Diode Circuits and Applications Questions and Answers

Q: What is the peak inverse voltage in a bridge rectifier?

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.

Q: What is the ripple frequency of a full-wave rectifier on a fifty hertz supply?

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.

Q: Using both half cycles instead, the ripple factor measured at the output is

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.

Q: A rectifier's output has a direct component of \( 20 \ \mathrm{V} \) and a ripple of \( 2 \ \mathrm{V} \) root-mean-square. Its ripple factor is

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.

Q: Why does a capacitor filter's ripple fall as the load resistance rises?

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.

Q: Why is an inductor input filter preferred for heavy loads?

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.

11.

What is the peak inverse voltage across the diode in a half-wave rectifier?

A.The peak of the input
B.Twice the peak of the input
C.Half the peak of the input
D.The peak minus one forward drop

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.

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12.

What is the peak inverse voltage in a centre-tapped full-wave rectifier?

A.The peak of the whole secondary less one drop
B.Twice the peak of half the secondary
C.The peak of half the secondary winding only
D.Half the peak of the whole secondary winding

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.

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13.

What is the peak inverse voltage in a bridge rectifier?

A.Half the peak of the secondary
B.Four times the peak of the secondary
C.The peak of the secondary
D.Twice the peak of the secondary

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.

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14.

What is the ripple frequency of a full-wave rectifier on a fifty hertz supply?

A.50 Hz
B.25 Hz
C.200 Hz
D.100 Hz

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.

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15.

A half-wave rectifier has a ripple factor of

A.1.21
B.0.48
C.0.707
D.1.57

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.

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16.

Using both half cycles instead, the ripple factor measured at the output is

A.1.11
B.0.48
C.1.21
D.0.707

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.

Discuss in Forum

17.

A rectifier's output has a direct component of \( 20 \ \mathrm{V} \) and a ripple of \( 2 \ \mathrm{V} \) root-mean-square. Its ripple factor is

A.10
B.0.05
C.0.1
D.0.2

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.

Discuss in Forum

18.

Why does a capacitor filter's ripple fall as the load resistance rises?

A.The rectifier conducts for a longer part of each cycle
B.The transformer's regulation improves at light load
C.The diode's forward drop becomes a smaller proportion
D.The capacitor discharges more slowly between peaks

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.

Discuss in Forum

19.

What is the chief drawback of a capacitor input filter at heavy load?

A.Large peak currents through the diodes
B.A ripple frequency twice the input frequency
C.A direct output far below the input's peak value
D.A conduction angle approaching the whole cycle

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.

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20.

Why is an inductor input filter preferred for heavy loads?

A.It removes the need for any capacitance at all
B.It smooths the current the diodes must carry
C.It gives a higher direct output than a capacitor filter
D.It reduces the peak inverse voltage the diodes see

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.

Discuss in Forum

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