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The CORRECT choice is (d) Peak forward voltage

The explanation: The voltage after which a diode increases rapidly is known as the offset voltage, THRESHOLD voltage, firing potential and cut-in voltage. Beyond this voltage, the forward bias voltage OVERCOMES the potential barrier and rapid CONDUCTION occurs.

The CORRECT answer is (a) Maximum reverse bias POTENTIAL which can be applied across a diode WITHOUT breakdown

Easiest explanation: PIV rating indicates the maximum ALLOWABLE reverse bias voltage which can be safely applied to a diode. If a reverse potential is greater than PIV rating then the diode will enter reverse breakdown region.

Right choice is (d) It is APPLICABLE in all bias STATES

The EXPLANATION: DIODE equation is I=IO(e^qV/kT – 1). It is applicable in all bias condition that is forward, reverse and unbiased states.

The correct option is (a) 1

Easiest EXPLANATION: Emission coefficient or ideality factor ACCOUNTS the effect of recombination taking place in the depletion region. The RANGE of factor is from 1 to 2. For Germanium it is 1.

Correct CHOICE is (B) 2

To elaborate: Emission COEFFICIENT or ideality factor accounts the effect of RECOMBINATION taking place in the depletion region. The range of factor is from 1 to 2. For silicon it is 2.

The correct option is (a) 26mV

The EXPLANATION is: VOLTAGE equivalent of TEMPERATURE VT is EQUAL to the product of Boltzman constant (J.K^-1) and temperature in Kelvin. At a temperature of 27°C, it’s value is VT=KT/q=26mV.

Correct choice is (B) Cut-in voltage decreases

Easiest explanation: As TEMPERATURE increases the conductivity of a semiconductor increases. The DIODE CONDUCTS smaller voltage at larger temperature. THEREFORE, cut-in voltage decreases.

The correct answer is (a) INCREASES

To explain: As TEMPERATURE increases the conductivity of a SEMICONDUCTOR increases. Reverse SATURATION CURRENT increases as temperature increases.

The correct option is (B) Forward breakdown OCCURS at VZ

To explain I would SAY: Voltage rating is the voltage at which reverse breakdown occurs. Voltage rating of a zener diode indicates the voltage BEYOND which current increases rapidly. Beyond this point increase in reverse bias voltage will only reflect on current which means voltage drop beyond this point is almost constant. This phenomenon holds the key role in the working of zener diode as a voltage regulator. Zener diode is a heavily doped diode because for zener breakdown we need a NARROW junction. It operates in reverse bias mode. In forward bias mode it operates as a normal diode.

Right choice is (a) Reverse breakdown VOLTAGE

To explain I would say: Voltage RATING of a zener diode indicates the voltage beyond which current increases rapidly. Beyond this point increase in reverse bias voltage will only REFLECT on current which means voltage drop beyond this point is ALMOST constant. This phenomenon holds the key role in the working of zener diode as a voltage REGULATOR.

Correct option is (B) Reverse BIAS current above which I-V characteristics is a straight line

Explanation: Knee current indicates the current above which reverse characteristics is a straight line. That is the RELATIONSHIP between current and voltage beyond this POINT will be directly proportional and this is very useful for solving REAL life problems.

Correct option is (d) Reverse scale current is in the order of nA

To ELABORATE: Cut-in VOLTAGE of Si diode is 0.7V but reverse breakdown voltage is almost 50V for the Si diode. So clearly cut-in voltage is much less than the reverse breakdown voltage. Reverse saturation current is in the order of NANO amperes. Scale current is another name of saturation current. It is named as scale current because saturation current is directly PROPORTIONAL to cross sectional area of a diode.

Right option is (a) 9.156mA

The explanation is: EQUATION for diode current

I=I0×(e^(V/ηVT)-1)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.

I0 = 1.17 X 10^-9A, VT = 0.0252V, η = 1, V = 0.4V

Therefore, I = 1.17×10^-9xe^0.4/0.025 -1 = 9.156mA.

Right option is (c) 5.5x 10^-7 A

To explain: Equation for diode current

I=I0×(e^(V/ηVT) – 1)where I0 = reverse saturation current

η = IDEALITY factor

VT = thermal voltage

V = applied voltage

Here, I = 0.1mA, η = 1.5, V = 0.2V, VT = TK/11600

Therefore, VT at T = 25+273=298 is 298/11600 = 0.0256V.

Therefore, reverse saturation current

IO=0.00055mA = 5.5×10^-7A.

Correct CHOICE is (b) 0V

Explanation: THERMAL VOLTAGE VT is given by k T/q

Where k is the boltzman constant and q is the charge of ELECTRON. This can be reduced to

VT = TK/11600

Therefore, VT = 298.15/11600 = 0.0257V.

Correct answer is (b) 0.143V

The best I can EXPLAIN: Equation for diode current

I=I0×(e^(V/ηVT)-1)where I0 = reverse SATURATION current

η = ideality factor

VT = thermal voltage

V = APPLIED voltage

η = 1.2, I2 = 10mA, I1 = 0.1mA and take VT = 0.026V

The CORRECT ANSWER is (a) 69.65°C

Easiest explanation: Equation for DIODE current

I=I0×(e^(V/ηVT)-1)where I0 = reverse saturation current

η = ideality factor

VT = thermal voltage

V = applied voltage

I0 = 10^-9A, η = 1.4, V = 0.6V, I = 2mA

We KNOW thermal voltage VT = TK/11600. Therefore, TK = VTx11600 = 0x11600 = 342.65K = 69.65°C.

