# Electronics Quiz On Diode Applications!

Reviewed by Matt Balanda
Matt Balanda, BS, Science |
Physics Expert
Review Board Member
Matt graduated with a Master's in Educational Leadership for Faith-Based Schools from California Baptist University and a Bachelor's of Science in Aerospace Engineering and Mathematics from the University of Arizona. A devoted leader, transitioned from Aerospace Engineering to inspire students. As the High School Vice-Principal and a skilled Physics teacher at Calvary Chapel Christian School, his passion is nurturing a love for learning and deepening students' connection with God, fostering a transformative educational journey.
, BS, Science
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Do you know about diodes and their applications? Play this electronics quiz based on diode applications to check your understanding of electronic devices. A diode is a two-terminal electrical component that allows the current to flow easily in one direction. Basically, it acts as a one-way switch for current. In the quiz below, you can learn and practice electronics questions mainly based on diodes and applications. So, what are you waiting for? Take the quiz, then.

## Diode Applications Questions and Answers

• 1.

### The average value of a half-wave rectified voltage with a peak value of 200V is

• A.

63.7V

• B.

127.3 V

• C.

141 V

• D.

0 V

A. 63.7V
Explanation
The average value of a half-wave rectified voltage can be calculated by taking the peak value and dividing it by Ï€. In this case, the peak value is 200V, so dividing it by Ï€ (approximately 3.14) gives us an average value of approximately 63.7V.

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• 2.

### When a 60Hz sinusoidal voltage is applied to the input of a half-wave rectifier, the output frequency is

• A.

120 Hz

• B.

30 Hz

• C.

60 Hz

• D.

0 Hz

C. 60 Hz
Explanation
The output frequency of a half-wave rectifier is the same as the input frequency. In this case, the input frequency is 60Hz, so the output frequency will also be 60Hz.

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• 3.

### The peak value of the input to the half-wave rectifier is 10V. The approximate peak value of the output is

• A.

10 V

• B.

3.18 V

• C.

10.7 V

• D.

9.3 V

D. 9.3 V
Explanation
The approximate peak value of the output for a half-wave rectifier is 0.7 times the peak value of the input. In this case, the peak value of the input is 10V, so the approximate peak value of the output would be 0.7 times 10V, which is 7V. However, since the question asks for the "approximate" peak value, the closest option to 7V is 9.3V.

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• 4.

### The diode, in a half wave rectifier with a peak input of 10V, must be able to withstand a reverse voltage of

• A.

10 V

• B.

5 V

• C.

20 V

• D.

3.18 V

A. 10 V
Explanation
In a half wave rectifier, the diode allows current to flow in only one direction, blocking the reverse current. The peak input voltage of 10V indicates the maximum voltage that the diode needs to withstand in the forward direction. Since the diode is not required to withstand any reverse voltage, the correct answer is 10V.

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• 5.

### The average value of a full-wave rectified voltage with a peak value of 75V is

• A.

53 V

• B.

47.8 V

• C.

37.5 V

• D.

23.9 V

B. 47.8 V
Explanation
The average value of a full-wave rectified voltage can be calculated by taking the peak value and dividing it by Ï€. In this case, the peak value is 75V, so the average value would be 75V/Ï€ = 47.8V.

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• 6.

### When a 60Hz sinusoidal voltage is applied to the input of a full-wave rectifier, the output frequency is

• A.

120 Hz

• B.

60 Hz

• C.

240 Hz

• D.

0 Hz

A. 120 Hz
Explanation
When a 60Hz sinusoidal voltage is applied to the input of a full-wave rectifier, the output frequency is 120 Hz. This is because a full-wave rectifier converts both the positive and negative halves of the input voltage into positive voltage pulses. As a result, the output waveform has twice the frequency of the input waveform. Since the input frequency is 60Hz, the output frequency will be 120Hz.

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• 7.

### The total secondary voltage in a center-tapped full-wave rectifier is 125 Vrms. Neglecting the diode drop, the rms output voltage is

• A.

125 V

• B.

177 V

• C.

100 V

• D.

62.5 V

D. 62.5 V
Explanation
In a center-tapped full-wave rectifier, the secondary voltage is divided equally between the two halves of the secondary winding. Since the total secondary voltage is given as 125Vrms, each half of the secondary winding will have a voltage of 125/2 = 62.5Vrms. The rms output voltage is equal to the voltage across one half of the secondary winding, so the correct answer is 62.5V.

