InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
In the given figure, which of the diodes are forward biased? A. `1,2,3`B. `2,4,5`C. `1,3,4`D. `2,3,4` |
|
Answer» Correct Answer - B In Figs. `2,4` and `5`. `P`-crystals are more positive as compared to `N`-crystals. |
|
| 152. |
Which of the following semiconductor diodes is reverse biased?A. B. C. |
|
Answer» Correct Answer - A Because `P`-side is more negative then `N`-side. |
|
| 153. |
Which of the following statement is not true ?A. The resistance of intrinsic of semiconductor decrease with increases of temperatureB. Doping pure `Si` with interval impurities give `P`-type semiconductorC. The mejority carriers in `N`-type semiconductors are holesD. A `PN`-junction can act as a semiconductor diode |
|
Answer» Correct Answer - C In `N`-type semiconductor majority change carriers are electrons. |
|
| 154. |
Assertion: A transistor can be used as an amplifier. Reason: A small change in input current can change output on a large scale. |
| Answer» Correct Answer - A | |
| 155. |
In the diagram, the input is across the terminals `A` and `C` and the output is across the terminals `B` and `D`, then the outputs is A. ZeroB. Same as inputC. Full wave rectifiedD. Half wave rectified |
| Answer» Correct Answer - C | |
| 156. |
Four equal resistors, each of resistance 10 ohm are connected as shown in the adjoining circuit diagram. Then the equivalent resistance between points A and B is-A. `5 Omega`B. `10 Omega`C. `20 Omega`D. `40 Omega` |
| Answer» Correct Answer - B | |
| 157. |
In a semiconducting material the mobilities of electrons and holes are `mu_(e)` and `mu_(h)` respectively. Which of the following is true?A. `mu_(e) gt mu_(h)`B. `mu_(e) lt mu_(h)`C. `mu_(e) =mu_(h)`D. `mu_(e) lt 0,mu_(h)gt0` |
| Answer» Correct Answer - A | |
| 158. |
In a semiconducting material the mobilities of electrons and holes are `mu_(e)` and `mu_(h)` respectively. Which of the following is true?A. `mu_(e) lt mu_(h)`B. `mu_(e) gt mu_(h)`C. `mu_(e)=mu_(h)`D. `mu_(e) lt 0 , mu_(h) gt 0` |
|
Answer» Correct Answer - B In a semiconductor, hole is equivalent to positively charged particle which moves in a direction opposite to that of an electron. Since electrons are lighter in mass compared to positively charged particles they move easily in the semiconducting material. Hence, the mobility of holes in `p`-type semiconductor is less than mobility of electrons in `n`-type semiconductor. i.e. `n_(e)gtn_(h)` |
|
| 159. |
which of the follwing is reverse biased?A. B. C. D. |
|
Answer» Correct Answer - C Only in option `(c)` , `P`-side is more negative as comapared to `N`-side. |
|
| 160. |
Which of the following is reverse biased diode?A. B. C. D. |
|
Answer» Correct Answer - B Because `P`-side is more negative as compared `N`-side. |
|
| 161. |
In short wave communication waves of which of the following frequeuncies will be reflected back by the ionospheric layer having electron density `10^(11)per m^(3)`?A. `18 MHz`B. `10 MHz`C. `12 MHz`D. `2 MHz` |
|
Answer» Correct Answer - D The perceived frequency of sky wave for reflection from an ionospheric layer is `v_(c)=9n^(1/2)` where `n` is the number density of electrons `//m^(3)`. Given, `n=10^(11)//m^(3)` `:. v_(c)=9xx(10^(11))^(1/2)=2.8 MHz~~2MHz` do not satisfy this condition. |
|
| 162. |
Assertion: At `0K` germanium is a superconductor. `Reason: At `0K` germanium offers zero resistance.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - D At `0K`, germanium offers infinite resistance, and it behaves as an insulator. |
|
| 163. |
A `0K` temperature, a `p`-type semiconductorA. has few holes and no few free electronsB. has equal numbers of holes and free electronsC. does not have any change carrierD. |
|
Answer» Correct Answer - A Conductor are those through which electric charge can easily flow, while insulators are those through which electric charge is difficult to flow. Solids whose electrical conductivity is intermediate between conductors and insulators are called semiconductors. At absolute zero a `p`-type semiconductor which has holes as charge carriers behaves like an ideal insulator and hence it has a few holes in valence band but no free electrons. |
|
| 164. |
A semiconducting device is connected in a series circuit with a battery and a resistance.A current is found to pass through the circuit .If the polarity of the battery is reversed, the current drops to almost zero.The device may beA. A `P`-type semiconductorB. An `N` -type semiconductorC. A `P-N` juctionD. An intrinsic semiconductor |
|
