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.
| 201. |
In the following circuits, Fig., which one of the two diodes is forward biased and which is reverse biased. |
|
Answer» (i) p-n junction is forward biased (ii) p-n junction is reverse biased |
|
| 202. |
In energy band diagram, the energy gap for carbon (diamond)is....... . |
| Answer» Correct Answer - 5.4 eV | |
| 203. |
Find the total current in the circuits shown in Fig. , , and . Each diode used is ideal |
|
Answer» In Fig. , both the diode `D_(1)` and `D_(2)` are forward biased. Their effective resistance in current is zero. Current in circuit is `I=3.0/10=0.3A` In Fig., diode `D_(1)` is forward bias and diode `D_(2)` is reverse biased. Resistance due to `D_(2)` will be infinity to the current flowing through it. So Current in the circuit, `I=(3.0)/(10+oo)=0` In Fig. , both the diode `D_(1)` and `D_(2)` are forward biased, Their effective resistance in the circuit is zero. So Current in the circuit is `I=3.0/10=0.3A` In Fig. , diode `D_(1)` is forward biased and diode `D_(2)` is reverse biased. No current will flow through `D_(2)` as its resistance become infinity being reverse biased. The resistance to current by `D_(1)` is zero. The current in the circuit is due to `D_(1)` only, which is given by `I=3.0/10=0.3A` |
|
| 204. |
In the Fig. find out the current passing through `R_(L)` and zener diode. |
|
Answer» Here, `V_(Z)=5V,` Voltage drop across R, `V_(R)= "input voltage"- V_(Z)=10-5=5V` Current though R, `I=V_(R)/R=(5V)/(80Omega)` `=6.25xx10^(-2)A.` Voltage drop across `R_(L) =V_(z)=5v`. Current through `R_(L)` is, `I_(L)=5/100=5xx10^(-2)A` Since `I=I_(L)+I_(z)`. so `I_(z)=I-I_(L).` Current through zener diode is, `I_(z)=I-I_(L)=6.25xx10^(-2)-5xx10^(-2)` `=1.25xx10^(-2)A` |
|
| 205. |
The maximum possible energy possessed by free electrons of a material at absolute zero temperature is equal to........... . |
| Answer» Correct Answer - fermi energy | |
| 206. |
In the circuit shown in Fig., find the value of `R_`. |
|
Answer» Let the current through the various arms be as shown in Fig. Follow from the solution of question 39, `I_B//I_C=.01//1=1//100` `I_(E)=I_(B)+I_(C)=I_C/100=I_C~~I_C` In closed circuit `AB_(1)CEDA` `I_(C)R_(C)+V_(CE)+I_(E)R_(E)=V_(C C) or I_(C)(R_C+R_(E))=V_(C C)-V_(CE)=12-3=9 ...(i)` In closed circuit AB_(1)GHJDA `(I_B+I)R_B+IR+IR=V_(C C)=12 or I_(B)I_(R)+I(R_(B)+R)=12` or `I_C/betaR_(B)+I(R_(B)+R)=12 ...(ii)` Let V be the potential difference across H and J, then `V=IR=V_(BE)+I_(E)R_(E)=V_(BE)+I_(C)R_(E)=0.5+I_(C)R_(E) or I=(0.5+I_(C)R_(E)//R` Putting this value of I in (ii), we get `I_C/betaR_(B)+1/R(1/2+I_(C)R_(E))(R_(B)+R)=12 or (I_C R_B)/beta+1/R[1/2R_B+1/2R+I_CR_(E)R_B+I_CR_(E)R]=12` or `I_C R_R + beta [1/2R_(B)+1/2R+I(C)_R_(E)R_(B)+I_(C)R_(E)R]=12beta R` or `I_C[R_(B)R+beta R_(E)(R_(B)+R)]+1/2beta(R_(B)+R)=12betaR` or `I_(C)[(100xx10^(3))(20xx10^(3))+100xx(10^(3))(100xx10^(3)+20xx10^(3))]+1/2xx100[100xx10^(3)+20xx10^(3)]=12xx100xx20xx10^(3)` or `I_(C)[2xx10^(9)+12xx10^(9)]+6xx10^(6)=24xx10^(6) or I_(C)=((24-6)10^(6))/((12+2)xx10^(9))=9/7xx10^(-3)A` Putting value of I_ in (i), we get `9/7xx10^(-3)(R_(C)+10^(3))=9 or R_(C)+10^(3)=7xx10^(3) or R_(C)=7xx10^(3)-10^(3)=6xx10^(3)Omega=6kOmega` |
|
| 207. |
Establish the Boolean identity `A.B+A.B.C+bar.B+A.bar.C=B+A.C` |
|
Answer» `A.B+A.B.C+barA.B+A.barB.C` `=A.B+barA.B+A.B.C+A.barB.C` `(A+barA).B+A.(B+barB).C=1.B+A.1.C` `=B+A.C` |
|
| 208. |
The semiconductors available in narural form are called.............. . |
| Answer» Correct Answer - element semiconductors | |
| 209. |
Explain that energy of a hole farther from the top of a valence band is high. |