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• 8.

### When the peak output voltage is 100V, the PIV for each diode in a center-tapped full-wave rectifier is (neglecting the diode drop)

• A.

100 V

• B.

200 V

• C.

141 V

• D.

50 V

B. 200 V
Explanation
In a center-tapped full-wave rectifier, each diode conducts during half of the input cycle. When the peak output voltage is 100V, the voltage across each diode will be equal to half of the peak output voltage, which is 50V. Since the diodes are connected in series, the total voltage across the two diodes will be 2 times the voltage across each diode, which is 2 * 50V = 100V. Therefore, the peak inverse voltage (PIV) for each diode in this rectifier is 100V.

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• 9.

### When the rms output voltage of a bridge full-wave rectifier is 20V, the peak inverse voltage across the diodes is (neglecting the diode drop)

• A.

20 V

• B.

40 V

• C.

28.3 V

• D.

56.6 V

C. 28.3 V
Explanation
The peak inverse voltage across the diodes in a bridge full-wave rectifier is equal to the peak voltage of the input AC signal. The peak voltage can be calculated by multiplying the rms voltage by the square root of 2. Therefore, the peak inverse voltage across the diodes in this case would be 20V multiplied by the square root of 2, which is approximately 28.3V.

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• 10.

### The ideal dc output voltage of a capacitor-input filter is equal to

• A.

The peak value of the rectified voltage

• B.

The average value of the rectified voltage

• C.

The rms value of the rectified voltage

• D.

Twice the value of the rectified voltage

A. The peak value of the rectified voltage
Explanation
The ideal dc output voltage of a capacitor-input filter is equal to the peak value of the rectified voltage because a capacitor-input filter is designed to smooth out the rectified voltage by storing charge during the peaks of the rectified waveform and releasing it during the troughs. As a result, the output voltage of the filter will closely resemble the peak value of the rectified voltage, providing a stable and continuous dc voltage.

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• 11.

### A certain power-supply filter produces an output with a ripple of 100mV peak-to-peak and a dc value of 20V. The ripple factor is

• A.

0.05

• B.

0.005

• C.

0.00005

• D.

0.02

B. 0.005
Explanation
The ripple factor is a measure of the amount of AC voltage present in the output of a power supply compared to the DC voltage. It is calculated by dividing the peak-to-peak ripple voltage by the DC voltage. In this case, the peak-to-peak ripple voltage is 100mV and the DC voltage is 20V. Dividing 100mV by 20V gives a ripple factor of 0.005.

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• 12.

### A 60V peak full-wave rectified voltage is applied to a capacitor-input filter. If f = 120 Hz, RL=10 kâ„¦ and C=10 ÂµF, the ripple voltage is

• A.

0.6 V

• B.

6 mV

• C.

5.0 V

• D.

2.88 V

C. 5.0 V
Explanation
The ripple voltage in a capacitor-input filter can be calculated using the formula Vr = (1/2) * (1/f) * (1/RL) * (Vp), where Vr is the ripple voltage, f is the frequency of the input voltage, RL is the load resistance, and Vp is the peak voltage of the input voltage. In this case, Vp is 60V, f is 120 Hz, and RL is 10 kâ„¦. Plugging these values into the formula, we get Vr = (1/2) * (1/120) * (1/10,000) * (60) = 0.6 V. Therefore, the correct answer is 0.6 V.

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• 13.

### If the load resistance of a capacitor-filtered full-wave rectifier is reduced, the ripple voltage

• A.

Increases

• B.

Is not affected

• C.

Decreases

• D.

Has a different frequency

A. Increases
Explanation
When the load resistance of a capacitor-filtered full-wave rectifier is reduced, it means that the load is drawing more current from the rectifier. This increased current causes the capacitor to discharge more quickly during the off cycles of the rectifier, resulting in a larger voltage drop across the load resistor. As a result, the ripple voltage, which is the variation in the output voltage, increases.

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• 14.

### Line regulation is determined by

• A.

• B.

• C.

• D.

