1. Calculate the forward bias current of a Si diode when forward bias voltage of 0.4V is applied, using characteristic equation of diode. The reverse saturation current is 1.17 x 10^{-9} A. Thermal voltage is 25.2 mV.

a) 9.156mA

b) 8.23mA

c) 1.256mA

d) 5.689mA

### View Answer

_{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}= 0V, η = 1, V = 0.4V Therefore, I= 1.17×10

^{-9}xe

^{0.4/0.025}-1 = 9.156mA

2. Calculate thermal voltage for a temperature of 25oC.

a) 0V

b) 0V

c) 0.026V

d) 0.25V

### View Answer

_{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}at T= 25+273=298 is 298/11600 = 0V

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

a) 5.06x 10^{-9} A

b) 5.06x 10^{-8} A

c) 5.06x 10^{-7} A

d) 5.06x 10^{-10} A

### View Answer

_{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 = 0V. Therefore, reverse saturation current

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

a) 0.658V

b) 0.726V

c) 0.526V

d) 0.618V

### View Answer

_{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 = 0V, η = 1.5, I = 1mA, I

_{0}= 10-10A

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

a) 69.65oC

b) 52.26oC

c) 25.23oC

d) 70.23oC

### View Answer

_{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 = 69.65oC.

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

_{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) 0V

d) 0.00728V

### View Answer

_{T}= 0.026V .

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

_{T}= 0.026V

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

a) 2.4mA

b) 0.9mA

c) 1mA

d) 4mA

### View Answer

_{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)}) = 0.560. Now the current becomes (3-0.560)/1000 = 2.44mA.

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 10-12A.

a) 0.845mA

b) 1.892mA

c) 2.359mA

d) 3.012mA

### View Answer

_{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 = 2.8mA. The V

_{D}for 2.8mA is 0.554V. Hence current is (3-(2×0.554))/1k =1.892mA.