## Electronic Devices and Circuits subject wise

Question 1 |

Fill in the Blank Type Question |

Question 2 |

^{+}p

^{++}configuration shown in the figure?

Question 3 |

-> As per the Question Structure shown is fabrication of npn transistor

-> Emitter highly doped (n

^{++})

-> Base lightly doped (p

^{+})

-> Collector (n

^{++})

-> CB junction (n

^{++}– p

^{+}junction)

Question 4 |

_{ms}= 0 V and no oxide charges. The flat-band, inversion, and accumulation conditions are represented, respectively, by the points

Q, R, P | |

Q, P, R | |

R, P, Q | |

P, Q, R |

Given C-V characteristics of MOS capacitor with p-type substrate for high frequencies.

Point-P possible in accumulation mode

Point-Q possible in flat band mode

Point-R possible in inversion mode

hence, option A is correct one

Question 5 |

**[k = 1.38 × 10 ^{-23} JK^{-1}, h = 6.625 × 10^{-34} J-s, q = 1.602 × 10^{-19} C]**

Fill in the Blank Type Question |

Question 6 |

2.4 and 2.4 | |

1.8 and 1.2 | |

1.8 and 2.4 | |

2.4 and 1.2 |

Question 7 |

Question 8 |

Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites | |

Silicon atoms act as n-type dopants in Arsenic sites and p-type dopants in Gallium sites | |

Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites | |

Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites |

**->**Substituting a Gallium site by a si atom produces a free electron

**so n-type**

**->**Substituting an Arsenic site by a si atom produces a hole

**so p-type.**

Question 9 |

^{+}-n Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of N

_{D1}=1×10

^{18}cm

^{-}

^{3}and N

_{D2}= 1×10

^{15}cm

^{-}

^{3}corresponding to the n

^{+}and n regions respectively. At the operational temperature T, assume complete impurity ionization, kT/q = 25 mV, and intrinsic carrier concentration to be n

_{i}= 1×10

^{10}cm

^{-}

^{3}. What is the magnitude of the built-in potential of this device?

0.748 V | |

0.460 V | |

0.288 V | |

0.173 V |

Question 10 |

_{E }for emitter, ∇p

_{B }for base. ∇n

_{C}for collector) normalized to equilibrium minority carrier concentration ( n

_{E0}for emitter, p

_{B0}for base, n

_{C0}for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in?

Forward active | |

Saturation | |

Inverse active | |

Cutoff |

so it works in

**Inverse active.**

Question 11 |

an increase in the low-frequency cutoff frequency | |

an increase in the high-frequency cutoff frequency | |

a decrease in the low-frequency cutoff frequency | |

a decrease in the high-frequency cutoff frequency |

Question 12 |

_{1}and M

_{2}are connected as shown below. The circuit is used as an amplifier with the input connected between G and S terminals and the output taken between D and S terminals, V

_{bias}and V

_{D}are so adjusted that both transistors are in saturation. The transconductance of this combination is defined as while the output resistance is where i

_{D}is the current flowing into the drain of M

_{2}. Let g

_{m1}, g

_{m2}be the transconductances and r

_{o1 }, r

_{o2}be the output resistance of transistors M

_{1}and M

_{2}respectively

Which of the following statements about estimates for g

_{m}and r

_{o}is correct?

and | |

and | |

and | |

and |

Question 13 |

_{1}= 156 nm (System 1) and another with a light source of wavelength λ

_{2}= 325 nm (System 2). Both photolithography systems are otherwise identical. If the minimum feature sizes that can be realized using System 1 and System 2 are L

_{min1}and L

_{min2}respectively, the ratio L

_{min1}/L

_{min2}(correct to two decimal places) is ____________

Fill in the Blank Type Question |

Question 14 |

Fill in the Blank Type Question |

Question 15 |

^{- }Si with doping density N

_{A1}= 10

^{15}cm

^{-3}and p Si with doping density N

_{A2}=10

^{17 }cm

^{-3}.

Given: Boltzmann constant k= 1.38 x 10^{-23 }J.K^{-1} , electronic charge q= 1.6 x 10^{-19} C.

Assume 100% acceptor ionization.

At room temperature (T= 300K), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be ____________.

