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Right answer is (c) DRIFT velocity

Best EXPLANATION: The carriers accelerate in the DIRECTION of electric field between COLLISIONS but for each time INTERVAL τc (collision time) there is a collision which randomises the velocity of the carrier. This average net velocity in the direction of electric field is Drift velocity.

The correct choice is (d) Because of difference in concentrations

The explanation: As there is a different level of doping in the p and N regions of a semi-conductor, the carriers (EITHER HOLES or electrons) move from a region of high concentration to a region of low concentration giving rise to diffusion CURRENT.

Right answer is (a) Ease of carrier DRIFT

Easiest explanation: Mobility is the ease with which carriers can drift. The HIGHER the COLLISION time, the greater is the mobility also the lighter is the carrier, the greater is its mobility. Thus on the APPLICATION of an electric FIELD it’s easier for carriers to drift.

The correct OPTION is (a) Area of cross section

To explain: Transition CAPACITANCE is the capacitance EXHIBITED by a DIODE due to the p-side and n-side of the diode is separated by a distance of depletion width same as in an electrolytic capacitor.

The equation of transition capacitance = ƸA/W

Where Ƹ = PERMITTIVITY of the material of diode, W = depletion width

A = area of cross section.

Correct OPTION is (a) The DIFFUSION causes the holes and electrons to COLLECT at the junction

To explain: The diffusion of carriers from one side to another makes the holes and electrons to collect on either side of the junction creating the depletion region. This is further widened or SHORTENED depending on the BIASING.

Correct option is (c) Nominal CAPACITANCE

Best explanation: Transition capacitance is the capacitance exhibited by a diode due to the p-side and n-side of the diode is separated by a distance of depletion width same as in an electrolytic capacitor. It occurs in a REVERSE biased diode. The other name for this is Junction capacitance, Space-CHARGE capacitance, BARRIER capacitance, Depletion region capacitance. Diffusion capacitance is due to the transport of charge carriers between the two TERMINALS of the device. It occurs in a forward biased diode.

The correct answer is (b) Width of depletion region

Best explanation: TRANSITION capacitance is the capacitance EXHIBITED by a diode DUE to the p-side and n-side of the diode is separated by a DISTANCE of depletion width same as in an electrolytic capacitor.

The EQUATION of transition capacitance = ƸA/W

Where Ƹ = permittivity of the material of diode, W = depletion width

A = area of cross section.

Right ANSWER is (b) DECREASES

The explanation: Transition capacitance is the capacitance exhibited by a diode due to the p-side and n-side of the diode is separated by a distance of depletion WIDTH same as in an electrolytic capacitor.

The equation of transition capacitance = ƸA/W

Where Ƹ = permittivity of the MATERIAL of diode, W = depletion width

A = area of CROSS section

As reverse bias increases depletion width also increases hence transition capacitance decreases.

The correct CHOICE is (b) DIRECTLY proportional to relative permittivity

For explanation: Transition capacitance is the capacitance EXHIBITED by a diode due to the p-side and n-side of the diode is separated by a distance of depletion width same as in an electrolytic capacitor.

The equation of transition capacitance = ƸA/W

Where Ƹ = permittivity of the material of diode, W = depletion width

A = area of CROSS section

Relative permittivity is directly proportional to the permittivity of the material.

Right CHOICE is (c) 1pF

Easy explanation: The equation of TRANSITION capacitance = ƸA/W

Where Ƹ = permittivity of the MATERIAL of diode, W = depletion WIDTH

A = AREA of cross section

Since depletion width increased 10 times and all other quantities are the same, the capacitance decrease by 10 times.

The correct option is (C) 10pF

Explanation: The equation of transition capacitance = ƸA/W

Where Ƹ = permittivity of the MATERIAL of diode, W = depletion width

A = area of CROSS section

Since permittivity becomes half capacitance also halves.

The correct answer is (b) 30pF

For explanation I would SAY: The EQUATION of transition capacitance = ƸA/W

Where Ƹ = permittivity of the material of diode, W = depletion width

A = area of cross section

Since A becomes 2A capacitance ALSO DOUBLES.

