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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 10-9A.
analog-circuits-questions-answers-freshers-q1
a) 4.8mA
b) 3.2mA
c) 2.5mA
d) 7mA

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Answer: a [Reason:] Since both diodes are identical VD = 0.6/2 = 0.3V Equation for diode current analog-circuits-questions-answers-freshers-q2a where I0 = 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 0.026 which is the standard value. Hence current through one diode is 10-9x(e0.3/0.026) = 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 25oC.
analog-circuits-questions-answers-freshers-q2
a) 0.235V
b) 0.3148V
c) 0.456V
d) 0.126V

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Answer: b [Reason:] Equation for diode current analog-circuits-questions-answers-freshers-q2a where I0 = 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. analog-circuits-questions-answers-freshers-q2b.

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.
analog-circuits-questions-answers-freshers-q3
a) 2.3mA
b) 3.2mA
c) 5.2mA
d) 4.6mA

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Answer: c [Reason:] Since voltage drop across diode is 0.4V current through resistor is 0.4/1k = 0.4mA Current through diode analog-circuits-questions-answers-freshers-q2a where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage Since ideality factor is not given take it as one. Current through diode I= 10-9 x (e0.4/0) = 4.8mA Total current =4.8mA+0.4mA = 5.2mA.

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.
analog-circuits-questions-answers-freshers-q3
a) 3.97mA
b) 4.51mA
c) 2.45mA
d) 3.28mA

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Answer: d [Reason:] Since voltage drop across diode is 0.5V current through resistor is 0.5/2k = 0.25mA Current through diode analog-circuits-questions-answers-freshers-q2a where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage Since ideality factor is not given take it as one. Current through diode I= 10-11 x (e0.5/0) = 3.03mA Total current =3.03mA+0.25mA = 3.28mA.

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.
analog-circuits-questions-answers-freshers-q5
a) 46.23oC
b) 50.47oC
c) 60.26oC
d) 56.89oC

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Answer: a [Reason:] Equation for diode current analog-circuits-questions-answers-freshers-q2a where I0 = 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. VD = 3-(2mAx1.3k) = 0.4V analog-circuits-questions-answers-freshers-q5a.

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

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Answer: b [Reason:] The equation relating final reverse saturation current (Io2 ) to initial reverse saturation current (Io1 ) is given by Io2 = 2(∆T/10)Io1 Where ∆T is temperature change Ratio will be 2(∆T/10) = 21 =2.

7. The reverse saturation current of a diode at 25oC is 1.5 x 10-9A and what will be reverse current at temperature 30oC
a) 3 x 10-9A
b) 2 x 10-9A
c) 2.12 x 10-9A
d) 1.5 x 10-9A

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Answer: c [Reason:] The equation relating final reverse saturation current (Io2) to initial reverse saturation current (Io1 ) is given by Io2 = 2(∆T/10)Io1 Where ∆T is temperature change Here ∆T = 5, Therefore, Io2 = 25/10 Io1 =1.414×1.5 x 10-9A = 2.121 x 10-9A.

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

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Answer: d [Reason:] The equation relating final reverse saturation current (Io2) to initial reverse saturation current (Io1) is given by Io2 = 2(∆T/10)Io1 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

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Answer: b [Reason:] Let VD be the voltage of diode, then by Kirchoff’s loop rule Vin = VD + IR1 This method of assumption contains small error but it is the simplest method. Let VD be 0.7V. Now the current I = (2-0.7)/1k = 1.3mA. Now the diode voltage for 1.3mA VD = VT ln⁡(I/I0 ) = 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 Vin be -5V and resistance R1 is 5K and the cut-in voltage of the diode is 0.7V. What will be the voltage Vout? Take reverse saturation current as 10-8A and operating temperature as 25oC.
a) 0V
b) -4.5V
c) -5V
d) -3.2V

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Answer: c [Reason:] Since diode is in reverse bias mode voltage across diode will be almost same as applied voltage. Since the current in the circuit is in micro amperes voltage drop at R1 is negligible.