Fill in the Blank Type Question |

Question 16 |

_{2}is twice that for M

_{1.}The two NMOS transistors are otherwise identical. The threshold voltage V

_{T}for both transistors is 1.0 V. Note that V

_{GS}for M

_{2}must be > 1.0 V.

Current through the nMOS transistors can be modeled as

for

for

The voltage (in volts, accurate to two decimal places) at V

_{x}is _______.

Fill in the Blank Type Question |

Question 17 |

^{2}, operating at 1.0 sun intensity, has a short circuit current of 20 mA, and an open circuit voltage of 0.65 V. Assuming room temperature operation and thermal equivalent voltage of 26 mV, the open circuit voltage (in volts, correct to two decimal places) at 0.2 sun intensity is __________.

Fill in the Blank Type Question |

Question 18 |

_{R}, V

_{G}and V

_{B}respectively. Assume donor and acceptor doping to be the same (N

_{A}and N

_{D}respectively) in the p and n sides of all the three diodes. Which one of the following relationships about the built-in voltages is TRUE?

V _{R }> V_{G} >V_{B} | |

V _{R} <_{ }V_{G}< V_{B} | |

V _{R }= V_{G} = V_{B} | |

V _{R }> V_{G} < V_{B} |

**->**For LED and emitting lights if Red, Blue and Green colours, doping concentration on both sides are equal.

**->**We know that energy band gap and built in potential of a p-n junction is inversely proportional to wavelength, i.e.

Question 19 |

the effective base width increases and common – emitter current gain increases | |

the effective base width increases and common – emitter current gain decreases | |

the effective base width decreases and common – emitter current gain increases | |

the effective base width decreases and common – emitter current gain decreases |

**"If the reverse bias voltage across the base collector junction is increased, then their effective base width will decrease and collector current will increase, therefore their common-emitter current gain increases."**

If reverse bias across base collector junction will increases, then

(i) Depletion width across base collector junction will increase.

(ii) Effective base width will decrease.

(iii) Recombination of carrier in base region will decrease.

(iv) Common emitter current gain ( β = I

_{C}/ I

_{B}) will increase

Question 20 |

µ

_{n}and oxide capacitance per unit area C

_{ox}. If gate-to-source voltage V

_{GS}=0.7V, drain-to- source voltage V

_{DS}=0.1V, ( µ

_{n}C

_{ox}) =100 µA/V

^{2}, threshold voltage V

_{TH}=0.3V and (W/L) =50, then the transconductance g

_{m}(in mA/V) is______

Fill in the Blank Type Question |

**Given**

V

V

_{GS}=0.7VAnswer Range : (0.45 to 0.55)

Question 21 |

_{C}-F

_{F}=0.9 eV, where E

_{C}and E

_{F}are the conduction band minimum and the Fermi energy levels of Si, respectively. Oxide ∈

_{r}=3.9, ∈

_{0 }=8.85 X 10

^{-14}F/cm. oxide thickness t

_{ox}=0.1 μm and electronic charge

q= 1.6 X 10

^{-19}C. If the measured flat band voltage of the capacitor is -1V, then the magnitude of the fixed charge at the oxide-semiconductor interface, in nC/cm

^{2}, is

Fill in the Blank Type Question |

=

**Answer Range : (6.85 to 6.95)**

Question 22 |

_{L}) is 2.5 mA/cm

^{2}and the open-circuit voltage (V

_{OC}) is 0.451 V. Consider thermal voltage (V

_{T}) to be 25mV. If the intensity of the incident light is increased by 20 times, assuming that the temperature remains unchanged. V

_{oc}(in volts) will be__________

Fill in the Blank Type Question |

Question 23 |

**->**IF Left side is p-region and right side is n-region then electric field triangle will be down warded

**->**If Left side is n-region and right side is p-region, then electric field triangle will be upward.

Question 24 |

Saturation, Saturation | |

Linear, Linear | |

Linear, Saturation | |

Saturation, Linear |

So, For transistor M

_{2}

Assume that M1 is working in saturation, so that

Now, for M1, transistor to work in saturation but it is not satisfied by M1 transistor and so, transistor M1 is ON but working in linear region.

Question 25 |

Intrinsic semiconductor doped with pentavalent atoms to form n-type semiconductor | |

Intrinsic semiconductor doped with trivalent atoms to form n-type semiconductor | |

Intrinsic semiconductor doped with pentavalent atoms to form p-type semiconductor | |

Intrinsic semiconductor doped with trivalent atoms to form p-type semiconductor |

Question 26 |

P: As channel length reduces, OFF-state current increases.

