This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Piecewise Linear Model of Diode-2”.

1. In the given circuit input voltage V_{in} is 3V and V_{B} is 1.5V. The resistance R is 1.5K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The Voltage V will be ____________

a) 2.7V

b) 3V

c) 0.8V

d) 1.5V

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Explanation: In piecewise linear model, we can consider diode as a voltage source of 0.5V along with an ideal diode. V

_{IN}being applied is 3V, which is greater than 1.5-0.5 V, and hence the diode is reverse biased and input appears at output.

2. In the circuit shown below voltage V_{in} is 3V and V_{B} is 2V. The resistor R is 1K. Cut-in voltage of diode is 0.7V. The voltage V is __________

a) 0.3V

b) 1V

c) 3V

d) 0V

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Explanation: In piecewise linear model, we can consider diode as a voltage source of 0.7V along with an ideal diode. V

_{IN}being applied is 3V, which is greater than 2-0.7 V, and hence the diode is reverse biased and considering it open, no output is obtained.

3. In the circuit shown below voltage V_{in} is -3V and V_{B} is -2V. The resistor R is 1K. Cut-in voltage of diode is 0.7V. Forward bias resistance is 10Ω. The voltage V is __________

a) -0.29V

b) -4.25V

c) 4.25V

d) 2.9V

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Explanation: Effective voltage across diode is one volt. Hence diode is in forward bias mode. So we can apply equivalent circuit of diode.

Net voltage through R and RD is -3-2+0.7 = -4.3V

Current through the circuit is -4.3/(R+RD) = -4.3/(1010) = -0.00425A=-4.25 mA

Hence voltage across R is 1Kx(-4.25mA) = -4.25V.

4. In the circuit shown below current I is 2mA and V_{B} is 1V. The resistor R is 1K. Cut-in voltage of diode is 0.7V. Forward bias resistance is 10Ω. The voltage V is __________

a) 3V

b) 2V

c) 1V

d) 0.3V

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Explanation: Since current source is reverse bias to the diode current passes through resistor R. Voltage across resistor R is 2mA x 1k =2V. Since resistor and diodes are parallel net output voltage V is V

_{B}+ voltage across resistor R = 1+2 = 3V.

5. In the circuit shown below current I is 2mA and V_{B} is -1V. The resistor R is 1K. Cut-in voltage of diode is 0.7V. Forward bias resistance is 10Ω. The voltage V is __________

a) 1.3V

b) 0.3V

c) 1V

d) 2V

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Explanation: Since current source is reverse bias to the diode current passes through resistor R. Voltage across resistor R is 2mA x 1k =2V. Since resistor and diodes are parallel net output voltage V is V

_{B}+ voltage across resistor R = -1+2 = 1V.

6. In the circuit shown below voltage V_{in} is 3V and V_{B} is 1V. The resistor R_{1} and R2 are 1K. Assume both diodes are identical. Forward bias resistance is 10Ω. Cut-in voltage of diode is 0.7V. The voltage V_{out} is __________

a) 1.235V

b) 0.234V

c) 1.314V

d) 1.564V

View Answer

Explanation: Since D1 and D

_{2}are forward biased we can replace them with their equivalent diagram.

Assume I be the current through the circuit. By kirchoff’s voltage rule,

V

_{out}= -V

_{D}+IRD+IR2 ————(1)

Current through R

_{1}(V

_{in}-V

_{out})/1k = (3-V

_{out})/1k

Current through diode is (V

_{in}-V

_{out}-V

_{D}-V

_{B})/RD = (0.3-V

_{out})/10

Total current I = (3-V

_{out})/1000+(1.3-V

_{out})/10 = 1.330-1.010V

_{out}

Substitute this in eq(1)

That is, V

_{out}= -0.7 + 1010(1.33-1.01V

_{out})

1021V

_{out}= 1342. Therefore, V

_{out}= 1.314V.

7. In the circuit shown below voltage V_{in} is -3V and V_{B} is -1V. The resistor R_{1} and R2 are 1K. Assume both diodes are identical. Forward bias resistance is 10Ω .Cut-in voltage of diode is 0.7V. The voltage V_{out} is __________

a) -2V

b) -3V

c) -1V

d) -.0.7V

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Explanation: Since both diodes are in reverse bias mode applied voltage V

_{in}will appear on V

_{out}. Diode D1 and D

_{2}disappears and leaves the terminal as open.

8. In the circuit shown below voltage V_{in} is 3V and V_{B} is 1V. The resistor R_{1} and R2 are 1K. Assume both diodes are identical. Forward bias resistance is 10Ω. Cut-in voltage of diode is 0.7V. The voltage V_{out} is __________

a) 1.14V

b) 1.23V

c) 0.32V

d) 1.34V

View Answer

Explanation:

Assume I be the current through the circuit. By kirchoff’s voltage rule,

V

_{out}= IR2 ————(1)

Current through R

_{1}(V

_{in}-V

_{out}-V

_{D})/1.01k = (2.3-V

_{out})/1010

Current through diode D

_{2}is 0 since D

_{2}is in reverse bias mode.

current I = (2.3-V

_{out})/1010

Substitute this in eq(1)

That is, V

_{out}=(2.3-V

_{out})/1010 x 1000 => 1.99V

_{out}= 2.27 => V

_{out}= 2.27/1.99 =1.144V.

9. In the circuit shown below voltage V_{in} is 3V and V_{B}1 is -1V and V_{B}2 is 1V. Assume both diodes are identical. Cut-in voltage of diode is 0.7V. Forward bias resistance is 10Ω. The voltage V_{out} is __________

a) 0.6V

b) 1V

c) 1.7V

d) 2V

View Answer

Explanation: In this condition both diodes are forward biased

This circuit can be further reduced to by assuming V

_{B}as 1V

By network analysis using kirchoff’s voltage rule current through RD

_{2}will be 0.09A.

The voltage in V

_{out}will be V

_{B}+V

_{D}-0.09×10 = 0.6V.

10. In the circuit shown below, cut-in voltage of diode is 0.7V. Forward bias resistance is 10Ω. The voltage V is?

a) 0.69V

b) 0.7V

c) 0.68V

d) 0.72V

View Answer

Explanation: Since diode is in forward bias we can replace diode with voltage source of 0.7V and resistor of resistance 10Ω.

V

_{out}will be IRD+V

_{D}= 2mAx10 + 0.7 = 0.72V.

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