The correct option is (c) 0.0718V

To explain I WOULD SAY: η = 1.2, I2 = 10mA, I1 = 1mA and take VT = 0.026V

.

Correct choice is (a) 2.4mA

The best explanation: Let VD be the VOLTAGE of diode, then by Kirchoff’s loop rule

3V = VD + IR1

This method of assumption contains small ERROR but it is the simplest method.

Let VD be 0.7V. Now the CURRENT I = (3-0.7)/1k = 2.3mA. Now the diode voltage for 2.3mA

VD = VT ln⁡(I/I^0) = 0.026 X ln((2.3×10^(-3))/10^(-12)) = 0.5864V.

Now the current becomes (3-0.5864)/1000 = 2.41mA.

The CORRECT choice is (b) 1.84mA

Best explanation: Let VD be the voltage of diode, then by Kirchoff’s loop rule

3V = 2VD + IR1

This method of assumption contains small ERROR but it is the simplest method.

Let VD be 0.6V. Now the CURRENT I = (3-1.2)/1k = 1.8mA.

The VD = VT ln((I/IO)+1) = 0.58V

Hence current is (3-(2×0.58))/1k = 1.84mA.

Correct answer is (b) 0.3148V

The explanation: Equation for diode current

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.

Voltage VD = ηVTln(I/IO + 1) = 1 x 0.0257ln (2 x 10^-3/(1.1 x 10^-9))+1) = 0.0257ln(1,818,182) = 0.3704 V.

Right answer is (a) 4.8mA

To explain: SINCE both diodes are identical VD = 0.8/2 = 0.4V

Equation for diode current

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(e^0.4/0.026) = 4.8mA.

Right answer is (a) 45.85°C

The EXPLANATION is: EQUATION for diode current

 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

VT = T/11600

Thus Temperature, T = 11600×0.0275 = 319 Kelvin = 45.85°C.

Correct OPTION is (C) 5.2mA

The explanation is: Since voltage drop across diode is 0.4V current through resistor is 0.4/1k = 0.4mA

Current through diode

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 (e^0.4/0.026) = 4.8mA

Total current = 4.8mA+0.4mA = 5.2mA.

Correct option is (d) 3.05mA

For explanation: Since voltage drop across diode is 0.5V current through resistor is 0.5/2k = 0.25mA

Current through diode

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 (e^0.5/0.0257) = 2.8mA

Total current = 2.8mA + 0.25mA = 3.05mA.

The CORRECT option is (b) 2

Explanation: 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) = 2^1 = 2.

Right option is (d) 2.82

The best explanation: 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 2^15/10 = 2.82.

Right choice is (c) 2.12 x 10^-9A

To explain I would SAY: 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 x 1.5 x 10^-9A = 2.121 x 10^-9A.

Right option is (b) 1.3V

To EXPLAIN: V1 = IR1 + VD

For silicon DIODE, cut-in VOLTAGE is 0.7V.

Hence IR1 = 2-0.7 = 1.3V

Drop across the RESISTOR is 1.3V.

Correct option is (C) -5V

To elaborate: Since diode is in reverse bias mode VOLTAGE across diode will be almost the same as the applied voltage. Since the CURRENT in the circuit is in micro AMPERES voltage drop at R1 is NEGLIGIBLE.

Correct ANSWER is (a) Operating point of DEVICE

The explanation: QUIESCENT point of a device represents operating point of a device. For a diode quiescent point is DETERMINED by constant DC current through the diode. The Q-point is the DC voltage or current at a specified terminal of an ACTIVE device with no input applied. A bias circuit is used to supply this steady voltage/current.

The correct answer is (d) Calculated from V-I graph

The explanation is: Average AC resistance and DYNAMIC or AC resistance are not EXACTLY the same. They both MEASURE the resistance in DIFFERENT WAYS. AC resistance is slope of the tangent of the curve of characteristic graph at Q-point. But average AC resistance is measured by measuring the slope of straight line between the limits of operation.

Right OPTION is (b) Maximum AC resistance

The explanation is: Static or DC resistance of a diode is the resistance OFFERED by a diode at its q-point. DC resistance represents steady state. That is, it is CALCULATED by treating current and voltage as constants.

Correct choice is (C) Q-point

Explanation: Static or DC RESISTANCE of a diode is the resistance offered by a diode at its q-point. DC resistance represents steady STATE. That is, it is calculated by treating current and VOLTAGE as constants.

Correct ANSWER is (b) DECREASES exponentially

The BEST I can explain: After cut-in voltage CURRENT exponentially increases with SMALL increase in voltage. This will considerably reduce resistance.

The CORRECT answer is (b) 10MΩ

The explanation: STATIC or DC RESISTANCE is the resistance of a diode at its operating point.

Static resistance at -10V = 10V/current at -10V = 10/1µA = 10MΩ.

Correct choice is (c) 0.7V

To explain I WOULD say: Static or DC RESISTANCE is the resistance of a DIODE at its operating point.

Resistance = voltage/CURRENT

Therefore, voltage = current X resistance = 10mA x 70 = 0.7V

The correct answer is (c) Perfect insulator

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.

Right choice is (a) 90 Ω

To ELABORATE: STATIC or DC resistance is the resistance of a diode at its operating point.

That is DC resistance = 0.9/10mA = 0.9×1000/10 = 90Ω.

The correct option is (d) 0V

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.

Correct answer is (a) 3mA

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 current

I = (3V/1K) = 3mA.

Correct choice is (a) -5V

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 reverse bias no current flows through the circuit so entire voltage appears on diode.

Correct choice is (a) 0A

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 diode is reverse bias no current will flow through the circuit.