Answer» Correct Answer - C When polarity of the battery is reversed, the `P-N` junction becomes reverse biased so no current flows. |
|
| 165. |
If a full wave reactifier circuit is operating from `50 Hz` mains, the fundamental frequency in the ripple will beA. `50 Hz`B. `70.7 Hz`C. `100 Hz`D. `25 Hz` |
|
Answer» Correct Answer - C In full wave rectifier, the fundamental frequency in ripple is twice that of input frequency. |
|
| 166. |
In `PN`-junction diode the reverse saturation curren is `10^(-5)` amp at `27^(@)C`. The forward current for a voltage of `0.2` volt isA. `2037.6xx10^(-3) amp`B. `203.76xx10^(-3) amp`C. `20.37xx10^(-3) amp`D. `2.0376xx10^(-3) amp` |
|
Answer» Correct Answer - C The forward current `i=i_(s)(e^(eV//kT)-1)=10^(-5)[e^((1.6xx10^(-19)xx0.2)/(1.4xx10^(-23)xx300))-1]` `=10^(-5)[2038.66-1]-20.376xx10^(-3)A` |
|
| 167. |
The approximate ratio of resistance in the forward and reverse biase of the `PN-` junction diode isA. `10^(7):1`B. `10^(-2):1`C. `1:10^(-4)`D. `1:10^(4)` |
|
Answer» Correct Answer - D Resistance in forward biasing `R_(f r)~~10^(5)Omega` and resistance in reverse biasing `Rw~~10^(5) Omega implies (R_(f r))/(R_(Rw)) =1/(10^(4))` |
|
| 168. |
The reverse biasing in a `PN` junction diodeA. Decreases the potential barrierB. Increases the potential barrierC. Increases the number of minority charge carriersD. Increases the number of majority charge carriers |
|
Answer» Correct Answer - B In reverse biasing, width of depletion layer increases. |
|
| 169. |
In the case of forward biasing of `PN`-junction, which one of the following figures correctely depicts the direction of flow of carriers?A. B. C. D. |
|
Answer» Correct Answer - C In forwards biasing both positive and negative charge carriers move towards the junction. |
|
| 170. |
Assertion: When `PN`-junction is forward biased then motion of charge carriers at junction is due to diffusion. In reverse biasing. The cause of motion of charge is drifting. Reason: In the following circuit emitter is reverse biased and collector is forward biased. A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - B In forward biasing of `PN` junction current flows due to diffustion of majority charge carriers. While in reverse biasing current flows due to drifting of minority charge carriers. The circuit given in the reason is a `PNP` transistor having emitter is more negative w.r.t. base so it is reverse biased and collector is more positive w.r.t. base so it is forward biased. |
|
| 171. |
A sinusoidal voltage of peak value `200` volts is connected to a diode and resistor `R` in the circuit shown so that half wave rectification occurs. If the forward resistance of the diode is negligible compared to `R` the `rms` voltage (in volt) across `R` is approximately A. `200`B. `100`C. `200/(sqrt(3))`D. `280` |
|
Answer» Correct Answer - B `V_(rms)=(V_(0))/2=200/2=100V` |
|
| 172. |
Assertion:The dominant mechanism for motion of charge carriers in forward and reverse biased silicon `P-N` junction are drift in both forward and reverse biase. Reason: In reverse biasing, no current flow through the junction.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - D In `PN`-junction, the diffusion of majority carriers takes place when junction is forward biased and drifting of minority carriers takes place across the function, when reverse biased. The reverse biase opposes the majority carriers but makes the minority carriers to cross the `PN`-junction. thus the small current in `muA` flows during reverse biase. |
|
| 173. |
Assertion: `NAND` or `NOR` gates are called digital building blocks. Reason: The repeated use of `NAND` (or `NOR`) gates can produce all the basic or complicated gates.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - A These gates are called digital building blocks because using these gates only (either `NAND` or `NOR` ) we can compile all other gates also (like `OR,AND,NOT,XOR`). |
|
| 174. |
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon `P-N` junction areA. Drift in forwards bias, diffusion in reverse biaseB. Diffusion in forward bias, drift in reverse biasC. Diffusion in both forwards and reverse biaseD. Drift in both forward and reverse biase |
|
Answer» Correct Answer - B In forward biasing the diffusion current increases and drift current remains constant so not current is due to the diffusion. In reverse biasing diffusion becomes more difficult so net current (very small) is due to the drift. |
|
| 175. |
