# Analog Circuits Questions and Answers – Constant Voltage Drop Model-2

This set of Analog Circuits Questions and Answers for Experienced people focuses on “Constant Voltage Drop Model-2”.

1. What is voltage across the resistor R if VA = -3V and VB = -5V is _________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 0V
b) -3V
c) -5V
d) -4V

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. In above circuit both the diodes are reverse biased and can be considered as open circuit. Hence output voltage is 0V.

2. What is current I if voltage V = 5V, VB = 2V, R1 & R2 = 2K. (Use constant voltage drop model assumption and take VD = 0.5V)

a) 1.25mA
b) 1mA
c) 2.75mA
d) 1.75mA

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.
Since Vin and VB are opposite net voltage is 3V. Voltage at R1 is 3V so current is 1.5mA. Voltage at R2 is 3-0.5 = 2.5V. So the current is 1.25mA. The net current is 2.75mA.

3. Current I if V = 5V and -5V when VB = 2V, R1 = 2K, R2 = 4K respectively are __________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 1.3mA, 0.23mA
b) 2.875mA, 0mA
c) 2mA, 0mA
d) 1.423mA, 0 mA

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.
When V=5V, diode is in forward bias and net total voltage becomes 4.5V across R1.
Current through branch 1 will be 4.5V / 2K = 2.25mA.
Current through branch 2 will be (4.5-2)/4K = 0.625mA.
So net current is sum of these two. Therefore, net current is 2.875mA.
When V = -5V, the diode is reverse biased and net current flowing is zero.

4. The output voltage V if Vin = 3V, R = 5K, VB = 2V is ___________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 0V
b) 2V
c) 1.5V
d) 3.5V

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.
Since Vin is reverse bias to the diode all voltage will appear across diode and no current flows. Thus, voltage across diode is Vin-VB = 1V. Net voltage V=3V.

5. In the circuit below VB = 2V, Vin = 5V. The voltage V across resistor R is ________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 0V
b) 2.5V
c) 1.5V
d) 3V

Explanation: The diode is in reverse bias and voltage across diode is -3V. Hence the voltage across the resistor is Vin+VD-VB=5-3-2=0V.

6. In the circuit Vin = 4V, VB = 3V, R = 5K. The voltage across diode V is _________
(Use constant voltage drop model assumption and take VD = 0.5V)

a) 0V
b) 0.5V
c) 1V
d) 1.5V

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. The diode here is in reverse bias and may be considered as an open circuit. Hence, the voltage across the diode is 4-3=1V.

7. In the circuit below Vin = 4V, R = 2K and VB = 2V. In these conditions the voltage across diode V is _________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 2.5
b) 4.5
c) 0.5
d) 1.5

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open. Since diode is forward biased it will produce a voltage drop of VD.

8. In the circuit shown in below I = 2mA, VB = 2V and R = 2K. The voltage V will be ________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 0V
b) 2.5V
c) 6V
d) 3.5V

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.
Since current source forward bases the diode voltage drop across diode is VD. So net voltage output is VD+VB.

9. In the circuit shown in below I = 2mA, VB = 2V and R = 2K. The voltage V will be ________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 4V
b) 3V
c) 6V
d) 5.5V

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.
Since current source reverse biases diode, all current pass through resistor R. So voltage across resistor is 4V. Since voltage source cannot produce current due to lack of closed circuit total voltage at the output is 6V.

10. For circuit shown below Vin = 3V, R1 = 6K, R2 = 2K. The voltage V will be ________ (Use constant voltage drop model assumption and take VD = 0.5V)

a) 2V
b) 3V
c) 3.5V
d) 2.5V

Explanation: In constant voltage drop model at forward bias diode can be replaced as a cell and in reverse bias diode can be avoided by considering the terminals are open.
Since diode is forward biased and parallel to resistor R1 voltage drop across diode is VD. So net voltage equals to Vin – VD.

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