| Answer» To understand it, consider an electron is removed from the filled valence band to the bottom of the empty conduction band. This creats a hole in the valence band. Since more energy is required to remove another electron which is farther from the top of the valence band, therefore, a hole in the valence state farther from the top of the valence band has higher energy, just as a conduction electron farther from the bottom of the conduction band has higher energy. | |
| 210. |
What is an ideal junction diode? |
| Answer» An ideal junction diode is one which acts as a perfect conductor when forward biased and perfect insulator when reverse biased. | |
| 211. |
What will be the input of `A` and `B` for the Boolean expression `bar((A+B)).bar((A.B))=1`?A. 0, 1B. 1, 0C. 0, 0D. 1, 1 |
| Answer» Correct Answer - C | |
| 212. |
Find the equivalent resistance of the network shown in Fig.1 between the point p and Q, when the p-n junction diode used ideal one. |
|
Answer» (i) When potential of P is higher than that of Q, the junction diode in circuit is forward biased. Its resistance becomes zero. Now we shall have two resistance in parallel each of `10 Omega` between points P and Q. Their effective resistance between P and Q is `R=(10xx10)/(10+10)=5Omega` (ii) When potential of P is lower than that of Q, the p-n junction in circuit is reverse biased. Its resistance becomes infinite. No current flows in the arm of p-n junction diode. The current will flow through 10Omega resistance connected in parallel arm to p-n junction diode. Therefor the resistance of circuit is 10Omega |
|
| 213. |
The output of NOT gate is 1(in level) if input is.......... |
| Answer» Correct Answer - zero (in level) | |
| 214. |
The gate for which output is high, if atleast one input low isA. NANDB. NORC. ANDD. OR |
| Answer» Correct Answer - A | |
| 215. |
To get an output `y=1` from the circuit shown below, the input must be A. `{:(A,B,C),(0,1,0):}`B. `{:(A,B,C),(0,0,1):}`C. `{:(A,B,C),(1,0,1):}`D. `{:(A,B,C),(1,0,0):}` |
|
Answer» Correct Answer - C In the figure A,B are inputs of OP gate whose output is `Y_(1)`. `|:(A,B,Y_(1)),(0,1,1),(0,0,0),(1,0,1),(1,0,1):}:|` Again `Y_(1)` and `C` are inputs of And gate `|{:(Y_(1),C,Y),(1,0,0),(0,1,0),(1,1,1),(1,0,0):}:|` To get an output `Y =1`, the inputs must be `(1,0,1)` |
|
| 216. |
The given electrical network is equivalent to: A. AND gateB. OR gateC. NOR gateD. NOT gate |
|
Answer» Correct Answer - C Output of 1st NOR gate is `y_(1)=bar(A+B)` Output of second NOR gate is `y_(2)=bar(y_(1)+y_(1))=bar(y_(1))=bar(bar(A+B))=A+B` Output of third NOT gate is `y_(3)=bar(y_(2))=bar(A+B)` Thus NOR gate is obtained. |
|
| 217. |
In the following combinations of logic gates, the outputs A, B and C are respectively A. 0, 1, 1B. 0, 1, 0C. 1, 1, 0D. 1, 0, 1 |
|
Answer» Correct Answer - C In logic gates , Output of NAND gate 0 and output of NOT is 1. The output of OR gate is 1. In logic gates , Output of NOT gate ia 1 and of other NOT gate is 0. The output of NAND gate is 1. In logic gate is 0 and output of AND gate is 0. |
|
| 218. |
(i) Identify the logic gates marked P and Q in the given logic circuit Fig. (ii) Write down the output at X for the inputs `A=0, B=0` and `A=1, B=1`. |
|
Answer» Logic gate P is NOT gate and logic gate Q is OR gate. The Boolean expression for this gate is `X=barA+B` The truth table for the given values of A and B of the above circuit is shown below. `|{:(A,barA,B,y=barA+B),(0,1,0,1),(1,0,1,1):}|` |
|
| 219. |
In n-type semiconductor when when all donor states are filled, then the net charge density in the donor states becomesA. 1B. `gt 1`C. lt 1, but not zeroD. zero |