The correct option is (b) 0.1 µm

Best explanation: The equation of TRANSITION capacitance = ƸA/W

Where Ƹ = PERMITTIVITY of the MATERIAL of DIODE, W = depletion width

A = area of cross section

Since depletion width and capacitance are inversely proportional

Depletion width decreases to 0.1 µm.

Correct answer is (b) Diffusion capacitance

The best I can explain: Transition capacitance is the JUNCTION capacitance in a reverse biased diode. The rate of CHANGE of immobile charges wrt a change in reverse bias voltage is called transition capacitance, or space charge or depletion LAYER capacitance.

Correct OPTION is (d) 5

The EXPLANATION: In A, CA ∝ \(\frac{1}{\sqrt{V}}\)

CB ∝ \(\frac{1}{\sqrt[3]{V}}\)

\(\frac{C_A}{C_B} = \frac{4}{8} = \frac{1}{2}\) = 0.5.

The correct answer is (c) 1.31

The best EXPLANATION: \(\FRAC{C_{T1}}{C_{T2}} = (\frac{V2}{V1})^{\frac{1}{2.5}} \)

\(\frac{C_{T1}}{C_{T2}}= 2^{\frac{1}{2.5}}\) = 1.31.

Right option is (d) Reverse resistance

The explanation: CD = \(\FRAC{εA}{W}\) which means CD ∝ \(\frac{1}{W}\). THUS, CD ∝ \(\frac{1}{\sqrt{DOPING}}\) concentration

CD = τg = \(\frac{τ}{r} = \frac{τI_F}{ηV_T}\)

where τ = carrier life time, g = dynamic CONDUCTANCE, IF = forward current.

Right answer is (a) The diode is a p^+N junction diode

The best I can EXPLAIN: Diffusion capacitance is proportional to the carrier lifetime of injected minority carriers / excess minority carriers. SINCE it is proportional to the lifetime of holes in N SIDE, this means the N side is the minority and P side is the majority. Hence it is a p^+n diode.

Right option is (B) False

For explanation I would say: Diffusion CAPACITANCE occurs in a forward BIASED DIODE, transition capacitance is EASY to see in reverse bias. CD > CT for a forward bias junction. In reverse bias though, CD may be neglected compared to CT.

Correct option is (d) 6nF

Easy EXPLANATION: Diffusion capacitance = \(\frac{dQ}{dV}\)

CD = \(\frac{1.2 x 10^{-8}}{2}\)

CD = 6 NF.

Right CHOICE is (b) 0.1μF

To EXPLAIN: Q = – \(\frac{V^{-9}}{9}\)

CD = \(\frac{DQ}{dV}\) = V^-10 = 1.024 x 10^-7

CD = 0.1μF.

The correct answer is (B) 0.01

Best explanation: C ∝ \(\frac{1}{\sqrt{DOPING}}\)

\(\frac{C1}{C2} = \sqrt{\frac{10^{16}}{10^{20}}} = \sqrt{\frac{1}{10^4}} = \frac{1}{100}\)

\(\frac{C1}{C2}\) = 0.01.

The correct ANSWER is (a) 4CD

For explanation I would SAY: CD = \(\frac{εA}{W}\) ∝ A.\(\sqrt{Doping}\)

CD‘ = \(\frac{ε2A}{W}\) ∝ 2A.2 2A.\(\sqrt{4Doping}\) = 4 CD

CD‘ = 4CD.

The CORRECT option is (a) 8.28μm

For EXPLANATION I would say: \(\frac{W1}{W2} = \SQRT{\frac{V1}{V2}} = \sqrt{\frac{0.7}{12}}\)

W2 = 4.14*2 = 8.28μm.

The correct CHOICE is (a) 0.78 V

Easy explanation: On DRAWING the LOAD line with the equation: VD = ED + IDRD, we get the operating point with the VALUE of at voltage at around 0.7-0.8 V. Hence, VDq=0.78 V.

