Test: Electronic Devices - 1

10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Electronic Devices - 1

Attempt Test: Electronic Devices - 1 | 10 questions in 30 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions

In a semiconductor which of the following carrier can contribute to the current?


In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons take part in the flow of the current.


The Fermi level in a semiconductor bar should:


Under equilibrium:

  • The fermi level across the entire material will be sum and does not vary with distance.
  • If there is any disturbance in the material, like junction contact, injection of impurities at any point, the charge carriers redistribute themselves such that the fermi-potential is same in entire material.

Under non-equilibrium:

  • The fermi level is uneven with gradient of charge distribution across distance, in material.
  • It can be studied using quasi-fermi states, related to charge distribution.
*Answer can only contain numeric values

Hole mobility in Ge at room temperature is 1900 cm2/V-sec. The diffusion coefficient is ________cm2/sec.
(Write answer to one decimal point.) 

(Take kT = 25 mV)


Using the relation, D = 0.025 × 1900 cm2/sec
= 47.5 cm2/sec


In the band diagram of a semiconductor, the fermi level is 0.3 eV above the intrinsic level. The energy level E in the diagram represents:


  • Given that fermi level is 0.3 eV above the intrinsic level, hence the type of semiconductor is n-type.
  • In n-type the donor level is just below the conduction band and above the fermi level.

The Difference between the donor energy level and fermi level in a n-type semiconductor in where 25% of the atoms are ionised at 300 k is:


As 25% of donor atoms are ionised, the occupation probability of donor level is 0.75.

ED - EF = -0.028 eV

*Answer can only contain numeric values

A semiconductor with intrinsic carrier concentration 1 × 1010 cm-3 at 300°K has both valence and conduction band effective densities of states NC and NV equal to 1019 cm-3. The band gap Eg is _____ eV.


The intrinsic concentration is related to bandgap by:

Eg = 0.0518 ln (108) eV

= 0.954 eV

*Answer can only contain numeric values

In a very long p-type Si bar with doping concentration Na = 1017 cm-3, excess holes are injected such that excessive concentration of holes at x = 0 is 5 × 1016 cm-3. The hole concentration at x = 1 μm is _____ × 1017 cm-3. Take μp = 500 cm2/v-s and recombination time constant τ = 10-8 s, kT = 0.0259 eV


Diffusion constant: 

Diffusion length: 

  • Hole concentration at any distance x is sum of equillibrium and excess hole concentration excess hole concentration varies with the distance x as Δp e-x/Lp
  • p = p + Δp e-x/Lp
    Where po = Doping concentration
    Δp = extra hole concentration
    = (1+.379) x 1017
    1.379 × 1017 cm-3

In an intrinsic semiconductor, the Fermi energy level EF doesn’t lie in the middle of the band gap cause:

  • In an intrinsic semi-conductor the Fermi energy:Where NC and NV are density of state in valence and conduction bond respectively.
  • Since NC ∝ mn, effective mass of electron and Nv ∝ mp, effective mass of hole. Since mn and mp are not equal: 

A silicon bar is doped with donor impurities ND = 2.25 × 1015 cm-3. If the electron mobility μn = 1000 cm2/v-s then the approximate value of resistivity of silicon bar assuming partial ionization of 55% is __________ (Ω - cm)


Given ND = 2.25 × 1015 cm-3

Partial ionization of 55%, number of electrons is:

⇒ n = 0.55 × 2.25 × 1015 cm-3

= 1.2375 × 1015 cm-3

Electron mobility μn = 1000 cm2/v-s

Resistivity of n-type silicon = 

= 5.05 Ω – cm


Holes are injected into n-type Ge so that the at the surface of the semiconductor hole concentration is 1014/cm3. If diffusion constant of a hole in Ge is 49cm2/sec and minority carrier lifetime is τp = 10-3 sec. Then the hole concentration Δp at a distance of 4mm from the surface is ______1014/cm3.



= 1.6 × 1013/cm3

= 0.16 × 1014/cm3

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