|
Answer» Correct Answer - B When all donor states in n-type semiconductor are filled, the number of electrons in donor states will increase. Due to it, the charge density in donor states will become more than one. |
|
| 220. |
Mobility of electron and holes in a sample of intristic germanium at room temperature are `0.36m^(2)V^(-1)s^(-1) and 0.17 m^(2)V^(-1)s^(-1)`. The electron and hole densities are each equal to `2.5xx10^(19) m^(-3)`. The electrical conductivity of germanium isA. `0.47Sm^(-1)`B. `1.09 Sm^(-1)`C. `2.12Sm^(-1)`D. `4.24 Sm^(-1)` |
|
Answer» Correct Answer - C `sigma=1/rho=e(mu_(e)n_(e)+mu_(h)n_(h))` `=1.6xx10^(-19)[0.36+0.17]xx(2.5xx10^(19))` `=2.12Sm^(-1)` |
|
| 221. |
Consider an n-p-n transistor with its base - emitter junction forward biased and collector base junction reverse biased . Which of the following statements are true?A. Electrons crossover from emitter to collectorB. holes move from base to collectorC. Electrons move from emitter to baseD. Electrons from emitter move out of base without going to the collector |
|
Answer» Correct Answer - A::C In an p-n transistor when base emitter junction is forward biased and collector-base junction is reverse biased, the majority charge carriers electrons move from emitter to base. In base few electrons get neutralised by electron-hole combination and the remaining electrons cross over the collector. |
|
| 222. |
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 the Reason is the correct explanation of Assertion.B. if both the Assertion and Reason are true but the reason in not a correct explanation of the Assertion.C. if the Assertion is true but the Reason is false.D. if both Assertion and Reason are false. |
|
Answer» Correct Answer - A Here, both Assertion and Reason are correct and Reason is correct explanation of Assertion. |
|
| 223. |
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon `P-N` junction areA. drift in forward bias and diffusion in reverse biasB. diffusion in forward bias and drift in reverse biasC. diffusion in both forward and the reverse biasD. drift in both forward and reverse bias |
|
Answer» Correct Answer - B Under the effect of electric field applied, the drift of charge carriers is from lower concentration to higher concentration and diffusion of charge carriers is from higher concentration to lower concentration. In forward biasing diffusion is more than drift and in reverse biasing, drift is more than diffusion. |
|
| 224. |
Statement-1 : The direction of diffusion current in a junction diode is from n-region to p-region. Statement-2 : The majority current carrier diffuse from a region of higher concentration to a region of lower concentration.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
|
Answer» Correct Answer - D The direction of diffusion current is that where positively charged particles move. Hence statement-1 is wrong. Statement-2 is true. |
|
| 225. |
Statement-1 : The temperature coefficient of resistance is positive for metals and negative for p-type semiconductor. Statement-2 : The effective charge carriers in metals are negatively charged whereas in p-type semiconductor, they are positively chargedA. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
|
Answer» Correct Answer - B Both statement-1 and satement-2 are correct but the statement-2 is not correct explanation of statement-1. |
|
| 226. |
In p-n junction, there is a .................of majority carriers across the junction in forward biading and..............of charge carriers in reverse biasing. |
| Answer» Correct Answer - diffusion; drifting | |
| 227. |