Changes in zener current and load current

D. Changes in zener current and load current
Explanation
Line regulation refers to the ability of a voltage regulator to maintain a stable output voltage despite fluctuations in the input voltage. In this case, changes in both the zener current and load current affect the line regulation. When the load current changes, it can cause variations in the output voltage. Similarly, changes in the zener current, which is the current flowing through the zener diode used in the voltage regulation circuit, can also impact the output voltage. Therefore, both changes in zener current and load current play a role in determining the line regulation of the voltage regulator.

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• 15.

### Load regulation is determined by

• A.

Changes in load current and input voltage

• B.

Changes in load current and output voltage

• C.

Changes in load resistance and input voltage

• D.

Changes in zener current and load current

B. Changes in load current and output voltage
Explanation
Load regulation refers to the ability of a power supply to maintain a stable output voltage despite variations in the load current. In other words, it measures how well the power supply can handle changes in the load current without significant fluctuations in the output voltage. Therefore, changes in load current and output voltage are the key factors that determine load regulation.

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• 16.

### A 10 V peak-to-peak sinusoidal voltage is applied across a silicon diode and series resistor. The maximum voltage across the diode is

• A.

9.3 V

• B.

5 V

• C.

0.7 V

• D.

10 V

• E.

4.3 V

D. 10 V
Explanation
The maximum voltage across the diode is 10 V because the peak-to-peak voltage applied across the diode is 10 V. Since the diode is forward biased, it conducts current and drops a constant voltage of approximately 0.7 V. Therefore, the remaining voltage of 9.3 V appears across the diode.

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• 17.

### In the input voltage to a voltage tripler has an rms value of 12 V, the dc output voltage is approximately

• A.

36 V

• B.

50.9 V

• C.

33.9 V

• D.

32.4 V

B. 50.9 V
Explanation
When a voltage tripler circuit is used, the output voltage is approximately three times the input voltage. In this case, the input voltage has an rms value of 12 V, so the dc output voltage would be approximately 3 times that value, which is 36 V. Therefore, the correct answer is 36 V.

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• 18.

### If one of the diode in a bridge full-wave rectifier opens, the output is

• A.

0 V

• B.

A half-wave rectified voltage

• C.

One-fourth the amplitude of the input voltage

• D.

A 120 Hz voltage

B. A half-wave rectified voltage
Explanation
If one of the diodes in a bridge full-wave rectifier opens, it means that the current can only flow through one half of the input AC cycle. As a result, the output will be a half-wave rectified voltage, where only the positive or negative half of the input waveform is present in the output. The other half will be blocked due to the open diode. Therefore, the correct answer is a half-wave rectified voltage.

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• 19.

### If you are checking a 60Hz full-wave bridge rectifier and observe that the output has a 60Hz ripple,

• A.

The circuit is working properly

• B.

The transformer secondary is shorted

• C.

There is an open diode

• D.

The filter capacitor is leaky

B. The transformer secondary is shorted
Explanation
If the output of a 60Hz full-wave bridge rectifier has a 60Hz ripple, it suggests that the transformer secondary is shorted. This means that there is a fault in the transformer, causing a direct connection between the primary and secondary windings. As a result, the rectifier is unable to properly convert the AC input into a smooth DC output, leading to the observed ripple.

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• 20.

### If one of the diode in a half-wave rectifier opens, the output is

• A.

0 V

• B.

A half-wave rectified voltage

• C.

One-fourth the amplitude of the input voltage

• D.

A 120 Hz voltage

A. 0 V
Explanation
If one of the diodes in a half-wave rectifier opens, it means that there is an open circuit in the circuitry. In this case, no current can flow through the circuit, and as a result, there will be no voltage output. Therefore, the output voltage will be 0 V.

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• 21.

### If one of the diodes in a bridge rectifier shorts,

• A.

The shorted diode may still function

• B.

The transformer will be damaged

• C.

The circuit will be half-wave rectifier

• D.

B. The transformer will be damaged
Explanation
If one of the diodes in a bridge rectifier shorts, it creates a direct short circuit across the transformer secondary winding. This causes a high current to flow through the transformer, which can lead to overheating and damage to the transformer. Therefore, the correct answer is that the transformer will be damaged.

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• 22.

### When the primary winding of the transformer in a power supply opens, output voltage is

• A.

0 V

• B.

A half-wave rectified voltage

• C.