A silicon specimen is made into a `P`-type semiconductor by dopping, on an average, one helium atoms per `5xx10^(7)` silicon atoms. If the number density of atoms in the silicon specimen is `5xx10^(28) at om//m^(3)` then the number of acceptor atoms in silicon per cubic centimeter will beA. `2.5xx10^(30) at oms//cm^(3)`B. `1.0xx10^(13) at oms//cm^(3)`C. `1.0xx10^(15) at oms//cm^(3)`D. `2.5xx10^(36) at oms//cm^(3)` |
|
Answer» Correct Answer - C Number density of a atoms in silicon specimen `=5xx10^(28) at ms//m^(3)=5xx10^(22)//cm^(3)` Since one atom of indium is dopped in `5xx10^(7) Si` atom. So number of indium atoms dopped per `cm^(3)` of silicon. `n=(5xx10^(22))/(5xx10^(7))=1xx10^(15) at om//cm^(3)` |
|
| 176. |
A `p-n` photodiode is made of a material with a band gap of `2.0 eV`. The minimum frequency of the radiation that can be absorbed by the material is nearlyA. `10xx10^(14) Hz`B. `5xx10^(14) Hz`C. `1xx10^(14) Hz`D. `10xx10^(14) Hz` |
|
Answer» Correct Answer - B `p-n` photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage. Energy of radiation=band gap energy That is `hv=2.0 eV` or `v=(2.0xx1.6xx10^(-19))/(6.6xx10^(-34))~~5xx10^(14)Hz` |
|
| 177. |
No biase is applied to a `P-N` junction, then the currentA. Is zero because the number of charge carriers flowing on both sides is sameB. Is zero because the charge carriers do not moveC. Is non zeroD. None of these |
|
Answer» Correct Answer - B In unbiased condition of `PN`- junction, depletion region is generated which stops the movement of charge carriers. |
|
| 178. |
Zener diode is used asA. Half wave rectifierB. Full wave rectifierC. `ac` voltage stabilizerD. `dc` voltage stabilizer |
|
Answer» Correct Answer - C For a wide range of values of load resistance, the current in the Zener diode may change but the voltage across it remains unaffacted. Thus the output voltage across the Zener diode is a regulated voltage. |
|
| 179. |
Zener breakdown takes place ifA. Doped impurity is lowB. Doped impurity is highC. Less impurity in `N`-partD. Less impurity in `P`-type |
|
Answer» Correct Answer - B Zener breakdown can occure in heavily doped diodes. In lightly doped diodes the necessary voltage is higher, and avalanche multiplication is then the cheif process involved. |
|
| 180. |
Symbolic representation of photodiode isA. B. C. D. |
|
Answer» Correct Answer - C In photodiode, it is illuminated by light radiations, which in turn produces electric current. |
|
| 181. |
The part of a transistor which is heavily doped to produce a large number of majority carriers, isA. BaseB. EmitterC. CollectorD. None of these |
|
Answer» Correct Answer - C Emitter is heavily dopped. |
|
| 182. |
The majority charge carriers in `P`-type semiconductors areA. ElectronsB. ProtonC. HolesD. Neutrons |
|
Answer» Correct Answer - C In `P`- type semiconductor, holes are the majority charge carriers. |
|
| 183. |
When forward bias is applied to a `P-N` junction, then what happence to the potential barrier `V_(B)`, and the width of charge depleted region `x`?A. `V_(B)` increase, `x` decreasesB. `V_(B)` decreases, `x` increasesC. `V_(B)` increase, `x` increasesD. `V_(B)` decrease, `x` decreases |
|
Answer» Correct Answer - D In forward biasing both `V_(B)` and `x` decreases |
|
| 184. |
In a transistor circuit shown here the base current is `35 muA`. The value of the resistor `R_(b)` is A. `123.5 kOmega`B. `257 kOmega`C. `380.05 k Omega`D. None of these |
|
Answer» Correct Answer - B `V_(b)=i_(b) R_(b) implies R_(b)=9/(35xx10^(-6))=257 kOmega`. |
|
| 185. |
If no external voltage is applied across `P-N` junction, there would beA. No elctric field across the junctionB. An electric field pointing from `N`-type to `P`-type side across the junctionC. An electric field pointing `P` -type to `N` -type side across the juncitonD. A temperary electric field during formation of `P-N` junction that would subsequently disappear |
|
Answer» Correct Answer - B Across the `P-N` junction, a barrier potential is developed whose direction is form `N` region to `P` region. |
|
| 186. |
Different voltages are applied across a `P-N` junction and the currents are measured for each value. Which of the following graphs is obtained between voltage and current?A. B. C. D. |
|
Answer» Correct Answer - C `PN` junction has low resistance in one direction potential difference `+V`, so a large current flows (forwards biasing). It has a high resistance in the opposite potential difference direction `-V`, so a very small current flows (Revese biasing). |
|
| 187. |
Consider the following statement `A` and `B` and identify the correct choice of the given answersA. `A` and `B` are correctB. `A` and `B` are wrongC. `A` is correct but `B` are wrongD. `A` is wrong but `B` is correct |