Q: As channel length reduces, output resistance increases.

R: As channel length reduces, threshold voltage remains constant.

S: As channel length reduces, ON current increases.

Which of the above statements are INCORRECT?

P and Q | |

P and S | |

Q and R | |

R and S |

Q: FALSE, As channel length reduces, output resistance reduces

R: FALSE: As channel length reduces, threshold voltage reduces

S: TRUE : As Channel reduces, ON current increases

If the channel length reduces, then threshold voltage also changes.

So option C is matching.

Question 27 |

^{17}cm

^{-3}on the p- side and a uniform donor doping concentration of 1016 cm-ion the n-side. No external voltage is applied to the diode. Given: kT/q = 26 mV,

The charge per unit junction area (nC cm-2) in the depletion region on the p-side is

Fill in the Blank Type Question |

Therefore, answer is -4.836 nC-cm

^{-2}

Question 28 |

_{N}C

_{OX }= 70 x 10

^{-6 }AV

^{-2}, the threshold voltage is 0.3V, and the channel length modulation parameter is 0.09 V

^{-1}. In the saturation region, the drain conductance (in micro siemens) is _____.

Fill in the Blank Type Question |

Question 29 |

Question 30 |

_{cc}= 15 V, —V

_{cc}= —15V, Load resistor. If Of power is incident on the photodiode, then the modulus of value of the photocurrent (in) through the load is _____.

Fill in the Blank Type Question |

Question 31 |

both the P-region and the N-region are heavily doped | |

the N-region is heavily doped compared to the P-region | |

the P-region is heavily doped compared to the N-region | |

an intrinsic silicon region is inserted between the P-region and the N-region |

Under normal forward bias operation as voltage begins to increase, electrons at first tunnel through the very narrow p-n junction barrier and fill electron states in the conduction band on the n-side, which becomes aligned with empty valence band hole states on the p-side of the p-n junction. As voltage increases further, these states become increasingly misaligned and the current drops. This is called negative resistance becomes current decreases with increasing voltage. p-n diode with high doping on both sides is called tunnel diode. Hence, the correct option is (A).

Question 32 |

^{16}/cm

^{3}. The electron and hole mobilities in the sample are 1200 cm

^{2}/V-s and 400 cm

^{2}/V-s respectively.

Assume complete ionization of impurities. The charge of an electron is 1.6 × 10

^{–19}C. The resistivity of the sample (in Ω-cm) is _________.

Fill in the Blank Type Question |

Question 33 |

_{J}) at a reverse bias (V

_{R}) of 1.25 V is 5 pF, the value of C

_{J}(in pF) when V

_{R}= 7.25 V is________.

Fill in the Blank Type Question |

Question 34 |

^{17}cm

^{–3}and 1 × 10

^{15}cm

^{–3}, respectively. The lifetimes of electrons in P region and holes in N region are both 100 µs. The electron and hole diffusion coefficients are 49 cm

^{2}/s and 36 cm

^{2}/s, respectively. Assume kT/q = 26 mV, the intrinsic carrier concentration is 1× 10

^{10}cm

^{–3}, and q = 1.6 × 10

^{–19}C. When a forward voltage of 208 mV is applied across the diode, the hole current density (in nA/cm

^{2}) injected from P region to N region is ______.

Fill in the Blank Type Question |

Where,

Q = charge on electron

n

_{i}= Intrinsic carrier concentration in silicon

N

_{D}= Donor doping

D

_{P}= Hole diffusion coefficient

L

_{p}= Mean diffusion length of hole

V

_{FB}= Forward voltage applied across diode

V

_{T}= kT/q = 26 mV

Using the above values, we get hole current density injected from P region to N region is = 28.617 nA/cm

^{2}

by substitute the values in the above equation

Question 35 |

only in active mode | |

only in active and saturation modes | |

only in active and cut-off modes | |

in active, saturation and cut-off modes |

Question 36 |

Widths of PMOS transistors should be doubled, while widths of NMOS transistors should be halved. | |

Widths of PMOS transistors should be doubled, while widths of NMOS transistors should not be changed. | |

Widths of PMOS transistors should be halved, while widths of NMOS transistors should not be changed. | |

Widths of PMOS transistors should be unchanged, while widths of NMOS transistors should be halved. |

Question 37 |

_{G}= 0.8 V is 2 x10

^{11}cm

^{-2}. For V

_{G}= 1.3V, the inversion carrier density is 4x10

^{11}cm

^{-2}. What is the value of the inversion carrier density for V

_{G}= 1.8 V?

