Piecewise Linear Model of Diode Questions and Answers

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
View Answer

Answer: b
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
View Answer

Answer: d
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
View Answer

Answer: b
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.

Equivalent circuit when effective voltage across forward bias mode diode is 1 volt

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
View Answer

Answer: a
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
View Answer

Answer: c
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.

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6. In the circuit shown below voltage V_in is 3V and V_B is 1V. The resistor R₁ 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

Answer: c
Explanation: Since D1 and D₂ are forward biased we can replace them with their equivalent diagram.

D1 & D2 are forward biased hence replaced 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 R₁ (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 R₁ 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
View Answer

Answer: b
Explanation: Since both diodes are in reverse bias mode applied voltage V_in will appear on V_out. Diode D1 and D₂ 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 R₁ 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

Answer: a
Explanation:

Current passing through diode in reverse biased mode D2 by kirchoff’s voltage rule

Assume I be the current through the circuit. By kirchoff’s voltage rule,
V_out = IR2 ————(1)
Current through R₁ (V_in-V_out-V_D)/1.01k = (2.3-V_out)/1010
Current through diode D₂ is 0 since D₂ 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_B1 is -1V and V_B2 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

Answer: a
Explanation: In this condition both diodes are forward biased

This circuit can be further reduced to by assuming V_B as 1V

Circuit reduced by assuming VB as 1V by network analysis using kirchoff’s voltage rule

By network analysis using kirchoff’s voltage rule current through RD₂ 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