This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Half-Wave Rectifier”.

1. The diode in a half wave rectifier has a forward resistance RF. The voltage is V_{m}sinωt and the load resistance is RL. The DC current is given by _________

a) V_{m}/√2R_{L}

b) V_{m}/(R_{F}+R_{L})π

c) 2V_{m}/√π

d) V_{m}/R_{L}

View Answer

Explanation: For a half wave rectifier, the I

_{DC}=I

_{AVG}=I

_{m}/π

I= V

_{m}sinωt/(R

_{F}+R

_{L})=I

_{m}sinωt

I

_{m}=V

_{m}/ R

_{F}+R

_{L}So, I

_{DC}=I

_{m}/π=V

_{m}/(R

_{F}+R

_{L}).

2. The below figure arrives to a conclusion that _________

a) for V_{i} > 0, V_{0}=-(R_{2}/R_{1})V_{i}

b) for V_{i} > 0, V_{0}=0

c) V_{i} < 0, V_{0}=-(R_{2}/R_{1})V_{i}

d) V_{i} < 0, V_{0}=0

View Answer

Explanation: The given op-amp is in inverting mode and this makes the output voltage to have a phase shift of 180°. The output voltage is now negative. So, the diode 1 is reverse biased and diode 2 is forward biased. Then output is clearly zero.

3. What is the output as a function of the input voltage (for positive values) for the given figure. Assume it’s an ideal op-amp with zero forward drop (D_{i}=0)

a) 0

b) -V_{i}

c) V_{i}

d) 2V_{i}

View Answer

Explanation: When the input of the inverted mode op-amp is positive, the output is negative.

The diode is reverse biased. The input appears at the output.

4. In a half wave rectifier, the sine wave input is 50sin50t. If the load resistance is of 1K, then average DC power output will be?

a) 3.99V

b) 2.5V

c) 5.97V

d) 6.77V

View Answer

Explanation: The standard form of a sine wave is V

_{m}sinωt. BY comparing the given information with this equation, V

_{m}=50.

Power=V

_{m}

^{2}/R

_{L}=50*50/1000=2.5V.

5. In a half wave rectifier, the sine wave input is 200sin300t. The average value of output voltage is?

a) 57.876V

b) 67.453V

c) 63.694V

d) 76.987V

View Answer

Explanation: Comparing with the standard equation, V

_{m}=200V.

Average value is given by, V

_{avg}=V

_{m}/π.

So, 200/π=63.694.

6. Efficiency of a half wave rectifier is

a) 50%

b) 60%

c) 40.6%

d) 46%

View Answer

Explanation: Efficiency of a rectifier is the effectiveness to convert AC to DC. For half wave it’s 40.6%. It’s given by, V

_{out}/V

_{in}*100.

7. If peak voltage for a half wave rectifier circuit is 5V and diode cut in voltage is 0.7, then peak inverse voltage on diode will be?

a) 5V

b) 4.9V

c) 4.3V

d) 6.7V

View Answer

Explanation: PIV is the maximum reverse bias voltage that can be appeared across a diode in the given circuit, If the PIV rating is less than this value of breakdown of diode will occur. For a rectifier, PIV=V

_{m}-V

_{d}=5-0.7=4.3V.

8. Transformer utilisation factor of a half wave rectifier is _________

a) 0.234

b) 0.279

c) 0.287

d) 0.453

View Answer

Explanation: Transformer utilisation factor is the ratio of AC power delivered to load to the DC power rating. This factor indicates effectiveness of transformer usage by rectifier. For a half wave rectifier, it’s low and equal to 0.287.

9. If the input frequency of a half wave rectifier is 100Hz, then the ripple frequency will be_________

a) 150Hz

b) 200Hz

c) 100Hz

d) 300Hz

View Answer

Explanation: The ripple frequency of the output and input is same. This is because, one half cycle of input is passed and other half cycle is seized. So, effectively the frequency is the same.

10. Ripple factor of a half wave rectifier is_________(I_{m} is the peak current and RL is load resistance)

a) 1.414

b) 1.21

c) 1.4

d) 0.48

View Answer

Explanation: The ripple factor of a rectifier is the measure of disturbances produced in the output. It’s the effectiveness of a power supply filter to reduce the ripple voltage. The ratio of ripple voltage to DC output voltage is ripple factor which is 1.21.

**Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.**

To practice all areas of Electronic Devices and Circuits, __here is complete set of 1000+ Multiple Choice Questions and Answers__.