4.5 x10 ^{11}cm^{-2} | |

6.0 x10 ^{11}cm^{-2} | |

7.2x x10 ^{11}cm^{-2} | |

8.4 x10 ^{11}cm^{-2} |

Question 38 |

^{+}n junction. The n-region is uniformly doped with a donor density N

_{D}. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field

**E**. Assume

_{crit}**E**to be independent of N

_{crit}_{D}. If the built-in voltage of the junction is much smaller than the breakdown voltage, V

_{BR}, the relationship between and is given by

= not constant | |

= constant | |

= constant | |

= constant |

Question 39 |

Given

**q =1.6 x10**, for silicon, the value of L in nm is_____

^{-19}coulombFill in the Blank Type Question |

Question 40 |

Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of V_{GS}, and V_{DS}. Given,

g_{m }= 0.5 μA/ V for V_{DS }= 50 mV and V_{GS }= 2V,

g_{d =}8 μA/ V _{for }V_{DS} = 0 mV and V_{GS }= 2V_{,}

The threshold voltage (in volts) of the transistor is ______

Fill in the Blank Type Question |

Question 41 |

no relationship among these band gaps exists |

Question 42 |

Fill in the Blank Type Question |

Question 43 |

*xy*-plane is bounded by the straight lines 2

*x*= 3

*y*,

*y*= 0 and

*x*= 3. The volume above the triangle and under the plane

*x*+

*y*+

*z*= 6 is ______

Fill in the Blank Type Question |

Question 44 |

*npn*BJT, biased in the active region, is linear, as shown in the figure. If the area of the emitter-base junction is 0.001 , in the base region and depletion layer widths are negligible, then the collector current Ic (in mA). at room temperature is __________ (Given: thermal voltage VT = 26 mV at room temperature, electronic charge)

Fill in the Blank Type Question |

Question 45 |

_{1}= 1 nm and dielectric constant ε

_{1}= 4) and Y (of thickness t

_{2}= 3 nm and dielectric constant ε

_{2}= 20). The capacitor in Figure II has only insulator material X of thickness t

_{Eq}. If the capacitors are of equal capacitance, then the value of t

_{Eq}(in nm) is ______

Fill in the Blank Type Question |

Question 46 |

D1 only | |

D2 only | |

Both D1and D2 | |

None of D1 and D2 |

Question 47 |

_{2}) of MOSFETs?

Sputtering | |

Molecular beam epitaxy | |

Wet oxidation | |

Dry oxidation |

**->**Dry oxidation has a lower growth rate than wet oxidation although the oxide film quality is better than the wet oxide film. Therefore thin oxides such as screen oxide, pad oxide, and especially gate oxide normally use the dry oxidation process. Dry oxidation also results in a higher density oxide than that achieved by wet oxide and so it has a higher breakdown voltage (5 to 10 MV/cm)

**->**Dry oxidation is better than wet oxidation, so dry oxidation is always preferred for gate dielectric because we want precision.

Hence, the correct option is (D).

Question 48 |

Current gain will increase | |

Unity gain frequency will increase | |

Emitter-base junction capacitance will increase | |

Early voltage will increase |

Question 49 |

_{EB}= 600 mV, the emitter-collector voltage V

_{EC}(in Volts) is _____.

Fill in the Blank Type Question |

Question 50 |

_{1}and D

_{2}are ideal. In the steady state condition, the average voltage V

_{ab}(in Volts) across the 0.5 μF capacitor is _______.

Fill in the Blank Type Question |

*D*

_{2}will be ON. So, peak voltage at point

*a*is

For negative half cycle, diode

*D*1 will be ON. So, peak voltage at point

*b*is

Hence

*V*=

_{ab}*V*−

_{a}*V*

_{b}= 50 −(−50) 50+50= 100 V

At steady state, voltage across capacitor remains same i.e. 100V.