One-fourth the amplitude of the input voltage

• D.

A 120 Hz voltage

A. 0 V
Explanation
When the primary winding of the transformer in a power supply opens, it means that there is no current flowing through the primary coil. As a result, there is no magnetic field generated, and therefore no induction of voltage in the secondary winding. This leads to an output voltage of 0 V.

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• 23.

### In a half-wave rectifier, the current through the load is approximately

• A.

The same as the input cycle

• B.

50% of the input cycle

• C.

One-fourth the amplitude of the input current

• D.

A 60Hz current

B. 50% of the input cycle
Explanation
In a half-wave rectifier, the current through the load is approximately 50% of the input cycle. This is because in a half-wave rectifier, only one half of the input cycle is allowed to pass through to the load. The other half is blocked by the rectifier. Therefore, the current through the load is only present during that allowed half of the input cycle, resulting in approximately 50% of the input cycle.

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• 24.

### Which of the following is true?

• A.

A full-wave voltage occurs on each half of the input cycle and has the frequency of twice the input frequency.

• B.

A half-wave voltage occurs once each input cycle and has the frequency equal to the half of the input frequency.

• C.

A full-wave voltage occurs on each half of the input cycle and has the period twice the input period.

• D.

A half-wave voltage occurs once each half of the input cycle and has the frequency equal to the half of the input frequency.

A. A full-wave voltage occurs on each half of the input cycle and has the frequency of twice the input frequency.
Explanation
The correct answer states that a full-wave voltage occurs on each half of the input cycle and has a frequency that is twice the input frequency. This means that during each cycle of the input signal, there are two full-wave voltages. This is because a full-wave voltage is produced during both the positive and negative half-cycles of the input signal. Additionally, the frequency of the full-wave voltage is twice the input frequency because it occurs twice within each cycle. Therefore, this answer accurately describes the characteristics of a full-wave voltage.

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• 25.

### Which type of full-wave rectifier has the greater output voltage for the same input voltage and transformer turns ratio?

• A.

Center-tapped full-wave rectifier

• B.

Bridge full-wave rectifier

• C.

Non-center-tapped full-wave rectifier

• D.

All (a), (b) and (c) have the same output voltage

B. Bridge full-wave rectifier
Explanation
The bridge full-wave rectifier has the greater output voltage for the same input voltage and transformer turns ratio compared to the center-tapped and non-center-tapped full-wave rectifiers. This is because the bridge rectifier uses four diodes in a bridge configuration, allowing it to utilize both halves of the input AC waveform. As a result, it can provide a higher output voltage compared to the other types of full-wave rectifiers.

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• 26.

### What is the average value of a full-wave rectified voltage with a peak value of 60V?

• A.

19.1 V

• B.

38.1 V

• C.

84.8 V

• D.

42.4 V

B. 38.1 V
Explanation
The average value of a full-wave rectified voltage can be found by taking the peak value and dividing it by Ï€. In this case, the peak value is given as 60V, so dividing it by Ï€ gives us an average value of approximately 19.1V. Therefore, the correct answer is 19.1V.

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• 27.

### What is the maximum voltage across an unbiased positive silicon diode limiter during the positive alternation of the input voltage?

• A.

+1.4 V

• B.

+0.7 V

• C.

-1.4 V

• D.

-0.7 V

B. +0.7 V
Explanation
During the positive alternation of the input voltage, the maximum voltage across an unbiased positive silicon diode limiter is +0.7 V. This is because a silicon diode has a forward voltage drop of approximately 0.7 V when it is conducting in the forward bias direction. Therefore, the diode will start conducting when the input voltage exceeds 0.7 V, limiting the voltage across it to +0.7 V.

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• 28.

### Diode _____ add a dc level to an ac voltage.

• A.

Limiters

• B.

Clampers

• C.

Voltage multipliers

• D.

Rectifiers

B. Clampers
Explanation
Clampers are used to add a DC level to an AC voltage. They shift the entire waveform up or down by a fixed amount, allowing the AC signal to be biased around a specific DC level. This is achieved by using a diode and a capacitor in series. The diode conducts during the positive half cycle of the AC signal, charging the capacitor to the peak voltage. During the negative half cycle, the diode is reverse biased and the capacitor holds the charge, effectively shifting the waveform. Therefore, clampers are the correct choice for adding a DC level to an AC voltage.

