Diode Circuits and Applications Questions and Answers

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.

Answer: B

An inductor opposes changes in current, so conduction spreads across the cycle and the peak currents fall - exactly the problem that limits capacitor filters at high power. Its direct output is lower rather than higher, the peak inverse voltage is unaffected, and a capacitor usually follows it in an LC arrangement. Choosing which element comes first decides whether the filter smooths voltage or current, and heavy loads care about the current. Most errors on this topic come from carrying a result across from a neighbouring scheme where it does not hold.

Answer: C

A clipper cuts off whatever lies beyond a reference level, leaving the rest untouched, which is how a waveform is limited or a peak protected. Shifting the direct level is a clamper's job, multiplying the peak is a voltage multiplier's, and pulse generation is something else again. Clipping and clamping are constantly confused because both use a diode and a reference, so it is worth fixing which one alters the shape and which one alters the position.

Answer: D

A clamper adds a direct offset so that the positive or negative peak rests at a chosen level, leaving the waveform's shape and amplitude intact. Removing part of the waveform is clipping, amplitude reduction is attenuation, and keeping one polarity is rectification. The distinction matters in practice because a clamper is used to restore a direct level that capacitive coupling has removed, which no clipper could do. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.