1. After cut-in voltage in piecewise linear model diode act as a

a) Resistor

b) Capacitor

c) Conductor

d) Insulator

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2. Reverse biased condition of a diode in piecewise linear model is equivalent to

a) Resistor

b) Capacitor

c) Conductor

d) Insulator

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3. Voltage drop produced by a diode in piecewise linear mode is

a) Constant and equal to knee voltage

b) Varying linearly with voltage

c) Varies exponentially with voltage

d) Constant and equal to twice of knee voltage

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4. In the given circuit voltage V = 2V.cut-in voltage of diode is 0.7V. The current I through the circuit is

(Assume piecewise linear model for diode)

a) 0.235mA

b) 1.29mA

c) 1.63mA

d) 2.27mA

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_{D})/R

_{1}+RD = (3-0.7)/1010 = 2.27mA.

5. In the given circuit input voltage Vin is 3V and V_{2} is 1V. The resistance R_{1} is 1.5K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The Voltage V will be

a) 2.03mA

b) 0.23mA

c) 1.58mA

d) 1.33mA

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_{1}-V

_{2}/R = 3-1/1.5k = 1.33mA.

6. In the given circuit input voltage V_{1} is -3V and V_{2} is -1V. The resistance R_{1} is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The Voltage V will be

(Use piecewise linear model of diode)

a) -1mA

b) -2mA

c) -0.2mA

d) -0.1mA

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7. In the given circuit input voltage V is 2V and V_{B} is 1V. The resistance R_{1} and R_{2} is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be

(Use piecewise linear model of diode)

a) 0.29mA

b) 0.21mA

c) 0.36mA

d) 0.15mA

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_{B}= 1V forward biases the diode, we can use equivalent circuit of diode as follows Current through R

_{1}, I1 = 1V/R

_{1}= 1mA. Current through R

_{2}, I2 = (1-V

_{D})/(R

_{2}¬+RD) = (1-0.7)/1010 = 0.297mA.

8. In the given circuit input voltage V is 3V and V_{B} is 1V. The resistance R_{1} and R_{2} is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be

(Use piecewise linear model of diode)

a) 0.96mA

b) 2.13mA

c) 1.56mA

d) 1.23ma

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_{0})/R

_{1}+(2-V

_{0}-V

_{B})/R

_{2}=(2-V

_{0})/1000+((1-V

_{0}))/1000 …………………….(1) V0 = V

_{D}+ IRD = 0.7+10I Put this value in eq(1) That is, I = (2-0.7-10I)/1000+((1-0.7-10I))/1000 => 1000I = 1.6 – 20I => 1020 I = 1.6 That is, I = 1.6/1020 = 1.56mA.

9. In the given circuit input voltage V is -3V and V_{B} is 1V. The resistance R_{1} and R_{2} is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be

(Use piecewise linear model of diode)

a) 1.2mA

b) 0mA

c) 0.8mA

d) 1mA

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

(Use piecewise linear model of diode)

a) 1V

b) 3V

c) 2.3V

d) 1.3V

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_{B}, the voltage Vin will appear on V. Hence V will be 3V.