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Correct answer is (d) 0.1mA

The best I can explain: In IDEAL DIODE model the diode is considered as a PERFECT conductor in forward BIAS and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and CURRENT through the diode at reverse bias is zero.

Since the diode is in reverse bias, it can be considered open and no current flows through it. Then effective voltage becomes 3-2 = 1V so current is 1/10K = 0.1mA.

Correct CHOICE is (B) 3V

To explain: In ideal diode model the diode is considered as a perfect conductor in FORWARD bias and perfect insulator in REVERSE bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero. SINCE first diode is in forward bias and second diode is in reverse bias. So Only first diode will pass the current.

The correct answer is (c) 3.25mA and 0A

Best EXPLANATION: In ideal diode model the diode is considered as a PERFECT conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero.

When V=5V, the diode is forward biased and can be considered as a SHORT circuit.

Current through resistor R1 = V/2k = 2.5mA.

Current through resistor R2 = (V – VB)/4k = 0.75mA.

So total current is3.25mA.

At V = -5V, diode is reverse bias So the current is zero.

Correct answer is (a) 0V

Easy explanation: In an ideal diode model, the diode is considered as a perfect CONDUCTOR in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is ZERO and current through the diode at reverse bias is zero.

Since both the diodes are in reverse bias mode, current through the diode is zero and we can consider the diode as an open CIRCUIT. So there is no voltage drop ACROSS resistor R.

The CORRECT OPTION is (d) 3mA

Best explanation: In IDEAL diode MODEL the diode is considered as a perfect CONDUCTOR in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero.

The diode is forward biased and can be considered as a short circuit.

 So voltage across R1 is V-VB.

 That is 3V. Therefore, current through R1 = 3V/2k = 1.5mA.

Current through R2 = 3/2K = 1.5mA.

Therefore, total current I = 1.5+1.5 = 3mA.

The correct CHOICE is (c) 3V

To explain I would say: In ideal diode model the diode is considered as a PERFECT conductor in forward bias and perfect insulator in reverse bias. That is VOLTAGE drop at forward bias is zero and CURRENT through the diode at reverse bias is zero.

The diode above is reverse biased and IDEALLY can be considered as an open circuit. So output is the voltage Vin = 3V.

Correct answer is (d) 0V

Easy explanation: In ideal diode model the diode is considered as a PERFECT conductor in forward BIAS and perfect INSULATOR in reverse bias. That is voltage drop at forward bias is zero and CURRENT through the diode at reverse bias is zero.

Since the diode is in reverse bias, no current flows through it and thus through the resistor too. Thus voltage ACROSS resistor is zero.

Correct option is (b) -2V

The best I can explain: In ideal diode model the diode is considered as a perfect CONDUCTOR in forward bias and perfect insulator in reverse bias. That is VOLTAGE drop at forward bias is zero and current through the diode at reverse bias is zero.

Assuming diode to be forward BIASED, KCL can’t be applied correctly at the node. Hence diode is reverse biased. The diode can be considered as an open circuit and the current flows through resistor R only. Hence voltage V = -2V.

Right answer is (d) 0V

Easiest explanation: In ideal diode model the diode is considered as a perfect conductor in FORWARD bias and perfect insulator in REVERSE bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero.

Since net voltage Vin – VB = 2V forward BIASES the diode, it can be considered as a short circuit and the voltage across diode is zero.

Right option is (a) 2V

For explanation: In IDEAL diode model the diode is considered as a perfect CONDUCTOR in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero.

Consider that diode is reverse BIASED. But, that does not satisfy KCL at the node. No incoming current is present. Hence diode is forward biased and short-circuited. Hence the output voltage = V = VB = 2V.

Right option is (a) 3V

The best explanation: In IDEAL diode model the diode is considered as a PERFECT conductor in FORWARD bias and perfect insulator in reverse bias. That is VOLTAGE drop at forward bias is zero and current through the diode at reverse bias is zero.

Since diode is forward biased ENTIRE voltage will appear across R2.

