1. Find the current I if each diodes are identical and assume two diodes are identical. Voltage V = 0.8V and let the reverse saturation current be 10^{-9}A.

a) 4.8mA

b) 3.2mA

c) 2.5mA

d) 7mA

### View Answer

_{D}= 0.6/2 = 0.3V Equation for diode current 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. Take VT as 0.026 which is the standard value. Hence current through one diode is 10

^{-9}x(e

^{0.3/0.026}) = 4.8mA.

2. Find voltage Vout if the reverse saturation current of the diode is 1.1×10^{-8A} and cut-in voltage of diode is 0.6V and assume the temperature as 25^{o}C.

a) 0.235V

b) 0.3148V

c) 0.456V

d) 0.126V

### View Answer

_{0}= reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage Since in this question ideality factor is not mentioned it can be taken as one. Take VT as its standard value 0.026V. .

3. The current Ix in the circuit is 1mA then find the voltage across diode D1. The resistance R1 is 1K. Assume the reverse saturation current is 10-9A. Voltage across resistor in this condition was 0.4V. Take VT of diode as0.026V.

a) 2.3mA

b) 3.2mA

c) 5.2mA

d) 4.6mA

### View Answer

4. Find the current Ix if the voltage across the diode is 0.5V. The reverse saturation current of diode 10-11 , Cut-in voltage of diode is 0.6V. Assume temperature at which diode operates is 25oC. The resistance R1=2K.

a) 3.97mA

b) 4.51mA

c) 2.45mA

d) 3.28mA

### View Answer

5. If the current I is 2mA then find the temperature at which diode operates. The cut-in voltage of diode is 0.6V. The reverse saturation current of diode is 10-9A. Resistance R is 1.3K.

a) 46.23oC

b) 50.47oC

c) 60.26oC

d) 56.89oC

### View Answer

6. If temperature increases 10oC the ratio of final reverse saturation current to initial reverse saturation current

a) 1

b) 2

c) 1.5

d) 3

### View Answer

_{o2}) to initial reverse saturation current (I

_{o1}) is given by I

_{o2}= 2

^{(∆T/10)}I

_{o1}Where ∆T is temperature change Ratio will be 2

^{(∆T/10)}= 2

^{1}=2.

7. The reverse saturation current of a diode at 25oC is 1.5 x 10^{-9}A and what will be reverse current at temperature 30oC

a) 3 x 10^{-9}A

b) 2 x 10^{-9}A

c) 2.12 x 10^{-9}A

d) 1.5 x 10^{-9}A

### View Answer

_{o1}) is given by I

_{o2}= 2

^{(∆T/10)}I

_{o1}Where ∆T is temperature change Here ∆T = 5, Therefore, I

_{o2}= 25/10 Io1 =1.414×1.5 x 10

^{-9}A = 2.121 x 10

^{-9}A.

8. How much times reverse saturation current will increase if temperature increases 15oC

a) 2.52

b) 4.62

c) 4.12

d) 2.82

### View Answer

_{o2}) to initial reverse saturation current (I

_{o1}) is given by I

_{o2}= 2

^{(∆T/10)}I

_{o1}Where ∆T is temperature change Ratio is 215/10 = 2.82.

9. If the Vin of the circuit the circuit is 2V and resistor have a resistance of 1K and cut-in voltage of diode is 0.7V. Reverse saturation current is 10-8A. Temperature at which diode operates is 30oC.The Vout is close to

a) 0.256V

b) 0.306V

c) 0.215V

d) 0.456V

### View Answer

_{in}= V

_{D}+ IR

_{1}This method of assumption contains small error but it is the simplest method. Let V

_{D}be 0.7V. Now the current I = (2-0.7)/1k = 1.3mA. Now the diode voltage for 1.3mA V

_{D}= V

_{T}ln(I/I

_{0 }) = 0.026 x ln((1.3×10

^{-3})/10

^{-8}) = 0.306. Therefore, voltage across diode will be close to 0.306V.

10. Let the V_{in} be -5V and resistance R_{1} is 5K and the cut-in voltage of the diode is 0.7V. What will be the voltage V_{out}? Take reverse saturation current as 10-8A and operating temperature as 25oC.

a) 0V

b) -4.5V

c) -5V

d) -3.2V

### View Answer

_{1}is negligible.