MCQOPTIONS
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MCQOPTIONS
This section includes 39 Mcqs, each offering curated multiple-choice questions to sharpen your Electronic Devices Circuits knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Built-in potential V0 of a junction depends on: |
| A. | Doping densities and Temperature |
| B. | Only Temperature |
| C. | Cross sectional area of the junction |
| D. | Doping densities only |
| Answer» B. Only Temperature | |
| 2. |
An example of trivalent impurity is: |
| A. | Aluminium |
| B. | Germanium |
| C. | Barium |
| D. | Chlorine |
| Answer» B. Germanium | |
| 3. |
If f(E) is Fermi dirac distribution function, then 1 – f(E) is the probability:(Where Ef is Fermi level) |
| A. | that a state is empty below Ef |
| B. | that a state is filled below Ef |
| C. | that a state is empty above Ef |
| D. | that a state is filled above Ef |
| Answer» B. that a state is filled below Ef | |
| 4. |
A silicon sample A is doped with 1018 atom/cm3 of Boron and another silicon sample B of identical dimensions is doped with 1018 atom/cm3 of Phosphorous. If the ratio of electron to hole mobility is 3, then the ratio of conductivity of sample A to that B is: |
| A. | \(\frac{3}{2}\) |
| B. | \(\frac{2}{3}\) |
| C. | \(\frac{1}{3}\) |
| D. | \(\frac{1}{2}\) |
| Answer» D. \(\frac{1}{2}\) | |
| 5. |
Mobilities of free electrons and holes in pure Ge are 0.38 and 0.18 m2/V s, respectively. Assume ni for Ge = 2.5 × 1019 m-3. Calculate the intrinsic resistivity of Ge. Take q = 1.6 × 10-19 Coulomb |
| A. | 0.786 Ωm |
| B. | 0.846 Ωm |
| C. | 0.446 Ωm |
| D. | 0.946 Ωm |
| Answer» D. 0.946 Ωm | |
| 6. |
Match the following list:List - IList – II(a) Resist on UV exposure undergoes cross-linking(i) +ve resist(b) Resist on UV exposure undergoes a decomposition reaction(ii) −ve photoresist(c) Space charge width at zero bias for M−S contact(iii) n.p = ni2(d) Law of mass action(iv) \({\left[ {\frac{{2{\epsilon_s}{V_{bi}}}}{{q{N_D}}}} \right]^{\frac{1}{2}}}\) Correct code are: |
| A. | (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii) |
| B. | (a) – (iii), (b) – (ii), (c) – (i), (d) – (iv) |
| C. | (a) – (ii), (b) – (iii), (c) – (iv), (d) – (i) |
| D. | (a) – (i), (b) – (ii), (c) – (iv), (d) – (iii) |
| Answer» B. (a) – (iii), (b) – (ii), (c) – (i), (d) – (iv) | |
| 7. |
A p-type silicon sample has an intrinsic carrier concentration of 1.5 × 1010 /cm3 and a hole concentration of 2.25 × 1015 /cm3. Then the electron concentration is |
| A. | 1.5 × 1025 /cm3 |
| B. | 105 /cm3 |
| C. | 1010 /cm3 |
| D. | 0 |
| Answer» C. 1010 /cm3 | |
| 8. |
In semiconductors, a donor may be |
| A. | a trivalent impurity |
| B. | a tetravalent impurity |
| C. | a pentavalent impurity |
| D. | a noble gas |
| Answer» D. a noble gas | |
| 9. |
In an intrinsic semiconductor, the intrinsic carrier density is: |
| A. | \({N_C}{N_V}{e^{\frac{{Eg}}{{2kT}}}}\) |
| B. | \(\sqrt {{N_C}{N_V}} {e^{\frac{{Eg}}{{kT}}}}\) |
| C. | \(\sqrt {{N_C}{N_V}} {e^{\frac{{-Eg}}{{2kT}}}}\) |
| D. | \(\sqrt {{N_C}{N_V}} {e^{ + \frac{{Eg}}{{2kT}}}}\) |
| Answer» D. \(\sqrt {{N_C}{N_V}} {e^{ + \frac{{Eg}}{{2kT}}}}\) | |
| 10. |
In an open circuited p-n junction, the contact difference of potential is |
| A. | \(\frac{kT}{q}\) |
| B. | \(\frac{kT}{q} ln\frac{(N_A N_D)}{(n_1^2 )}\) |
| C. | \(\frac{kT}{q} ln\frac{(N_D)}{(n_1^2 )}\) |
| D. | \(\frac{kT}{q}ln\frac{N_A}{(N_D,n_1^2 )}\) |
| Answer» C. \(\frac{kT}{q} ln\frac{(N_D)}{(n_1^2 )}\) | |
| 11. |
A. 0.2 μm long semiconductor sustains a voltage of 1 V. If the low field mobility is 1350 cm2/V.s and the saturation velocity of the carriers 107 cm/s, the effective mobility will be: |
| A. | 270 cm2 /V.S |
