Electronic Devices and Circuits subject wise
Question 1 |

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Question 2 |

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Question 3 |
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-> 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 |
Q, R, P | |
Q, P, R | |
R, P, Q | |
P, Q, R |
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 |
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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 an Arsenic site by a si atom produces a hole
so p-type.
Question 9 |
0.748 V | |
0.460 V | |
0.288 V | |
0.173 V |

Question 10 |

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 |



Which of the following statements about estimates for gm and ro is correct?
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Question 13 |
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Question 14 |
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Question 15 |
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 |

Current through the nMOS transistors can be modeled as




The voltage (in volts, accurate to two decimal places) at Vx is _______.
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Question 17 |
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Question 18 |
VR > VG >VB | |
VR < VG< VB | |
VR = VG = VB | |
VR > VG < VB |
->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 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 = IC / IB) will increase
Question 20 |
µn and oxide capacitance per unit area Cox. If gate-to-source voltage VGS=0.7V, drain-to- source voltage VDS=0.1V, ( µnCox) =100 µA/V2, threshold voltage VTH=0.3V and (W/L) =50, then the transconductance gm (in mA/V) is______
Fill in the Blank Type Question |
VGS=0.7V

Question 21 |
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/cm2, is
Fill in the Blank Type Question |


=







Answer Range : (6.85 to 6.95)
Question 22 |
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Question 23 |
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->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 M2


Assume that M1 is working in saturation, so that



Now, for M1, transistor to work in saturation


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




So option C is matching.
Question 27 |

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 |
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Question 29 |




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Question 30 |




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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 |
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 _________.
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Question 33 |
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Question 34 |
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Where,
Q = charge on electron
ni = Intrinsic carrier concentration in silicon
ND = Donor doping
DP = Hole diffusion coefficient
Lp = Mean diffusion length of hole
VFB = Forward voltage applied across diode
VT = kT/q = 26 mV


Using the above values, we get hole current density injected from P region to N region is = 28.617 nA/cm2
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 |
4.5 x1011cm-2 | |
6.0 x1011cm-2 | |
7.2x x1011cm-2 | |
8.4 x1011cm-2 |

Question 38 |



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Question 39 |


Given q =1.6 x10-19 coulomb,

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Question 40 |
Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of VGS, and VDS. Given,
gm = 0.5 μA/ V for VDS = 50 mV and VGS = 2V,
gd =8 μA/ V for VDS = 0 mV and VGS = 2V,
The threshold voltage (in volts) of the transistor is ______
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Question 41 |

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no relationship among these band gaps exists |

Question 42 |



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Question 43 |
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Question 44 |




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Question 45 |

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Question 46 |

D1 only | |
D2 only | |
Both D1and D2 | |
None of D1 and D2 |
Question 47 |
Sputtering | |
Molecular beam epitaxy | |
Wet oxidation | |
Dry oxidation |
->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 |

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Question 50 |

Fill in the Blank Type Question |

For negative half cycle, diode D1 will be ON. So, peak voltage at point b is

Hence Vab= Va− Vb
= 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 |

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 1015 cm–3, then the doping concentration in the n-type region will be 1016 cm–3. |

In option (C) it is given as 700 mV, so it is not true.
Question 52 |
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Question 53 |

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Question 54 |

Then the value of a is__________.
Fill in the Blank Type Question |
Energy of signal x1[n] is


Again, x2[n] =

So, energy of signal x2[n] is

Given that
Energy of signal x1[n] = Energy of signal x2 [n]

Since

α = 1.5
Question 55 |

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 |
4 | |
14 | |
2 | |
12 |
->At steady state (at the end of lifetime) t = 1μsec, concentration of hole-electron pair in 1 μsec is
= 1020 ×10–6 = 1014
So, x = 14
Question 57 |


100 | |
200 | |
300 | |
400 |

Given,

= 104 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 1016 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
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Question 60 |
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Question 61 |



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Question 62 |

25.40 | |
28.32 | |
30.66 | |
32.42 |

Question 63 |



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Question 64 |
The minority carrier mobility | |
The minority carrier recombination lifetime | |
The majority carrier concentration | |
The excess minority carrier concentration |
nno and Pno = Electron and hole concentrations respectively under thermal equilibrium
nn and pn =Excess elements and hole concentrations respectively
Question 65 |
1 x 1016 cm-3 and 5 x 1018 cm-3, respectively. Assume that the intrinsic carrier concentration in silicon ni = 1.5 x1010 cm-3 at 300K, kT/q =26 mV and the permittivity of silicon

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 |




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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 |
22.52 | |
245.2 | |
255.2 | |
265.2 |

Question 70 |

If



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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 |
107 | |
100 | |
10 | |
0 |
A = 1 × 10-12 m2
d = 10-6 m
V = Ad = 10-18 m3
= 10-12 cm3
ND = n = 1019 / cm3

= 10 / cm3
So holes in volume V is
H = pV = 10-11
Since holes number cannot be a decimal number so H = 0.
Question 75 |

The gate-source overlap capacitance is approximately
0.7 fF | |
0.7 pF | |
0.35 fF | |
0.24 pF |

∈r1 = 3.9 as between gate and source there is SiO2
A1 = 1μm × δ = 1 × 10-6 × 20 × 10-9
= 2 × 10-14 m2
d1 = 1nm = 10-9 m
So,

Cg = 69.42 × 10-17 F
Cg = 0.69 × 10-15
Cg = 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 SiO2 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 |
0.48 ![]() | |
0.35 ![]() | |
0.16 ![]() | |
1.25 ![]() |
conductivity of extrinsic semiconductor(



here , n and p are no. of electrons and protons respectively
q=1.6



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

Using mass action law , n=

Then ,



,

Resistivity (


=1.25

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 kE 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 10-8wΩk-2
Refer the Topic Wise Question for Conductive Materials and Superconductors Electronic Devices and Circuits
It’s very useful.
super. kindly update till 2023