Statement-1 : When the temperature of a semiconductor is increased, then its resistance decreases. Statement-2 : The energy gap between conduction band and valence band is very small for semiconductor.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
|
Answer» Correct Answer - A When temprature of semiconductor is raised, more covalent bonds are broken. Due to it, large number of electrons and holes are produced in semicoductor. Some electrons move from valence band to conduction band crossing over a very small energy gap and become free electrons. As a result of which the conductivity of semiconductor increases and hence resistance of semiconductor decreases. |
|
| 228. |
Statement-1 : At a fixed temperature, silicon will have a minimum conductivity when it has a smaller acceptor doping. Statement-2 : The conductivity of an intrinsic semiconductor is slightly higher than that of a lightly doped p-type semiconductor.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
|
Answer» Correct Answer - C Here Assertion is correct but Reason is wrong, as conductivity of an intrinsic semiconductor is less than of a lightly doped p-type semiconductor. |
|
| 229. |
Statement-1 : A p-type semiconductor has ney positive charge on it. Statement-2 : Holes are majority carriers in it.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
|
Answer» Correct Answer - D Here, statement-1 is wrong, as p-type semiconductor has no net posotive or negative charge on it. Statement-2 is true. |
|
| 230. |
An electric field us applied to a semiconductor.Let the number of charge carriers be n and the average drift speed be v.If the temperature is increased,A. both n and v will decreaseB. both n and v will increaseC. n will increase but v will decreaseD. v will increase but n will decrease |
|
Answer» Correct Answer - C When temperature of a semiconductor is increased, more covalent bonds get broken. As a result of it, the number of charge carriers will increase. Due to rise in temperature, the atoms/ions of the semiconductor, will start vibrating with more amplitude and frequency. Due to it, the electrons drifting towards the positive end of semiconductor will suffer more rapid collisions, consequently drift velocity will decrease. |
|
| 231. |
Find the average value of d.c. voltage that can be obtained from the half wave rectifier of Fig. assuming that the diode is ideal one. |
|
Answer» Max. value of the input voltage,`V_(0)=sqrt(2)V_(rms)=sqrt(2)xx200=282.8V` to primary of transformer Max. value of output voltage across secondary of transformer `V_(s)=n_(s)/n_(p)xxV_(p)=1/10xx282.8=28.28V` d.c. voltage across the load is `V_(dc)=V_(s)/pi=28.28/3*142=9.0V` |
|
| 232. |
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 be |
|
Answer» No density of silicon `=5xx10^(28) "atoms"//m^(3)` `=5xx10^(22) "atoms"//cm^(3)`. No. of acceptor `"atoms"//cm^(3) =5xx10^(22)//(5xx10^(7))` `=10^(15)//cm^(3)` |
|
| 233. |
A semiconductor has equal electron and hole concertration `6xx10^(8) m^(-3)`. On doping with a certain impurity electron concertration increase to `8xx10^(12) m^(-3)`. Identify the type of semiconductor after doping. |
| Answer» The resulting semiconductor is n-type as electrons are majority charge carriers in it. | |
| 234. |
In half wave rectification, if the input frequency is `50 Hz`, what is the output frequency? What is the output frequency of a full wave rectifier for the same frequency? |
| Answer» Since the output voltage obtained in half wave rectification is once in one cycle of input voltage, hence the output ripple frequency after half wave rectification `= 50Hz`. In a full wave rectification, we get the output voltage twice in the same direction for one cycle of input voltage. Hence the output ripple frequency after full wave recitification `= 100Hz`. | |