The correct answer is (a) Resistor

The explanation: After cut –in VOLTAGE diode ACT as a resistor in piecewise linear mode. In normal operation diode CURRENT is EXPONENTIALLY RELATED to voltage.

Right choice is (d) 1.33mA

Easy explanation: Since both VOLTAGE sources are reverse BIAS to the DIODE, diode in the circuit disappears and EQUIVALENT circuit becomes as follows

So current I = V1-V2/R

= 3-1/1.5k = 1.33mA.

Right answer is (B) Varies linearly with voltage after knee voltage

The explanation is: Voltage drop produced by diode in piecewise LINEAR model is not CONSTANT. Since it contains EFFECT of resistor, the diode voltage linearly increases as input voltage increases.

The CORRECT answer is (d) Insulator

The best explanation: For a diode in reverse bias mode current through the diode is in micro amperes or nano amperes. Hence we can assume it as ZERO. In PIECEWISE linear model reverse current is assumed to zero. That is, as an insulator.

Correct answer is (b) 3V

For explanation: In PIECEWISE LINEAR model, we can consider diode as a voltage source of 0.5V along with an IDEAL diode. VIN being applied is 3V, which is greater than 1.5-0.5 V, and HENCE the diode is reverse biased and input appears at output.

The correct choice is (b) 3V

To explain: If we consider diode as a SHORT CIRCUIT, the voltage in circuit is thus 3-1=2V and CURRENT FLOWS from TOP to bottom across diode. But that only happens in a reverse bias of the diode. Hence the diode is in reverse bias and open. Output voltage V=3V.

The correct option is (d) 0V

The best explanation: In piecewise LINEAR model, we can consider diode as a VOLTAGE source of 0.7V along with an IDEAL diode. VIN being applied is 3V, which is greater than 2-0.7 V, and HENCE the diode is reverse biased and considering it OPEN, no output is obtained.

Correct answer is (a) 3V

For explanation: Since current source is reverse bias to the diode current passes through resistor R. Voltage across resistor R is 2MA X 1K = 2V. Since resistor and diodes are parallel NET output voltage V is VB + voltage across resistor R = 1+2 = 3V.

The correct CHOICE is (b) -4.25V

For explanation I WOULD say: Effective voltage ACROSS diode is one volt. Hence diode is in forward bias MODE. So we can apply equivalent circuit of diode.

Net voltage through R and RD is -3-2+0.7 = -4.3V

Current through the circuit is -4.3/(R+RD) = -4.3/(1010) = -0.00425A=-4.25 mA

Hence voltage across R is 1Kx(-4.25mA) = -4.25V.

Right answer is (B) -3V

Easy explanation: Since both diodes are in reverse bias mode APPLIED voltage Vin will appear on Vout. Diode D1 and D2 disappears and leaves the terminal as OPEN.

Correct answer is (a) 1.14V

Explanation: Assume I be the CURRENT through the circuit. By kirchoff’s VOLTAGE rule,

Vout = IR2————(1)

Current through R1 (Vin-Vout-VD)/1.01k = (2.3-Vout)/1010

Current through diode D2 is 0 SINCE D2 is in reverse BIAS mode.

current I = (2.3-Vout)/1010

Substitute this in EQ(1)

That is, Vout = (2.3-Vout)/1010 x 1000=> 1.99Vout = 2.27=> Vout = 2.27/1.99 = 1.144V.

Correct choice is (a) 0.6V

To explain: In this CONDITION both diodes are FORWARD biased

This circuit can be further REDUCED to by assuming VB as 1V

By NETWORK analysis using kirchoff’s voltage rule CURRENT through RD2 will be 0.09A.

The voltage in Vout will be VB+VD-0.09×10 = 0.6V.

The correct ANSWER is (d) 0.72V

The BEST explanation: SINCE diode is in forward bias we can REPLACE diode with voltage source of 0.7V and resistor of resistance 10Ω.

Vout will be IRD+VD = 2mAx10 + 0.7 = 0.72V.