| B. | 50 cm2 / V.S |
| C. | 7.75 cm2/ V. S |
| D. | 174 cm2 / V.S |
| Answer» E. | |
| 12. |
Discrete energy level formed due to doping in n-type material is called: |
| A. | Acceptor energy level |
| B. | Mid energy level |
| C. | Donor energy level |
| D. | Intermediate energy level |
| Answer» D. Intermediate energy level | |
| 13. |
At very high temperature, an n-type semiconductor behaves like |
| A. | a p-type semiconductor |
| B. | an intrinsic semiconductor |
| C. | a superconductor |
| D. | an n-type semiconductor |
| Answer» C. a superconductor | |
| 14. |
Assume that the values of mobility of holes and that of electrons in an intrinsic semiconductor are equal and the values of conductivity and initrinsic electron density are 2.32 /Ω m and 2.5 × 1019 /m3 respectively. Then, the mobility of electron/hole is approximately: |
| A. | 0.3 m2/Vs |
| B. | 0.5 m2/Vs |
| C. | 0.7 m2/Vs |
| D. | 0.9 m2/Vs |
| Answer» B. 0.5 m2/Vs | |
| 15. |
A sample of germanium is made p-type by addition of indium at the rate of one indium atom for every 2.5 × 108 germanium atoms. Given, ni = 2.5 × 1019 / m3 at 300 K and the number of germanium atoms per m3 = 4.4 × 1028. What is the value of np? |
| A. | 3.55 × 1018/m3 |
| B. | 3:76 × 1018/ m3 |
| C. | 7.87 × 1018 / m3 |
| D. | 9.94 × 1018/ m3 |
| Answer» B. 3:76 × 1018/ m3 | |
| 16. |
For intrinsic gallium arsenide, conductivity at room temperature is 10-6 (Ω-m)-1, the electron and hole motilities are, respectively 0.85 and 0.04 m2/V-s. The intrinsic carrier concentration n at room temperature is |
| A. | 7.0 × 1012 m-3 |
| B. | 0.7 × 1012 m-3 |
| C. | 7.0 × 10-12 m-3 |
| D. | 0.7 × 10-12 m-3 |
| Answer» B. 0.7 × 1012 m-3 | |
| 17. |
A crystalline silicon is doped uniformly with phosphorus atoms with doping density 1.92 ×1016 atoms/cm3. The hole density in this material at room temperature is: |
| A. | 1016 holes/cm3 |
| B. | 1.17 × 104 holes/cm3 |
| C. | 1.17 holes/cm3 |
| D. | 104 holes/cm3 |
| Answer» C. 1.17 holes/cm3 | |
| 18. |
Direction: Question consists of two statements, one labeled as the 'Assertion (A)' and the other as 'Reason (R)'. Examine these two statements carefully and select the answer to this question using the codes given below:Assertion (A): In an intrinsic semiconductor, the concentration of electrons and holes increases with an increase in temperature.Reason (R): Law of mass action holds good in the case of semiconductors. |
| A. | Both A and R are individually true and R is the correct explanation of A |
| B. | Both A and R are individually true but R is not the correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» C. A is true but R is false | |
| 19. |
Direction: Question consists of two statements, one labeled as the 'Assertion (A)' and the other as 'Reason (R)'. Examine these two statements carefully and select the answer to this question using the codes given below:Assertion (A): At low temperature, the conductivity of a semiconductor increases with an increase in temperature.Reason (R): The breaking of the covalent bonds increases with an increase in temperature, generating an increasing number of electrons and holes. |
| A. | Both A and R are individually true and R is the correct explanation of A |
| B. | Both A and R are individually true but R is not the correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» B. Both A and R are individually true but R is not the correct explanation of A | |
| 20. |
For a semiconductor with electrons and holes as carriers, the resistivity is given by: |
| A. | \(\frac{1}{{{\mu _n}n + {\mu _p}p}}\) |
| B. | \(\frac{1}{{q\left( {{\mu _n}n + \mu_p p\;} \right)}}\) |
| C. | q (μnn+μpp) |