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• 29.

### Which of the following is true?

• A.

The smaller the ripple voltage, the better the filter.

• B.

The higher the ripple voltage, the better the filter.

• C.

The quality of the filter does not depend on the ripple voltage.

• D.

High ripple voltage is caused by the power dissipated in the load resistor.

A. The smaller the ripple voltage, the better the filter.
Explanation
A smaller ripple voltage indicates that the filter is more effective at reducing the fluctuations in the output voltage. This means that the filter is able to smooth out the voltage waveform more efficiently, resulting in a more stable and consistent output. Therefore, a smaller ripple voltage is indicative of a better filter.

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• 30.

### Diode _______ cut off voltage above or below specified levels.

• A.

Limiters

• B.

Clampers

• C.

Voltage multipliers

• D.

Rectifiers

A. Limiters
Explanation
Diodes can be used as limiters to restrict the voltage above or below certain specified levels. Limiters are used to protect electronic circuits from excessive voltage, preventing damage to the components. By allowing current flow only within a certain range, diode limiters ensure that the voltage does not exceed the desired limits.

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• 31.

### A capacitor in a power supply used to reduce the variation of the output voltage from a rectifier.

• A.

Filter

• B.

Mica

• C.

Ceramic

• D.

Mylar

A. Filter
Explanation
A capacitor in a power supply is used as a filter to reduce the variation of the output voltage from a rectifier. It helps smooth out the pulsating DC voltage produced by the rectifier, by storing and releasing electrical energy as needed. This helps to provide a more stable and consistent output voltage, which is important for many electronic devices that require a steady power supply.

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• 32.

### A circuit that converts an ac sinusoidal input voltage into a pulsating dc voltage with two output pulses occurring for each input cycle.

• A.

Half-wave rectifier

• B.

Full-wave rectifier

• C.

Clipper

• D.

Clamper

B. Full-wave rectifier
Explanation
A full-wave rectifier is a circuit that converts an AC sinusoidal input voltage into a pulsating DC voltage with two output pulses occurring for each input cycle. Unlike a half-wave rectifier that only uses half of the input waveform, a full-wave rectifier utilizes both halves of the waveform to produce a smoother and more continuous output. This is achieved by using a bridge rectifier configuration that consists of four diodes. The diodes conduct in alternate pairs, allowing the positive and negative halves of the input voltage to be rectified. As a result, the full-wave rectifier provides a more efficient conversion of AC to DC compared to a half-wave rectifier.

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• 33.

### A circuit that converts an ac sinusoidal input voltage into a pulsating dc voltage with one output pulse occurring for each input cycle.

• A.

Half-wave rectifier

• B.

Full-wave rectifier

• C.

Clipper

• D.

Clamper

A. Half-wave rectifier
Explanation
A half-wave rectifier is a circuit that converts an AC sinusoidal input voltage into a pulsating DC voltage with one output pulse occurring for each input cycle. It does this by allowing only the positive half of the input waveform to pass through while blocking the negative half. This results in a pulsating DC waveform with only one output pulse for each input cycle.

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• 34.

### The maximum value of reverse voltage which occurs at the peak of the input cycle when the diode is reversed-biased.

• A.

Rms voltage

• B.

Peak voltage

• C.

Average voltage

• D.

Peak inverse voltage

D. Peak inverse voltage
Explanation
The peak inverse voltage refers to the maximum reverse voltage that occurs at the peak of the input cycle when a diode is reversed-biased. This is the highest voltage that the diode can withstand in the reverse direction without getting damaged. It is an important parameter to consider when selecting a diode for a particular application, as exceeding the peak inverse voltage can lead to failure of the diode.

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Matt Balanda |BS, Science |
Physics Expert
Matt graduated with a Master's in Educational Leadership for Faith-Based Schools from California Baptist University and a Bachelor's of Science in Aerospace Engineering and Mathematics from the University of Arizona. A devoted leader, transitioned from Aerospace Engineering to inspire students. As the High School Vice-Principal and a skilled Physics teacher at Calvary Chapel Christian School, his passion is nurturing a love for learning and deepening students' connection with God, fostering a transformative educational journey.

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• Mar 01, 2024
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• Jun 17, 2009
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