| Answer» Correct Answer - C | |
| 188. |
The electrical resistance of depletion layer is large becauseA. It has no charge carriersB. It has a large number of charge carriersC. It contains electrons as a charge carriersD. It has holes as charge carriers |
|
Answer» Correct Answer - A Depletion layer consists of mainly stationary ions. |
|
| 189. |
In `P-N` junction, avalanche current flows in circuit when biassing isA. ForwardB. ReverseC. zeroD. Excess |
|
Answer» Correct Answer - B At a particular reverse voltage in `PN`- junction diode, current mainly flows due to the diffusion of majority charge carriers. |
|
| 190. |
Avalanche breakdown is due toA. Collision of minority charge carriersB. Increases in depletion layer thicknessC. Decreases in depletion layer thicknessD. None of these |
|
Answer» Correct Answer - A At high reverse voltage, the minority charge carriers, acquires very high velocities. These by collision break down the covalent bonds, generating more carriers. This mechnism is called Avalanche breakdown. |
|
| 191. |
Assertion: Zener diode works on aa principle of of breakdown voltage. Reason: Current increases suddenly after breakdown voltage.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - A When the reverse voltage across the Zener diode is equal to or more than the breakdown voltage, the reverse current increases sharply. |
|
| 192. |
Zener breakdown in a semi-conductor diode occurs whenA. Forward current exceeds certain valueB. Reverse biase exceeds certain valueC. Forwads biase exceeds certain valueD. Potential barrier is reduced to zero |
|
Answer» Correct Answer - B When reverse biase is increased, the electric field at the junction also increases. At some stage the electric fields breaks the covelent bond, thus the large number of charge carriers are generated. This is called Zener breakdown. |
|
| 193. |
The temperature coefficient of resistance of a semi conductor isA. Is always positiveB. Is always negativeC. Is zeroD. May be positive or negative or zero |
|
Answer» Correct Answer - B The temperature coeffiecient of resistance of a semiconductor is always negative. |
|
| 194. |
The exerted energy of the electrons at absolute zero is calledA. Fermi energyB. Emission energyC. Work functionD. Potential energy |
|
Answer» Correct Answer - A The highest energy level which an electrons can occupy in the valence band at `0 K`, is called Fermi energy level. |
|
| 195. |
Metallic solids are always opaque becauseA. Solids effect the incident lightB. Incident light is readially absorbed by the free electron in a metalC. Incident light is scattered by solids moleculesD. Energy band traps the incident light |
|
Answer» Correct Answer - B Metallic solids are opaque because incident light is absorbed by the free electrons in a metal. |
|
| 196. |
Assertion: Diode lasers are used as optical sources in optical communication. Reason: Diode layers consume less energy.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
|
Answer» Correct Answer - B A laser diode is a laser where the active medium is a semiconductor similar to that found in a light emitting diode. Laser diode is formed by doping a very thin layer on the surface of a crystal water. They find wide use in optical communication as easily coupled light sources. Diode lasers are effiecient source of high power radiation and with suitable beam shaping technology the output can be easily launched into the cladding of a suitably designed fibre. Hence, preferred in optical communication. They also consume less power. |
|
| 197. |
The relation between `alpha` and `beta` parameters of current gains for a transistors is given byA. `alpha=(beta)/(1-beta)`B. `alpha=(beta)/(1+beta)`C. `alpha=(1-beta)/(beta)`D. `alpha=(1+beta)/(beta)` |
|
Answer» Correct Answer - B `i_(e)=i_(b)+i_(c) implies (i_(e))/(i_(c))=(i_(b))/(i_(c))+1 implies 1/(alpha)=1/(beta)+1implies alpha=(beta)/((1+beta))`. |
|
| 198. |
The output of the given logica gate is A. `Y=bar(A)B+bar(B)`B. `Y=bar(A)B+bar(B)A`C. Y=1D. `Y=(bar(A)+bar(B))bar(B)` |
| Answer» Correct Answer - C | |
| 199. |
In the given circuit , the voltage across the base emitter junction is A. 2VB. 1VC. 3VD. 4V |
| Answer» Correct Answer - B | |
| 200. |
Two identical p-n junctions may be connected in series in which a battery in three ways , fig . The potential drops across the two p - n junctions are equal in A. First and second circuitsB. Second and third circuitsC. Third and first circuitsD. All of these |
| Answer» Correct Answer - B | |