| D. | \(\frac{1}{{q\left( {{\mu _n}n - {\mu _p}p} \right)}}\) |
| Answer» C. q (μnn+μpp) | |
| 21. |
An n-type of silicon can be formed by adding impurity of:1. Phosphorous2. Arsenic3. Boron4. AluminiumWhich of the above are correct? |
| A. | 1 and 2 |
| B. | 2 and 3 |
| C. | 3 and 4 |
| D. | 1 and 4 |
| Answer» B. 2 and 3 | |
| 22. |
Consider the following statements regarding the formation of P-N junctions:1. Holes diffuse across the junction from P-side to N-side.2. The depletion layer is wiped out.3. There is a continuous flow of current4. A barrier potential is set up across the junction. Which of the above statement are correct? |
| A. | 1 and 3 |
| B. | 2 and 3 |
| C. | 1 and 4 |
| D. | 2 and 4 |
| Answer» D. 2 and 4 | |
| 23. |
If the forward bias is applied to the diode, holes are injected from p side to n side, the concentration Pn of holes in the n side above its thermal equilibrium value Pn(0) is given by |
| A. | Pn0 – Pn(0) exp (-x / Lp) |
| B. | Pn0 + Pn(0) exp (+x / Lp) |
| C. | Pn0 + Pn(0) exp (-x / Lp) |
| D. | Pn0 – Pn(0) exp (+x / Lp) |
| Answer» D. Pn0 – Pn(0) exp (+x / Lp) | |
| 24. |
A given semiconductor is doped by phosphorous at different doping concentrations (given below).(a) 2×1012 cm3(b) 4×1014 cm3(c) 3×1013 cm3(d) 2×1015 cm3Sequence them in terms of decreasing conductivity: |
| A. | (a), (b), (c), (d) |
| B. | (d), (b), (c), (a) |
| C. | (b), (a), (d), (c) |
| D. | (d), (c), (a), (b) |
| Answer» C. (b), (a), (d), (c) | |
| 25. |
Consider the following statements:The intrinsic carrier concentration of a semiconductor1. Depends on doping2. Increase exponentially with a decrease of the bandgap of the semi-conductor.3. Increase non-linearly with an increase of temperature4. Increases linearly with increase of temperatureWhich of the above statements are correct? |
| A. | 1, 2 and 3 |
| B. | 1 and 2 only |
| C. | 2 and 3 only |
| D. | 2 and 4 only |
| Answer» D. 2 and 4 only | |
| 26. |
A current of 1A flows towards the right in a conductor. The number of electrons passing per second through any cross-section of the conductor and their direction is about |
| A. | 6 × 1018 towards right |
| B. | 6 × 1018 towards left |
| C. | 6 × 1016 towards right |
| D. | 6 × 1016 towards left. |
| Answer» C. 6 × 1016 towards right | |
| 27. |
How many electrons are there in the valence shell of a pure semiconductor? |
| A. | 1 |
| B. | 3 |
| C. | 4 |
| D. | 6 |
| Answer» D. 6 | |
| 28. |
In a semiconductor, the concentration of holes in the valence band can be given as |
| A. | NV exp[-(EF – EV)/kT] |
| B. | NV exp [(EF – EV)/kT] |
| C. | \(\frac{{kT}}{q}{N_V}\;exp\;\left[ {\left( {{E_V}\;-\;{E_F}} \right)} \right]\) |
| D. | \(\frac{{kT}}{q}\;exp\;\left[ { - \left( {{E_F}\;-\;{E_V}} \right)} \right]\) |
| Answer» B. NV exp [(EF – EV)/kT] | |
| 29. |
Doped silicon has a Hall-coefficient of 3.68 × 10-4 m3C-1 and then its carrier concentration value is: |
| A. | 2.0 × 1022 m-3 |
| B. | 2.0 × 10-22 m-3 |
| C. | 0.2 × 1022 m-3 |
| D. | 0.2 × 10-22 m-3 |
| Answer» B. 2.0 × 10-22 m-3 | |
| 30. |
Ge and Si have: |
| A. | Negative temperature coefficient of resistivity |
| B. | Positive temperature coefficient of resistivity |
| C. | High resistance |
| D. | Low resistance |
| Answer» B. Positive temperature coefficient of resistivity | |
| 31. |
A particular sample of n-type Germanium has a resistivity of 0.1 Ωm at 300 K. Calculate the donor concentration. Take μn = 0.38 and q = 1.6 × 10-19 Coul. |
| A. | 1.24 × 1020 m-3 |
| B. | 1.64 × 1020 m-3 |
| C. | 1.84 × 1020 m-3 |
| D. | 1.44 × 1020 m-3 |
| Answer» C. 1.84 × 1020 m-3 | |
| 32. |
