V_m/\sqrt2\) |
| Form factor | 1.571 | 1.111 | 1.111 |
| Ripple factor \(\gamma\) | 1.211 | 0.483 | 0.483 |
| Ripple frequency | 50 Hz | 100 Hz | 100 Hz |
| Efficiency \(\eta_{\max}\) | 40.5 % | 81.1 % | 81.1 % |
| PIV per diode | \(V_m\) | \(2V_m\) | \(V_m\) |
| Diodes in series with load | 1 (\(0.7\) V lost) | 1 (\(0.7\) V lost) | 2 (\(1.4\) V lost) |
| Average current per diode | \(I_{dc}\) | \(I_{dc}/2\) | \(I_{dc}/2\) |
| TUF | 0.287 | 0.692 | 0.811 |
Take a 24 V RMS secondary (\(V_m = 33.94\) V), a 100 \(\Omega\) load and silicon diodes with \(V_F = 0.7\) V.
- Half-wave: peak output \(33.94-0.7 = 33.24\) V, \(V_{dc} = 33.24/\pi = 10.58\) V, \(I_{dc} = 105.8\) mA, ripple 12.82 V RMS.
- Centre-tapped (using the same 24 V per half): peak output 33.24 V, \(V_{dc} = 2\times33.24/\pi = 21.16\) V, \(I_{dc} = 211.6\) mA. But the transformer is now a 24-0-24 unit, twice the winding.
- Bridge: peak output \(33.94-1.4 = 32.54\) V, \(V_{dc} = 2\times32.54/\pi = 20.72\) V, \(I_{dc} = 207.2\) mA, from the plain 24 V transformer.
The bridge gives 2.1 % less output than the centre-tapped circuit from half the secondary copper, and its diodes need a 50 V rating against 100 V. The PIVs are 33.9 V for half-wave and bridge and 67.9 V for the centre-tapped circuit, all before allowing margin for mains transients — specify 400 V parts in every case, since they cost the same.
When to choose which. Use a bridge by default: best transformer utilisation, lowest PIV, cheapest transformer, and available as a single package. Use a centre-tapped circuit when the output voltage is low enough that a second diode drop is a significant fraction of it (below about 10 V), when you need a split \(\pm\) supply from one winding, or when the diodes must be Schottky devices whose reverse leakage makes the 2\(V_m\) rating awkward but whose forward drop you cannot afford to double. Use a half-wave circuit only where the current is very small and cost dominates absolutely — a few milliamperes to run an indicator or a mains-referenced timing circuit — or where a transformerless supply makes the DC magnetisation problem irrelevant because there is no transformer.
None of these circuits yet produces a usable DC output. A bridge delivering 20.72 V DC also delivers 10.01 V RMS of 100 Hz ripple, which no logic circuit and no amplifier will tolerate. Chapter 10 adds the reservoir capacitor that turns this pulsating output into a nearly steady one, and shows what that costs in diode current, conduction angle and regulation.
8 Summary and Key Results
| Quantity | Centre-tapped | Bridge |
|---|---|---|
| Diodes / transformer | 2 diodes, tapped secondary wound for \(2V_m\) | 4 diodes, plain secondary wound for \(V_m\) |
| \(V_{dc}\) | \(2V_m/\pi = 0.6366V_m\) | \(2V_m/\pi\), less \(2V_F\) at the peak |
| \(V_{rms}\) | \(V_m/\sqrt2 = 0.7071V_m\) | \(V_m/\sqrt2\) |
| Form factor | \(\pi/2\sqrt2 = 1.1107\) | 1.1107 |
| Ripple factor | \(\sqrt{\text{FF}^2-1} = 0.4834\) | 0.4834 (quoted 0.482) |
| Peak factor | \(\sqrt2 = 1.414\) | 1.414 |
| Ripple frequency | 100 Hz | 100 Hz |
| Efficiency | \(8/\pi^2 = 81.1\ \%\) | 81.1 % (quoted 81.2 %) |
| PIV per diode | \(2V_m\) — the whole secondary | \(V_m\) |
| Diode drops in the path | 1 (\(V_m - 0.7\) V at the peak) | 2 (\(V_m - 1.4\) V at the peak) |
| Diode average current | \(I_{dc}/2\) | \(I_{dc}/2\) |
| TUF | 0.5732 secondary, 0.8106 primary, 0.6919 mean | 0.8106 |
| Transformer VA per DC watt | 1.45 | 1.23 |
| Lowest ripple harmonic | 100 Hz at \(0.471V_{dc}\) RMS | same — no 50 Hz component at all |
| Switch-on surge | \((V_m-2V_F)/R_s\) | 42.4 A for 24 V RMS and \(R_s\) = 0.8 \(\Omega\); 0.33 \(\Omega\) limits it to 30 A |
9 Common Mistakes
The blocking diode's cathode is held at \(+V_m\) by the conducting diode while its anode sits at \(-V_m\), so it withstands \(2V_m\), not \(V_m\). Designing a 24-0-24 supply with 50 V diodes puts 67.9 V across them and they fail on the first cycle. The mistake comes from thinking of each half-winding in isolation; the correct method is always to draw the whole loop at the peak instant and apply Kirchhoff's voltage law.
