Diode Circuits and Applications Questions and Answers

Answer: D

The minority carrier charge stored during conduction must be swept out before the diode can block, and that reverse recovery time sets the speed. Depletion formation is fast, the package's thermal constant is far slower but irrelevant to switching, and establishing the forward drop is quick. Stored charge is why Schottky diodes, which store almost none, switch so much faster, and it is the property to look for in any fast switching application. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: A

With no load current flowing on the reverse half cycle, the diode stands off the full input peak. Twice the peak is the figure for a centre-tapped full-wave rectifier, where the non-conducting diode sees both halves of the winding. Half the peak and the peak less a drop arise from neither topology. Peak inverse voltage is the rating that most often decides a diode choice, and it differs between topologies that look superficially similar. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: B

The idle diode has one half-winding's peak on its cathode and the other's negative peak on its anode, so it stands off twice the peak of half the secondary. That is the topology's chief disadvantage against a bridge, which needs only the single peak. The remaining options understate it. Working out what the non-conducting device sees, rather than what the conducting one carries, is the habit this rating rewards. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: C

Each diode in a bridge stands off only the secondary's peak, because the two non-conducting diodes share the reverse voltage rather than one taking it all. That halving of the requirement, together with the absence of a centre tap, is why the bridge is the standard choice despite needing four diodes instead of two. Counting devices is a poor guide to cost when the ratings differ by a factor of two. The shared properties explain why these schemes form a family, and the differences explain why more than one exists.

Answer: D

Both half cycles are used, so the output has two humps per input cycle and a ripple frequency of 100 Hz. A half-wave rectifier would give 50 Hz. The remaining figures follow from neither. The doubled frequency is the full-wave rectifier's second advantage after efficiency: a filter has half as long to sag between peaks, so the same capacitor leaves half the ripple.

Answer: A

For a half-wave output the ripple factor follows from the form factor:

\[ \begin{aligned} r &= \sqrt{\left(\tfrac{I_{rms}}{I_{dc}}\right)^{2} - 1} \\ &= \sqrt{1.57^{2}-1} = 1.21 \end{aligned} \]

so 1.21, meaning the ripple exceeds the direct component. 0.48 is the full-wave figure. 0.707 and 1.57 are the root-mean-square factor and the form factor themselves. A ripple factor above unity is the clearest statement of why a half-wave rectifier needs heavy filtering. Working one numerical case through both schemes side by side fixes the distinction better than any rule of thumb.

Answer: B

Using the full-wave form factor:

\[ r = \sqrt{1.11^{2}-1} = 0.48 \]

giving 0.48, less than half the half-wave value even before any filtering. 1.21 is the half-wave figure and 1.11 is the form factor used in the calculation. The improvement comes from using both half cycles, which raises the direct component and halves the gap the ripple has to bridge - two effects working the same way.

Answer: C

The definition is the ratio of the two:

\[ r = \frac{2}{20} = 0.1 \]

so 0.1, or ten per cent - respectable for a simple capacitor filter. Ten inverts the ratio, and the remaining values double or halve it. Expressing ripple as a fraction of the direct output rather than as an absolute voltage is what makes two supplies of different voltages comparable.

Answer: D

A larger load resistance means a longer discharge time constant, so the capacitor sags less between one peak and the next. The conduction angle actually shortens at light load rather than lengthening. Transformer regulation and the diode drop affect the output level rather than the ripple. Ripple in a capacitor filter is always a story about how far the capacitor discharges, and every factor that matters acts through that. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.

Answer: A

The capacitor is recharged in a short burst near each peak, so the diodes and transformer carry peak currents many times the average, which is what limits the topology at high power. The doubled ripple frequency is a benefit. The output falling below the peak is expected and modest. And a wide conduction angle would actually reduce the peak currents. A filter that stores energy must refill its store quickly, and the peak current is the price.