Correct choice is (b) Perfect conductor

Easy explanation: In constant VOLTAGE drop model the diode is considered as a perfect conductor in forward bias and perfect insulator in REVERSE bias. That is voltage drop at forward bias is ZERO and current through the diode at reverse bias is zero. In this 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 correct option is (b) 2.5mA

Best 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 D1 is in forward biased there will be a voltage drop of 0.5V. So net voltage will be 2.5V and HENCE current is 2.5mA.

Correct OPTION is (c) Perfect insulator

Easiest explanation: In constant voltage drop the diode is CONSIDERED as a perfect CONDUCTOR in forward BIAS and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and CURRENT through the diode at reverse bias is zero. In this 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.

Correct answer is (c) -5V

To explain I would SAY: 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 reverse BIASED ENTIRE voltage will appear across diode.

Right choice is (d) 0mA

The best I can explain: In constant voltage drop model, in forward bias, the diode can be replaced as a cell and a SHORT circuit, and in reverse bias as an open circuit. In above circuit, the diode is reverse biased, hence when open CIRCUITED, no current FLOWS.

The correct option is (c) 0.5V

The explanation is: 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 is forward biased and drop across it is 0.5V. MAXIMUM CURRENT FLOWS through the diode and thus the voltage across diode is same as that across the resistor = 0.5V.

The CORRECT ANSWER is (d) 0.1mA

For explanation: In ideal diode MODEL the diode is considered as a perfect conductor in forward bias and perfect insulator in reverse bias. That is voltage drop at forward bias is zero and current through the diode at reverse bias is zero.

Since both voltage is reverse bias to diode the diode will disappear from the CIRCUIT. Then effective voltage BECOMES 3-2 = 1V so current is 1/10K = 0.1mA.

The correct answer is (b) 0.25mA

For 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 both voltage sources are forward biased the diode V2 and diode can be replaced by a single cell of VD. So NET voltage in the circuit is 2.5V. So the current will be 0.25mA.

Correct choice is (b) 2.875mA, 0mA

Easy 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.

Correct choice is (a) 2.5

The best 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 first diode is forward biased and second diode reverse bias, we can ignore the second diode. Drop ACROSS the first diode is 0.5V and hence net voltage at output is 3-0.5=2.5V.

The correct option is (a) 0V

For EXPLANATION I would say: 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.

Right option is (c) 2.75mA

For 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.

Correct CHOICE is (d) 3.5V

The best 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.

Correct ANSWER is (a) 0V

To explain: 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.

Right option is (C) 0.5

For 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.

The CORRECT choice is (d) 2.5V

The explanation is: 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.

The CORRECT OPTION is (a) Applied ELECTRIC field over a given distance

To elaborate: Drift current is a type of electric current due to the movement of charge carriers which OCCURS because of applied electric field across the p-n junction often STATED as electromotive force over a given distance.

Right option is (b) Variation in carrier concentration

The EXPLANATION is: Diffusion CURRENT is due to the actual movement of carrier charges from one side to ANOTHER. The DIRECTION of diffusion DEPENDS on the slope of the carrier concentration that is the gradient of density of carriers.

The correct CHOICE is (c) Drift+diffusion current

To explain: In an UNBIASED semi-conductor the drift current is balanced by diffusion current and hence there is no current flowing in the semi-conductor in this condition, but in biased condition both these currents are unbalanced and hence the TOTAL current flowing is the vector sum of both drift and diffusion current.

The correct choice is (a) True

The explanation is: In good CONDUCTORS there are many electrons moving freely in the conduction band and thus on application of an electric field these free electrons move whereas in semi-conductors DRIFT CURRENT flows because of LESS NUMBER of free electrons.

The correct choice is (B) DRIFT current density

Easy explanation: Here ‘Q’ is the charge on carrier, ‘n’ is the number of CARRIERS, ‘µn’ is the mobility constant and ‘E’ is the electric field intensity. These are the factors which comprise the drift current and hence the equation represents the drift current density.