So average value=100V

**Another Approach**Question 51 |

**NOT TRUE**?

The left side of the junction in n-type and the right side is p-type | |

Both the n-type and p-type depletion regions are non-uniformly doped | |

The potential difference across the depletion region is 700 mV | |

If the p-type region has a doping concentration of 10 ^{15} cm^{–3}, then the doping concentration in the n-type region will be 10^{16} cm^{–3}. |

In option (C) it is given as 700 mV, so it is not true.

Question 52 |

^{–1}) is ____.

Fill in the Blank Type Question |

Question 53 |

_{n}= m

_{n}C

_{ox}(W/L) = 1 mA/V

^{2}; V

_{TN}= 1V. Assume that the channel length modulation parameter l is zero and body is shorted to source. The minimum supply voltage V

_{DD}(in volts) needed to ensure that transistor M

_{1}operates in saturation mode of operation is _________

Fill in the Blank Type Question |

Question 54 |

_{1}[n] and x

_{2}[n] have the same energy. Suppose x

_{1}[n] = a 0.5

^{n}u[n], where a is a positive real number and u[n] is the unit step sequence. Assume

Then the value of a is__________.

Fill in the Blank Type Question |

_{1}[n]= -(0.5)

^{n}u[n]

Energy of signal

*x*

_{1}[

*n*] is

Again,

*x*

_{2}[

*n*] =

So, energy of signal

*x*

_{2}[

*n*] is

Given that

Energy of signal

*x*

_{1}[

*n*] = Energy of signal

*x*

_{2}[

*n*]

Since is a positive real number, so we have

α = 1.5

Question 55 |

^{16}/cm

^{3}. The expected value of mobility versus doping concentration for silicon assuming full dopant ionization is shown below. The charge of an electron is C.

The conductivity (in S cm

^{–1}) of the silicon sample at 300 K is____

Hole and Electron Mobility in Silicon at 300 K

1.82 | |

1.92 | |

192 | |

150.2 |

Question 56 |

^{20}electron hole pairs per cm

^{3}second. The minority carrier lifetime in the sample is 1 μs. In the steady state, the hole concentration in the sample is approximately 10

^{x}, where x is an integer. The value of x is___

4 | |

14 | |

2 | |

12 |

**->**Rate of generation =10

^{20}electron-hole pairs per cm

^{3}per second.

**->**At steady state (at the end of lifetime) t = 1μsec, concentration of hole-electron pair in 1 μsec is

= 10

^{20}×10

^{–6}= 10

^{14}

So,

*x*= 14

Question 57 |

_{D}and the mobility of electrons are 10

^{16}cm

^{-3}and 1000 cm

^{2}V

^{–1}s

^{–1}, respectively. The average time (in μs) taken by the electrons to move from one end of the bar to other end is__________

100 | |

200 | |

300 | |

400 |

Given,

= 10

^{4}cm/s

Question 58 |

2 | |

4 | |

6 | |

8 |

Question 59 |

The doping concentrations on the p-side and n-side of a silicon diode are
** 1 X 10 ^{16} CM^{-3} and 1 X 10 ^{17} CM ^{-3} ** , respectively.
A forward bias of 0.3 V is applied to the diode. At T = 300K, the intrinsic carrier concentration of silicon
The electron concentration at the edge of the depletion region on the p-side is

Question 60 |

_{D}= 2.25 x 10

^{15}atoms / cm

^{3}. Given the intrinsic carrier concentration of silicon at T = 300 K is n

_{i}= 1.5 x 10

^{10}cm

^{-3}. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are

Question 61 |

Question 62 |

_{n}on the N-side of the junction is 0.2 μm and the permittivity of silicon

**(ε**is

_{si})**1.044x10**At the junction, the approximate value of the peak electric field (in kV/cm) is _________.

^{-12}F/cm25.40 | |

28.32 | |

30.66 | |

32.42 |

Question 63 |

Question 64 |

The minority carrier mobility | |

The minority carrier recombination lifetime | |

The majority carrier concentration | |

The excess minority carrier concentration |

_{no}+ n

_{n}) (P

_{no}+ P

_{n})

n

_{no}and P

_{no}= Electron and hole concentrations respectively under thermal equilibrium

n

_{n}and p

_{n}=Excess elements and hole concentrations respectively

Question 65 |

**1 x 10**respectively. Assume that the intrinsic carrier concentration in silicon n

^{16}cm^{-3}and 5 x 10^{18}cm^{-3},_{i}= 1.5 x10

^{10}cm

^{-3}at 300K,

**kT/q =26 mV**and the permittivity of silicon The built-in potential and the depletion width of the diode under thermal equilibrium conditions, respectively, are

