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

_{in}is reverse bias to diode it will appear in output. V

_{in}will appear across the diode and also V

_{B}is less than V

_{in}so V

_{in}will appear across 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

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) -0.23V

b) 0.29V

c) -2.72V

d) -1.3V

### View Answer

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

_{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

_{B}+ voltage across resistor R = -1+2 = 1V.

6. In the circuit shown below voltage V_{in} is 3V and V_{B} is 1V. The resistor R_{1} 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

_{2}are forward biased we can replace them 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 R

_{1}(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 R_{1} 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

_{in}will appear on V

_{out}. Diode D1 and D

_{2}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 R_{1} 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

_{out}= IR2 ————(1) Current through R

_{1}(V

_{in}-V

_{out}-V

_{D})/1.01k = (2.3-V

_{out})/1010 Current through diode D

_{2}is 0 since D

_{2}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_{B}1 is -1V and V_{B}2 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

_{B}as 1V By network analysis using kirchoff’s voltage rule current through RD

_{2}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

_{out}will be IRD+V

_{D}= 2mAx10 + 0.7 = 0.72V.