Questions on rectification and smoothing

Question 1

State the function of a diode in a circuit and explain why a single diode in series with an alternating supply produces half-wave rectification.

Accepted Answer

A diode conducts in one direction only (1). Only the half-cycles that drive current in the conducting direction pass, so the load receives humps of one polarity with gaps where the reversed half-cycles are blocked (1).

Question 2

Sketch the output of a half-wave rectifier fed with a sinusoidal input, and the output of a full-wave rectifier fed with the same input. Label the difference between them.

Accepted Answer

Half-wave: positive humps separated by flat intervals at zero, one hump per input cycle (1). Full-wave: the negative half-cycles are inverted rather than removed, giving two humps per input cycle with no flat gaps (1).

Question 3

State one disadvantage of half-wave rectification compared with full-wave rectification, and one advantage in terms of components.

Accepted Answer

Disadvantage: half the input power is discarded (or the output has longer gaps, making it harder to smooth) (1). Advantage: only one diode is needed instead of four (1).

Question 4

A sinusoidal 50 Hz supply is full-wave rectified. State the number of voltage peaks per second in the output.

Accepted Answer

100 per second: full-wave rectification gives one peak every half-cycle (1).

Question 5

A sinusoidal 50 Hz supply is half-wave rectified, with no smoothing. State the time between successive peaks of the output, and the length of time in each cycle for which the output is zero.

Accepted Answer

Peaks arrive once per 20 ms cycle, so 20 ms apart (1); the output is zero for the blocked half-cycle, 10 ms in each cycle (1).

Question 6

State what is meant by the ripple of a smoothed supply and what causes it.

Accepted Answer

Ripple is the small periodic variation that remains in the smoothed output voltage (1); it is caused by the capacitor discharging through the load between the peaks that recharge it (1).

Question 7

A bridge rectifier contains four diodes. Explain how it produces full-wave rectification. Refer to which diodes conduct in each half-cycle.

Accepted Answer

The diodes conduct in opposite pairs (1). In one half-cycle one pair is forward biased and steers the current through the load in a particular direction; in the other half-cycle the supply reverses and the other pair conducts (1). The current through the load is in the same direction in both half-cycles, so the output has two humps per input cycle (1).

Question 8

Explain how a capacitor connected across the load smooths the output of a full-wave rectifier.

Accepted Answer

Near each peak the rectified voltage charges the capacitor (1). As the rectified voltage falls, the capacitor discharges through the load, holding the output voltage up until the next peak recharges it (1). The output is therefore a nearly steady voltage with a small ripple, instead of falling to zero between humps (1).

Question 9

A smoothing capacitor of 2200 μF discharges through a load of 500 Ω. Calculate the time constant of the discharge.

Accepted Answer

τ = RC = 500 × 2200 × 10⁻⁶ (1) τ = 1.1 s (1)

Question 10

State and explain the effect on the ripple of a smoothed supply of increasing the capacitance of the reservoir capacitor.

Accepted Answer

The ripple decreases (1); a larger capacitance gives a longer time constant RC, so the capacitor voltage falls less in the fixed time between successive peaks (1).

Question 11

A smoothed power supply works well with a 2.0 kΩ load. Explain why the ripple increases when the load is changed to 200 Ω.

Accepted Answer

The smaller resistance makes the time constant RC ten times shorter (a larger current is drawn) (1); the capacitor discharges much further between peaks, so the sag in voltage between top-ups, the ripple, is larger (1).

Question 12

An oscilloscope connected across the load of a rectifier circuit shows positive humps 10 ms apart. The ac supply has a frequency of 50 Hz. Deduce whether the rectification is half-wave or full-wave.

Accepted Answer

The supply period is 1/50 = 20 ms, so humps every 10 ms means one per half-cycle (1); that requires the negative half-cycles to be folded up rather than blocked, so the rectification is full-wave (half-wave would give humps 20 ms apart) (1).

Question 13

A smoothed full-wave supply holds its output near 15 V across a load that draws an almost steady current of 60 mA. The reservoir capacitor is 1000 μF and peaks arrive every 10 ms. Estimate the fall in the output voltage between successive peaks.

Accepted Answer

Charge lost by the capacitor: Q = IΔt = 0.060 × 0.010 = 0.60 × 10⁻³ C (1) ΔV = Q/C (1) ΔV = (6.0 × 10⁻⁴)/(1000 × 10⁻⁶) = 0.60 V (1)

Question 14

A student smooths a half-wave rectified 50 Hz supply and a full-wave rectified 50 Hz supply using identical capacitors and identical loads. Explain which output shows the larger ripple.

Accepted Answer

With half-wave rectification the capacitor must supply the load for 20 ms between peaks, but only 10 ms with full-wave (1); with the same time constant RC, the voltage falls further during the longer discharge (1); so the half-wave output has the larger ripple (1).

Question 15

A full-wave rectified 50 Hz supply is smoothed by a 470 μF capacitor across a 1.0 kΩ load. Calculate the time between successive peaks and the discharge time constant, and use the two values to justify whether the smoothing is effective.

Accepted Answer

Full-wave rectification gives a peak every half-cycle, so every 10 ms (1)
τ = RC = 1000 × 470 × 10⁻⁶ = 0.47 s (1)
τ is 47 times the interval between peaks (1)
The capacitor loses only a small fraction of its charge before each top-up, so the ripple is small and the smoothing is effective (1).

Question 16

An unsmoothed full-wave rectifier is fed from a supply of rms voltage 12 V. State the peak output voltage, and describe how the smoothed output would sit relative to the unsmoothed humps if both were drawn on the same axes.

Accepted Answer

V₀ = √2 × 12 = 17 V (1). The smoothed trace touches each hump near its peak, then falls only slightly before the next peak recharges the capacitor (1); it stays close to 17 V, always at or above the unsmoothed curve at the same instant (1).

Question 17

A 100 μF capacitor smooths a full-wave rectified 50 Hz supply across a 2.0 kΩ load. Calculate the fraction of the peak voltage remaining after one 10 ms interval between peaks, and comment on the size of the ripple.

Accepted Answer

RC = 2000 × 100 × 10⁻⁶ = 0.20 s (1)
V/V₀ = e^−t/RC = e^{−0.010/0.20} (1)
V/V₀ = 0.951, so about 5% of the peak voltage is lost between top-ups (1)
The ripple is only a few per cent of the peak, so the smoothing is acceptable for most purposes (1).

Question 18

A full-wave rectified 50 Hz supply feeds a 500 Ω load. The design requires the output to stay above 95% of the peak voltage at all times. Three reservoir capacitors are available: 100 μF, 220 μF and 470 μF. Using V = V₀e^−t/RC, deduce the smallest of the three capacitors that meets the requirement.

Accepted Answer

Full-wave rectification: the capacitor discharges for up to 10 ms between peaks (1)
Time constants: RC = 0.050 s, 0.11 s and 0.235 s (1)
Fractions remaining after 10 ms: e^−t/RC = 0.82, 0.91 and 0.96 (1)
Only the 470 μF capacitor stays above 0.95 of the peak (1)
So the 470 μF capacitor is the smallest that meets the requirement (1)

Question 19

One of the four diodes in a bridge rectifier fails so that it no longer conducts. Deduce and explain the effect on the smoothed output across the load.

Accepted Answer

In one half-cycle the conducting path through the bridge is broken, so no current reaches the load; in the other half-cycle the remaining pair still conducts (1). The output becomes half-wave rectified, with peaks 20 ms apart instead of 10 ms (1). The capacitor must supply the load for twice as long between top-ups, so the ripple of the smoothed output increases (1).