0.7 V and 1 x 10 ^{-4} cm | |

0.86 V and 1 x 10 ^{-4} cm | |

0.7 V and 3.3 x 10 ^{-5} cm | |

0.86 V and 3.3 x 10 ^{-5} cm |

Question 66 |

^{6}cm

^{-3}respectively. In order to generate electron hole pairs in GaAs, which one of the wavelength ranges of incident radiation, is most suitable? (Given that: Plank’s constant is velocity of light is and charge of electron is

Now, since energy required is more than band gap and energy is inversely proportional to wavelength.

Therefore, λ<0.87μm

Question 67 |

The slope of the line can be used to estimate

Band gap energy of silicon (Eg) | |

Sum of electron and hole mobility in silicon | |

Reciprocal of the sum of electron and hole mobility in silicon | |

Intrinsic carrier concentration of silicon |

Question 68 |

0.85 μm | |

1.125 μm | |

1.450 μm | |

2.250 μm |

Question 69 |

_{D}= 1×10

^{15}/cm

^{3 }donor atoms. Assume that the intrinsic carrier concentration n

_{i}= 1.5×10

^{10}/cm

^{3}

_{.}If the sample is additionally doped with N

_{A}= 10×10

^{18}/cm

^{3}acceptor atoms, the approximate number of electrons/cm

^{3}in the sample, at T=300 K, will be _________________.

22.52 | |

245.2 | |

255.2 | |

265.2 |

Question 70 |

If is the lowest energy level of the conduction band, is the highest energy level of the valance band and is the Fermi level, which one of the following represents the energy band diagram for the biased N-type semiconductor?

_{f}lies near to the conduction Band and since it is positively biased on right side so energy band diagram will be tilted downwards on right side, hence option D is the correct answer.

Question 71 |

injection, and subsequent diffusion and recombination of minority carriers | |

injection, and subsequent drift and generation of minority carriers | |

extraction, and subsequent diffusion and generation of minority carriers | |

extraction, and subsequent drift and recombination of minority carriers |

(i) holes will be injected from p-side.

(ii) holes will be diffused in n-side.

(iii)Recombination will take place.

Question 72 |

superior quality oxide with a higher growth rate | |

inferior quality oxide with a higher growth rate | |

inferior quality oxide with a lower growth rate | |

superior quality oxide with a lower growth rate |

- In IC technology, The electrical properties of dry oxidation is better than wet oxidation. But it is a low processor
- Dry oxidation as compared to wet oxidation produces superior quality oxide with a lower growth rate

Question 73 |

The current in the circuit is

10 mA | |

9.3 mA | |

6.67 mA | |

6.2 mA |

10- (1000) i – v = 0 ...(1)

since 10 V is greater than 0.7 v so current will flow through diode and is

from (1) and (2) we have

Question 74 |

_{i}= 10

^{10}per cm

^{3}) n-channel MOS transistor has an area of 1 sq pm and a depth of 1 pm. If the dopant density in the source is 10

^{19}/cm

^{3}, the number of holes in the source region with the above volume is approximately

10 ^{7} | |

100 | |

10 | |

0 |

_{i}= 10

^{10}/ cm

^{3}

A = 1 × 10

^{-12}m

^{2}

d = 10

^{-6}m

V = Ad = 10

^{-18}m

^{3}

= 10

^{-12}cm

^{3}

N

_{D}= n = 10

^{19}/ cm

^{3}

= 10 / cm

^{3}

So holes in volume V is

H = pV = 10

^{-11}

Since holes number cannot be a decimal number so H = 0.

Question 75 |

_{2}, respectively, are 11.7 and 3.9. and ε

_{0}= 8.9 x 10

^{–12}F/m.

The gate-source overlap capacitance is approximately

0.7 fF | |

0.7 pF | |

0.35 fF | |

0.24 pF |

_{g}is

∈

_{r1}= 3.9 as between gate and source there is SiO

_{2}

A

_{1}= 1μm × δ = 1 × 10

^{-6}× 20 × 10

^{-9}

= 2 × 10

^{-14}m

^{2}

d

_{1}= 1nm = 10

^{-9}m

So,

C

_{g}= 69.42 × 10

^{-17}F

C

_{g}= 0.69 × 10

^{-15}

C

_{g}= 0.7fF

Question 76 |

1) Lower diffusion capacitance

2) Smaller parasitic delay and lower dynamic power consumption

3) Lower threshoid voltages

Select the correct answer using the code given below.