Right choice is (C) R1 < R2

Best explanation: Here, the y-intercept=E/R hence, lower the y-intercept, higher the VALUE of R, Hence, R1

The CORRECT choice is (c) 9.3 V

For EXPLANATION I would SAY: We know that VR=IDR = 9.3 x 1 = 9.3 V.

The correct OPTION is (a) 0 V

For EXPLANATION I would SAY: The value of voltage CALCULATED across the resistor is calculated by E-VD, which is constant for both cases.

Correct option is (C) 0.7 V

To explain I WOULD say: At the point of intersection, the VALUE of DIODE VOLTAGE is approximately 0.7 V.

Correct choice is (a) 0.7 V

Best explanation: In the approximate equivalent model, the diode voltage is fixed at the forward BIAS THRESHOLD voltage, which for a SILICON diode is EQUAL to 0.7 V

Right option is (a) 9.25 mA

The explanation: In the approximate equivalent model, the characteristic is ASSUMED to be a vertical UPWARD line at V=0.7 V. Hence, the current at point of intersection is DETERMINED to be 9.25 mA

Right CHOICE is (b) 0 V

The BEST I can explain: In ideal diode model, we take the FORWARD threshold voltage to be zero. Hence the diode characteristic is represented as the UPPER half of the y-axis. Hence, VD = 0 V.

The correct answer is (b) 10 mA

For explanation: In IDEAL diode model, we take the forward THRESHOLD voltage to be zero. HENCE thee diode characteristic is REPRESENTED as the upper half of the y-axis. Hence, ID = E/R = 10 mA.

The correct choice is (a) True

Explanation: In the IDEAL DIODE we assume the diode voltage as zero whereas in the approximate equivalent CIRCUIT model, we assume VD = 0.7 V and the actual value is somewhere about 0.75-0.8 V. HENCE, approximate equivalent model is more accurate.

Correct choice is (b) 1.21

The explanation: RIPPLE factor of a rectifier is the measure of the EFFECTIVENESS of a power supply filter

in reducing the ripple voltage. It is calculated by taking ratio of RMS AC COMPONENT of output voltage to DC component of output voltage.

r = √Irms^2 – IDC^2/IDC

For a half wave rectifier, it is 1.21.

Right answer is (b) 50Hz

Easy EXPLANATION: The ripple frequency of output and input is the same since one HALF CYCLE of input is passed and other half cycle is blocked. So EFFECTIVELY frequency is the same.

The correct choice is (b) 0.287

Explanation: Transformer utilization factor is the ratio of DC output power to the AC RATING of the secondary WINDING. AC rating is the product of RMS voltage ACROSS winding and RMS current through the winding, expressed in volt-amp. This factor indicates the effectiveness of transformer usage by a rectifier. For half WAVE rectifier it is low and equal to 0.287.

Right OPTION is (c) 40.6%

BEST explanation: Efficiency of a rectifier is a measure of the ABILITY of a rectifier to convert input power into DC power. Mathematically it is equal to the ratio of DC output power to the total input power and efficiency = 40.6xRL/RF+RL%. Its MAXIMUM value is 40.6 %.

Correct option is (d) 63.661V

Explanation: The equation of SINE wave is in the form Em sin wt.

Therefore, Em=200

Hence output VOLTAGE is Em/ᴨ. That is 200/ᴨ = 63.6619V.

The correct OPTION is (c) 50Hz

The best explanation: The EQUATION of SINE wave is in the form EM sin wt.

On comparing equation w = 100 ᴨ rad/s

We know w=2 ᴨ x frequency.

Therefore, frequency = 50 Hz.

Ripple frequency and input frequency are the same.

The correct choice is (c) 0.693

The explanation: Transformer UTILIZATION factor is the ratio of DC POWER supplied to the AC rating of the primary winding. The factor indicated the effectiveness of transformer usage by the RECTIFIER. For a center tapped full wave rectifier, it is equal to 0.81 w.r.t the primary winding, 0.57 w.r.t the secondary winding (DOUBLE of that of a half wave rectifier) and the average VALUE is 0.69.