This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Characteristic Equation of Diode 1”.

1. Calculate the forward bias current of a Si diode when forward bias voltage of 0.4V is applied, the reverse saturation current is 1.17×10^{-9}A and the thermal voltage is 25.2mV.

a) 9.156mA

b) 8.23mA

c) 1.256mA

d) 5.689mA

View Answer

Explanation: Equation for diode current

I=I

_{0}×(e

^{(V/ηVT)}-1) where I

_{0}= reverse saturation current

η = ideality factor

V

_{T}= thermal voltage

V = applied voltage

Since in this question ideality factor is not mentioned it can be taken as one.

I

_{0}= 1.17 x 10

^{-9}A, V

_{T}= 0.0252V, η = 1, V = 0.4V

Therefore, I= 1.17×10

^{-9}xe

^{0.4/0.025}-1 = 9.156mA.

2. Calculate the thermal voltage when the temperature is 25°C.

a) 0V

b) 0V

c) 0.026V

d) 0.25V

View Answer

Explanation: Thermal voltage V

_{T}is given by k T/q

Where k is the boltzman constant and q is the charge of electron. This can be reduced to

V

_{T}= T

_{K}/11600

Therefore, V

_{T}= 298.15/11600 = 0.0257V.

3. Calculate the reverse saturation current of a diode if the current at 0.2V forward bias is 0.1mA at a temperature of 25°C and the ideality factor is 1.5.

a) 5.5x 10^{-9} A

b) 5.5x 10^{-8} A

c) 5.5x 10^{-7} A

d) 5.6x 10^{-10} A

View Answer

Explanation: Equation for diode current

I=I

_{0}×(e

^{(V/ηVT ) }-1) where I

_{0}= reverse saturation current

η = ideality factor

V

_{T}= thermal voltage

V = applied voltage

Here, I = 0.1mA, η = 1.5, V= 0.2V, V

_{T}= T

_{K}/11600

Therefore, V

_{T}at T= 25+273=298 is 298/11600 = 0.0256V.

Therefore, reverse saturation current

I

_{O}=0.00055mA = 5.5×10

^{-7}A.

4. Find the applied voltage on a forward biased diode if the current is 1mA and reverse saturation current is 10^{-10}. Temperature is 25°C and take ideality factor as 1.5.

a) 0.658V

b) 0.726V

c) 0.526V

d) 0.618V

View Answer

Explanation: Equation for diode current

I=I

_{0}×(e

^{(V/ηVT)}-1) where I

_{0}= reverse saturation current

η = ideality factor

V

_{T}= thermal voltage

V = applied voltage

V

_{T}at T= 25+273=298 is 298/11600 = 0.0256V, η = 1.5, I = 1mA, I

_{0}= 10

^{-10}A

5. Find the temperature at which a diode current is 2mA for a diode which has reverse saturation current of 10^{-9} A. The ideality factor is 1.4 and the applied voltage is 0.6V forward bias.

a) 69.65°C

b) 52.26°C

c) 25.23°C

d) 70.23°C

View Answer

Explanation: Equation for diode current

I=I

_{0}×(e

^{(V/ηVT)}-1) where I

_{0}= reverse saturation current

η = ideality factor

V

_{T}= thermal voltage

V = applied voltage

I

_{0}= 10-9A, η = 1.4, V =0.6V, I = 2mA

We know thermal voltage V

_{T}= T

_{K}/11600 .Therefore, T

_{K}= V

_{T}x11600 = 0x11600 = 342.65K = 69.65°C.

6. Consider a silicon diode with η=1.2. Find change in voltage if the current changes from 0.1mA to 10mA.

a) 0.154V

b) 0.143V

c) 0.123V

d) 0.165V

View Answer

Explanation: Equation for diode current

I=I

_{0}×(e

^{(V/ηVT)}-1) where I

_{0}= reverse saturation current

η = ideality factor

V

_{T}= thermal voltage

V = applied voltage

η = 1.2, I2 = 10mA, I1 = 0.1mA and take V

_{T}= 0.026V

7. If current of a diode changes from 1mA to 10mA what will be the change in voltage across the diode. The ideality factor of diode is 1.2.

a) 0.718V

b) 7.18V

c) 0.0718V

d) 0.00728V

View Answer

8. What will be the ratio of final current to initial current of a diode if voltage of a diode changes from 0.7V to 872.5mV. Take ideality factor as 1.5.

a) 90.26

b) 52.36

c) 80.23

d) 83.35

View Answer

9. What will be the current I in the circuit diagram below. Take terminal voltage of diode as 0.7V and I_{0}as 10^{-12}A.

a) 2.4mA

b) 0.9mA

c) 1mA

d) 4mA

View Answer

Explanation: Let V

_{D}be the voltage of diode, then by Kirchoff’s loop rule

3V = V

_{D}+ IR1

This method of assumption contains small error but it is the simplest method.

Let V

_{D}be 0.7V. Now the current I = (3-0.7)/1k = 2.3mA. Now the diode voltage for 2.3mA

V

_{D}= V

_{T}ln(I/I

^{0}) = 0.026 x ln((2.3×10

^{(-3)})/10

^{(-12)}) = 0.5864V.

Now the current becomes (3-0.5864)/1000 = 2.41mA.

10. Find current I through the circuit using characteristic equation of diode. The terminal voltage of each diode is 0.6V. Reverse saturation current is 10^{-12}A.

a) 0.84mA

b) 1.84mA

c) 2.35mA

d) 3.01mA

View Answer

Explanation: Let V

_{D}be the voltage of diode, then by Kirchoff’s loop rule

3V = 2V

_{D}+ IR1

This method of assumption contains small error but it is the simplest method.

Let V

_{D}be 0.6V. Now the current I = (3-1.2)/1k = 1.8mA.

The V

_{D}= V

_{T}ln((I/I

_{O})+1) = 0.58V

Hence current is (3-(2×0.58))/1k =1.84mA.

**Sanfoundry Global Education & Learning Series – Analog Circuits.**

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