Match the following list:List - IList – II(a) Boltzman Approximation gives(i) \({\left[ {\frac{{KT}}{{{q^2}n}}{\varepsilon _s}{\varepsilon _0}} \right]^{\frac{1}{2}}}\)(b) Total number of holes in the valence band(ii) \(\mathop \smallint \limits_{ - \infty }^{{E_v}} {g_v}\left( E \right){f_h}\left( E \right)dE\)(c) Position of Fermi level(iii) \(\frac{{{E_c} + {E_v}}}{2}KT{\left[ {\frac{{m_h^*}}{{m_e^*}}} \right]^{\frac{3}{4}}}\)(d) Debye screening length(iv) \({f_e}\left( E \right) = \exp \left[ { - \frac{{E - {E_f}}}{{kT}}} \right]\) Correct code is: |
| A. | (a) – (iii), (b) – (ii), (c) – (i), (d) – (iv) |
| B. | (a) – (iv), (b) – (ii), (c) – (iii), (d) – (i) |
| C. | (a) – (iii), (b) – (i), (c) – (iv), (d) – (i) |
| D. | (a) – (iv), (b) – (iii), (c) – (ii), (d) – (i) |
| Answer» C. (a) – (iii), (b) – (i), (c) – (iv), (d) – (i) | |
| 33. |
A block of Silicon is doped with a donor atom density ND = 3 × 1014 atoms/cm3 and with an acceptor density of NA = 0.5 × 1014 atoms/cm3. Find the resultant density of electrons. |
| A. | 4.5 × 1014 electrons/cm3 |
| B. | 3.5 × 1014 electrons/cm3 |
| C. | 2.5 × 1014 electrons/cm3 |
| D. | 5.5 × 1014 electrons/cm3 |
| Answer» D. 5.5 × 1014 electrons/cm3 | |
| 34. |
An n-type silicon bar 0.1 cm long and 100 μm2 cross-sectional area has a majority carrier concentration of 5 x 1020 /m3 and the carrier mobility is 0.13 m2/V-s at 300 K. If a charge of an electron is 1.6 X 10-19 coulomb, the resistance of the bar is |
| A. | 106 ohm |
| B. | 107 ohm |
| C. | 105 ohm |
| D. | 104 ohm |
| Answer» B. 107 ohm | |
| 35. |
Match the two lists and choose the correct answer from the code given below:List IList II(a) Concentration of holes in an n-type semiconductor \(\left( i \right)\frac{{n_i^2}}{{{N_A}}}\)(b) Concentration of electrons in a p-type semiconductor \(\left( {ii} \right)\frac{{kT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)(c) Fermi Level in n-type material \(\left( {iii} \right)\frac{{n_i^2}}{{{N_D}}}\)(d) Contact Potential \(\left( {iv} \right){E_C} - kTln\left( {\frac{{{N_C}}}{{{N_D}}}} \right)\) Code: |
| A. | (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) |
| B. | (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) |
| C. | (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) |
| D. | (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) |
| Answer» C. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv) | |
| 36. |
A Ge sample at room temperature has intrinsic carrier concentration, ni = 1.5 ×1013 cm-3, and is uniformly doped with an acceptor of 3 × 1016 cm-3 and a donor of 2.5 × 1015 cm-3. Then, the minority charge carrier concentration is: |
| A. | 0.918 × 1010 cm-3 |
| B. | 0.818 ×1010 cm-3 |
| C. | 0.918 ×1 012 cm-3 |
| D. | 0.818 ×1012 cm-3 |
| Answer» C. 0.918 ×1 012 cm-3 | |
| 37. |
Arrange the following materials in increasing mobility of holes (At 300K)(a) C(b) Ge(c) SiC(d) GaAsCode: |
| A. | (a) < (b) < (c) < (d) |
| B. | (d) < (c) < (b) < (a) |
| C. | (c) < (a) < (d) < (b) |
| D. | (a) < (c) < (d) < (b) |
| Answer» E. | |
| 38. |
If density of states in conduction band and density of states in valence band are not equal, then the Fermi level in intrinsic semiconductor is |
| A. | \(\frac{{{E_c} - {E_v}}}{2} + \frac{{{N_C} + {N_v}}}{2}\) |
| B. | \(\frac{{{E_C} + {E_v}}}{2}\) |
| C. | \(\frac{{{E_C} + {E_v}}}{2} - \frac{{kT}}{2}\ln \frac{{{N_C}}}{{{N_v}}}\) |
| D. | \(\frac{{{E_C} + {E_v}}}{2} + \frac{{kT}}{2}\ln \frac{{{N_C}}}{{{N_v}}}\) |
| Answer» D. \(\frac{{{E_C} + {E_v}}}{2} + \frac{{kT}}{2}\ln \frac{{{N_C}}}{{{N_v}}}\) | |
| 39. |
A 1m long metal wire of radius 1 cm has a resistance of 1.6 × 10-4 Ω. The resistivity of the metal is |
| A. | 5.0 × 10-8 Ωm |
| B. | 1.0 × 10-8 Ωm |
| C. | 1.6 × 10-8 Ωm |
| D. | 3.2 × 10-8 Ωm |
| Answer» B. 1.0 × 10-8 Ωm | |