A 24-0-24 transformer and a 24 V transformer are not the same component: the first has twice the secondary copper and costs accordingly. Quoting "the centre-tapped circuit gives 21.2 V and the bridge only 20.7 V" is true and misleading, because the bridge did it with half the winding. The honest comparison is at equal DC output, where the bridge needs a smaller and cheaper transformer, which is what the TUF figures of 0.811 against 0.692 are telling you.
A supply drawing 300 mA in normal operation can pull 42 A for a few milliseconds at switch-on, because the discharged reservoir capacitor is a short circuit and only the winding and diode resistance limit the current. The relevant rating is \(I_{FSM}\), not \(I_{F(AV)}\), and the failure is intermittent because it depends on the phase of the mains at the moment the switch closes. If the fuse blows "sometimes when you turn it on", this is why.
10 Chapter Review
1. A bridge rectifier is fed from an 18 V RMS secondary into a 220 \(\Omega\) load. Taking \(V_F = 0.7\) V per diode, find the peak output, \(V_{dc}\), \(I_{dc}\), the ripple voltage and the PIV.
\(V_m = 18\sqrt2 = 25.456\) V at the secondary. Two diodes conduct in series at all times, so the peak at the load is \(25.456 - 1.4 = 24.056\) V. Then \(V_{dc} = 2\times24.056/\pi = 15.315\) V and \(I_{dc} = 15.315/220 = 69.6\) mA. The RMS output is \(24.056/\sqrt2 = 17.010\) V, so the alternating content is \(\sqrt{17.010^2-15.315^2} = 7.404\) V RMS and \(\gamma = 7.404/15.315 = 0.483\), as expected. The PIV is the full secondary peak, 25.46 V, so 50 V diodes would work but 400 V parts (a 1N4004 bridge or equivalent) cost the same and survive mains transients. Note that the two diode drops have cost 5.5 % of the output; a centre-tapped circuit would have delivered 15.76 V from the same 18 V per half, at the price of a 18-0-18 transformer.
2. Derive the ripple factor and efficiency of a full-wave rectifier and explain why each is exactly twice as good as the half-wave value — or is it?
\(V_{dc} = (1/\pi)\int_0^\pi V_m\sin\theta\,d\theta = 2V_m/\pi\) and \(V_{rms}^2 = (1/\pi)\int_0^\pi V_m^2\sin^2\theta\,d\theta = V_m^2/2\), so \(V_{rms} = V_m/\sqrt2\). The form factor is \(\pi/(2\sqrt2) = 1.1107\), and \(\gamma = \sqrt{1.1107^2-1} = 0.4834\). The efficiency is \(\eta = (2I_m/\pi)^2R/(I_m/\sqrt2)^2R = 8/\pi^2 = 0.8106\). The efficiency is exactly twice the half-wave value of \(4/\pi^2\), because \(V_{dc}\) doubles (squared: ×4) while \(V_{rms}^2\) also doubles (×2), giving a net factor of two. The ripple factor is not exactly half: 1.2114/0.4834 = 2.506, not 2. The reason is that \(\gamma\) involves a square root of a difference, \(\sqrt{\text{FF}^2-1}\), which is not a linear function of the form factor. So the common statement "full-wave rectification halves the ripple" understates the improvement by 25 %.