1, 2 and 3 | |

1 and 2 only | |

1 and 3 only | |

2 and 3 only |

1. Lower diffusion capacitance

2. Lower parasitic delay

3. Lower threshold voltage.

4. Lower dynamic power consumption.

Question 77 |

mask programming | |

one-time programming | |

avalanche injection or Fowler-Nordheim tunneling | |

erasing |

Applying a ‘high voltage’ to the control gate causes ‘Avalanche Injection’ in the EPROMS. Due to this electrons overcome the resistance offered by SiO

_{2}layer and gets trapped on the floating gate.

Question 78 |

**Statement (I):**The external surface of a crystal is an imperfection in itself as the atomic bonds do not extend beyond the surface.

**Statement (II):**The external surfaces have surface energies that are related to the number of bonds broken at the surface.

Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) | |

Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) | |

Statement (I) is true but Statement (II) is false | |

Statement (I) is false but Statement (II) is true |

Question 79 |

**Statement (I):**By organizing various ‘optical functions into an ‘array structure’ via nano-pattern replication, ‘spatial integration’ is established.

**Statement (II):**By adding a nano-optic layer or layers to functional optical materials, the ‘hybrid integration’ is possible to be achieved!

Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) | |

Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) | |

Statement (I) is true but Statement (II) is false | |

Statement (I) is false but Statement (II) is true |

- Nano-optic elements consists of numerous nano scale structure created by replicating nano pattern masters, with spatial integration method is established.
- In hybrid integration method, discreate nono-optic devices are produced by adding a nano-optic layer or layers.
- Hybrid integration can be achieved by adding a nano-optic layer or layers to functional optical materials. The process by which nano-optic elements are fabricated is flexible and robust, allowing for cost-saving manufacturing process integration in creating hybrid optics.

Question 80 |

GaAs | |

GaP | |

GaSe | |

GaN |

GaSe compound does not exist .

Refer the Topic Wise Question for Basic Electronics Engineering Electronic Devices and Circuits

Question 81 |

0 mA | |

200 mA | |

100 mA | |

2 mA |

if a diode is conducting in forward bias and immediately switched to reverse bias voltage , the diode will allow to flow the current in reverse bias for short time so that the forward voltage bleeds off, (or we can say for the time which diode takes to be in complete reverse bias, it act as a short circuit). This is called reverse recovery time.

I =

= 200mA.

Refer the Topic Wise Question for Basics of Semiconductors Electronic Devices and Circuits

Question 82 |

^{16}atoms/cm

^{3}at 300K temperature. Determine the approcimate resistivity of the slab. (Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 x 10

^{10}/ cm

^{3}Hole Mobility =500 cm

^{2}/Vs at 300 K; Electron Mobilty = 1300 cm

^{2}/Vs at 300 K).

0.48 -cm | |

0.35 -cm | |

0.16 -cm | |

1.25 -cm |

conductivity of extrinsic semiconductor()= nq+pq

here , n and p are no. of electrons and protons respectively

q=1.6

=1300 cm

^{2}/Vs

500 cm

^{2}/Vs

As it is Boron doped Silicon semiconductor so it is p-type.

Therefore p==10

^{16}atoms/cm

^{3}

Using mass action law , n=

Then , = 1.6 (n)

,

Resistivity () =

=1.25 -cm

Refer the Topic Wise Question for Carrier Transport Electronic Devices and Circuits

Question 83 |

L is known as

Lattice constant | |

Lorenz number | |

Lanevin Function | |

Larmor number |

Here, L=Lorenz number

Lorenz number is widely used for the determination of k

_{E}in case of σ measurements. The given equation is called as Widemann-Franz law and it states that the ratio of the contribution(electrical) done by the thermal conductivity(k) to conductivity(σ) in a certain metal is directly proportional to the temperature 'T'.

L=2.44 x 10

^{-8}wΩk

^{-2}

Refer the Topic Wise Question for Conductive Materials and Superconductors Electronic Devices and Circuits

It’s very useful.