NTA UGC NET/JRF Exam, August -2024 (Electronic Science)

Total Questions: 100

1. The peak concentration in the lateral autodoping profile is a function of the following parameters

(a) Temperature
(b) Surface concentration in the buried layer
(c) Applied voltage
(d) Growth rate

Correct Answer: D. (a), (b) and (d) only
Solution:

The peak concentration in the lateral autodoping profile (i.e., unintended doping in an epitaxial layer from the buried-layer or substrate doping) depends primarily on:
(a) Temperature: Affects diffusion and vapour-phase transport of dopants.
(b) Surface concentration in the buried layer: Higher buried-layer doping leads to higher autodoping drive.
(d) Growth rate: During epitaxial growth, a slower growth rate gives dopants more time to diffuse/laterally redistribute, affecting the peak concentration.
Typically, "(c) Applied voltage" is not a usual factor in standard thermal/chemical epitaxy autodoping. Hence, the correct combination is (a), (b), (d).

2. A parallel - plate capacitive transducer having air as dielectric between the plates, plate area is 50 mm × 50 mm and plate spacing is 0.5 mm. The displacement causes the capacitance to change by 10 pF. The sensitivity of the capacitive displacement transducer is:

Correct Answer: B. 66.67 pF/mm

3. In a charge coupled device, the charge storage and transfer action is controlled by one of the following:

Correct Answer: B. Gate Electrodes
Solution:In a Charge-Coupled Device (CCD), charge packets are created and transferred under the control of clocked gate electrodes (MOS gates). Source/drain electrodes do not perform the storage/transfer function in the same manner; rather, CCD shifting of charge is governed by sequentially pulsed gate electrodes creating potential wells/shifts.

4. Following set of instructions in 8086 searches a table of 100 bytes for OAН. Select the correct option:

MOV CX, 100
MOV AL, OAH
(a) STC
(b) CLC
(c) CLD
(d) REPNE SCASB
(e) JCXZ NOT FOUND
NOT FOUND : END

Correct Answer: C. (c), (d), (a), (e), (b)

5. The band gap of a semiconductor material that can be considered for making solar cell will be:

Correct Answer: C. Between 1 & 2 eV
Solution:

For a single-junction photovoltaic (solar cell), the most useful bandgap range typically lies roughly between 1 eV and 2 eV (e.g., Si ~1.12 eV, GaAs~1.43 eV). Too small a gap (<1 eV) wastes much energy as heat (low open-circuit voltage), and too large a gap (>3 eV) captures too little of the solar spectrum.

6. Arrange the following in the decreasing order of their Bandgap (eV) at 300K.

(a) GaAs
(b) Si
(c) Ge
(d) ZnO
(e) InSb

Correct Answer: D. (d), (a), (b), (c), (e)
Solution:

Which is
ZnO > GaAs > Si > Ge > InSb
ZnO≈3.37 eV
GaAs≈ 1.43 eV
Si≈1.12 eV
Ge≈ 0.66 eV
InSb≈0.17 eV
Hence, in decreasing order:
ZnO > GaAs > Si > Ge > InSb.

7. A thyristor having the equivalent capacitance of the depletion layer of reverse biased junction as 20 picofarad can be fired with a dv/dt of 120 V/usec. The capacitive current flowing through the junction will be:

Correct Answer: B. 2.4 mA
Solution:

A thyristor can be inadvertently triggered ("fired") by the capacitive current through its depletion-layer junction if the voltage across it rises too fast. The current is given by

8. The circular loop of radius 'a' as shown in the figure carries a current I. The magnetic field (in the spherical coordinate system) at the observation point is:

 

Correct Answer: A.

9. APMMC instrument has a three - resister Ayrton shunt connected in parallel with it to make an ammeter as shown in the figure. The current range of the ammeter when connected to C is:

Correct Answer: C. 100 mA
Solution:The PMMC (APMMC) movement has Rₘ = 1 k and a full-scale current of 100 μA, so the meter drops about
Vₘ = (100 μΑ)(1 ΚΩ)
= 0.1 V
at full deflection.
When the ammeter switch is placed to "C", the circuit routes the meter in series with R₃ = 9 as one branch and R₂ = 0.95 as the parallel shunt branch. For the meter to carry its full-scale 100 μA, the voltage across the parallel combination (meter + R₃ and R₂) must be the same:

Hence the total current through that node is about
Iₜₒₜₐₗ ≈0.106 A + 0.0001 A
= 0.1061 A (≈ 106 mA).
A small series resistor R₁ = 0.05 2 adds only a negligible drop. Thus the overall ammeter current range on this tap is about 100 mA.

10. Match the List-I with List-II.

Choose the correct answer from the options given below:

(a)(b)(c)(d)
A.IIIIVIII
B.IIIIVIII
C.IIIIIVII
D.IIIVIIII
Correct Answer: B.