3. Why is the transformer utilisation factor of a centre-tapped rectifier 0.692 when its secondary alone gives only 0.573?
Each half of the secondary conducts for only half of each cycle, so it carries a half-sinusoid of current with RMS \(I_m/2\) at \(V_m/\sqrt2\) RMS. Summing both halves, the secondary VA is \(2\times(V_m/\sqrt2)(I_m/2) = V_mI_m/\sqrt2\), and with \(P_{dc} = 4V_mI_m/\pi^2\) the secondary TUF is \(4\sqrt2/\pi^2 = 0.5732\). The primary, however, sees a continuous sinusoidal current — the two half-cycles of secondary current appear in the primary as one uninterrupted alternating current — so its VA is \((V_m/\sqrt2)(I_m/\sqrt2) = V_mI_m/2\), giving \(\text{TUF}_{\text{pri}} = 8/\pi^2 = 0.8106\), the same as a bridge. The transformer must be designed to satisfy both windings, and the conventional figure of merit takes the arithmetic mean, \((0.5732+0.8106)/2 = 0.6919\). It is a convention rather than a derivation, and when specifying a real transformer you should size the secondary from 0.5732 and the primary from 0.8106 separately.
4. A 30 V RMS secondary with a total series resistance of 0.5 \(\Omega\) feeds a bridge and a 4700 µF capacitor. Find the worst-case switch-on surge and choose a limiting resistor for diodes rated \(I_{FSM} = 50\) A.
\(V_m = 30\sqrt2 = 42.43\) V, less the two diode drops, 41.03 V. With the capacitor discharged the load is irrelevant and the current is limited only by \(R_s\): \(I_{\text{surge}} = 41.03/0.5 = 82.1\) A, which is 164 % of the 50 A rating. To hold the surge to 50 A the path resistance must be at least \(41.03/50 = 0.821\ \Omega\), so \(R_{\text{lim}} \ge 0.321\ \Omega\); a standard 0.33 \(\Omega\) gives \(41.03/0.83 = 49.4\) A, on the limit, and 0.47 \(\Omega\) gives 42.3 A with a comfortable 15 % margin. Choose 0.47 \(\Omega\). Its power rating follows from the steady-state RMS current in the rectifier branch, which for a capacitor-input filter is typically two to three times \(I_{dc}\); at \(I_{dc} = 1\) A that is about 2.5 A RMS, so \(P = 2.5^2\times0.47 = 2.9\) W and a 5 W wirewound part is needed. At that dissipation an NTC thermistor, or a relay that shorts the resistor after 200 ms, becomes the better engineering answer.
5. A designer needs a 5 V, 3 A DC supply and is choosing between a bridge and a centre-tapped rectifier. Which would you recommend, and what changes if the requirement is 300 V at 50 mA?
At 5 V, choose the centre-tapped circuit, or better still a centre-tapped circuit with Schottky diodes. The bridge puts two forward drops in series with the load: at 3 A a silicon rectifier drops nearly 1.0 V, so two of them cost 2.0 V out of a peak of about 8 V — a quarter of the available voltage, and 6 W of dissipation needing a heat sink. The centre-tapped circuit loses one drop instead of two, halving both figures; with Schottky diodes at 0.4 V the loss falls to 1.2 W. The penalty, a secondary wound for twice the voltage, is trivial at 8 V. At 300 V and 50 mA the reasoning reverses completely. The 1.4 V lost in a bridge is 0.5 % of the output and irrelevant, while the centre-tapped circuit would need a 600 V PIV rating and a secondary wound for 600 V peak — more insulation, more copper and a more expensive transformer. Choose the bridge, in a single package rated at 1000 V. The general rule is that diode drops matter at low voltage and insulation matters at high voltage, and the crossover is around 10–15 V.