Physics 9702 · Paper 4

Rectification

Electronics runs on d.c., so the mains' to-and-fro must be made one-way. One diode does it crudely; four do it well.

Half-wave and full-wave

is the conversion of a.c. to d.c. — the banked one-liner (9702/41/O/N/24 Q6(a)(i)) (9702/41/M/J/25 Q6(a)(i)). The two flavours differ in one word each: half-wave rectification removes the voltage in one direction; full-wave rectification reverses it (9702/41/O/N/24 Q6(a)(ii)).

One diode blocks the negative half-cycles: half-wave rectification, with zero output half the time. A bridge of four diodes flips them instead: full-wave rectification — twice the hump rate, no flat gaps, and much more usable power.

Half-wave needs one diode in series with the load — adding that diode to a circuit was a mark of its own (9702/41/M/J/23 Q5(a)(i)). The price: zero output half the time, so the mean power is one quarter of the peak (half from the sine curve's shape, half again from the missing half-cycles) (9702/42/M/J/25 Q8(b)(iii)) — a 16 W supply delivers 8.0 W full-wave rectified but only 4.0 W half-wave (9702/42/F/M/24 Q4(d)).

The bridge rectifier in the half-cycle when the left supply terminal is positive: two opposite diodes conduct (highlighted), two block. Next half-cycle the other pair takes over — but both routes send the current the SAME way through the load.

The bridge rectifier: four diodes in a diamond. Each half-cycle, one opposite pair conducts while the other pair blocks — and both routes send the current the same way through the load. Papers test it forwards (draw the four diodes (9702/41/M/J/25 Q6(b)(iii))) and backwards (three diodes missing, orientations marked one by one, plus the current-direction arrow in $R$ (9702/42/M/J/23 Q7(a))).
A subtle favourite: full-wave rectification does not change the r.m.s. value. Flipping the negative half-cycles leaves the power–time graph identical, so the mean power — and hence $V_{rms}$ — is unchanged (9702/42/M/J/23 Q7(b)(iii)) (9702/41/M/J/24 Q7(c)).

Worked example

Half-wave by the power route (2025 paper)

A half-wave rectifier feeds a 45 Ω resistor from a supply of r.m.s. voltage 6.0 V. Find the peak power and the r.m.s. value of the output (9702/42/M/J/25 Q8).

Peak voltage: $V_0 = 6.0\sqrt{2} = 8.5\ \text{V}$ , so peak power $P_0 = V_0^2/R = 8.5^2/45 = 1.6\ \text{W}$ .
Half-wave mean power: $P = P_0/4 = 0.40\ \text{W}$ .
Work the r.m.s. output from the power: $V_{rms} = \sqrt{PR} = \sqrt{0.40 \times 45} = 4.2\ \text{V}$ .

Answer

P_0 = 1.6\ \text{W};\quad V_{rms}(\text{out}) = 4.2\ \text{V} = V_0/2

The half-wave output's r.m.s. is $V_0/2$ , not $V_0/\sqrt{2}$ — only the power route gets there safely.

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