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Rectification and smoothing questions
The mains supply is alternating and almost every device runs on direct current, so the conversion happens inside every charger. Diodes force the current one way, half-wave with one and full-wave with four in a bridge, and a capacitor across the output fills in the gaps between peaks. This is examined directly by CIE.
19 original questions · 50 marks · the rectification and smoothing notes · Magnetic fields
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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.
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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).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.
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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).State one disadvantage of half-wave rectification compared with full-wave rectification, and one advantage in terms of components.
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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).A sinusoidal 50 Hz supply is full-wave rectified. State the number of voltage peaks per second in the output.
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100 per second: full-wave rectification gives one peak every half-cycle (1).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.
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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).State what is meant by the ripple of a smoothed supply and what causes it.
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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).A bridge rectifier contains four diodes. Explain how it produces full-wave rectification. Refer to which diodes conduct in each half-cycle.
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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).Explain how a capacitor connected across the load smooths the output of a full-wave rectifier.
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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).A smoothing capacitor of 2200 μF discharges through a load of 500 Ω. Calculate the time constant of the discharge.
State and explain the effect on the ripple of a smoothed supply of increasing the capacitance of the reservoir capacitor.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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