| 235. |
The applied input a.c. power to a half wave retifier I 100 watt. The d.c. output power obtained is 40 watt. (i) What is the rectification efficiency and (ii) What is the power efficiency? |
|
Answer» Rectification efficiency `=(d.c. "output power")/(a.c. "input power")xx100` `40/100xx100=40%` Power efficiency `=(d.c. "output power")/(a.c. "input power for half cycle")` `=40/((100//2))xx100=80%` |
|
| 236. |
Find the current produced at room temperature in a pure germanium plate of area `2 xx 10^(-4) m^(2)` and of thickness `1.2 xx 10^(-3) m` when a potential of `5 V` is applied across the faces. Concentration of carries in germanium at room temperature is `1.6 xx 10^(6)` per cubic metre. The mobilities of electrons and holes are `0.4 m^(2) V^(-1) s^(-1)` and `0.2 m^(2) V^(-1) s^(-1)` respectively. The heat energy generated in the plate in `100` second is. |
|
Answer» `sigma=n_(i)e(mu_(e)+mu_(h))` `=1.6xx10^(6)xx1.6xx10^(-19)(0.4+0.2)` `=1.53xx10^(-13)` Xcurrent produced in germanium plate `I=JA=sigmaEA=sigma(V/d)A` `=1.53xx10^(-13)xx(5)/((1.2xx10^(-3)))xx2xx10^(-4)` `=1.28xx10^(-13)A` Heat generated in plate, `H=VIt` `=5xx1.28xx10^(-13)xx100` `6.4xx10^(-11)J` |
|
| 237. |
Suppose a pure Si-crystal has `5xx10^(28) "atoms" m^(-3)`. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Give that `n_(i)=1.5xx10^(16)m^(-3)`. |
|
Answer» `1"ppm"=1` part per million `=1/10^(6)`. so no. of pentavalent As atoms doped in given Si crystal, `=(5xx10^(28))/10^(6)=5xx10^(22)m^(-3)`. As one pentavalent atoms donates one free electron to the crystal structure, so no. of free electrons in the given si crystal, `n_(e)=5xx10^(22)m^(-3)` Number of holes, `n_(h)=n_(i)^(2)/n_(e)=(1.5xx10^(16))^(2)/(5xx10^(22))` `4.5xx10^(9)m^(-3)` |
|
| 238. |
Carbon , silicon and germanium have four valence elcectrons each . These are characterised by valence and conduction bands separated by energy band - gap respectively equal to ` (E_g)_(c) (E_g)_(si) ` and ` (E_g)_(Ge) `. Which of the following statements ture ?A. `(E_(g))_(Si) lt (E_(g))_(Ge) lt(E_(g))_`B. `(E_(g))_ lt (E_(g))_(Ge) lt (E_(g))_(Si)`C. `(E_(g))_ gt (E_(g))_(Si) gt (E_(g))_(Ge)`D. `(E_(g))_ =(E_(g))_(Si) =(E_(g))_(Ge)` |
| Answer» The energy band gap is maximum for carbon, less for silicon and least for germanium out of the given three elements. Hence | |
| 239. |
Assertion : The energy gap between the valence band and conduction band is greater in silicon than in germanium. Reason : Thermal energy produces fewer minority carriers in silicon than in germanium.A. if both the Assertion and Reason are true and the Reason is the correct explanation of Assertion.B. if both the Assertion and Reason are true but the reason in not a correct explanation of the Assertion.C. if the Assertion is true but the Reason is false.D. if both Assertion and Reason are false. |
|
Answer» Correct Answer - B Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. |
|
| 240. |
On doping germanium with donor atoms of density `10^(17) cm^(-3)`, find its conductivity in mho/cm, if `mu=3800 cm^(2)//V-s`. |
|
Answer» `sigma=n e mu_(e)=10^(17)xx1.6xx10^(-19)xx3800` `=60.8 mho//cm`. |
|
| 241. |
The mean free path of conduction electrons in copper is about `4.0xx10^(-8) m`. The electric field which can give on the average `2.4 eV` energy to conduction electron in copper block is `axx10^(7) V m^(-1)`. What is the integer value of a? |
|
Answer» Here, mean free path, `x=400xx10^(-8)m`. Energy acquired by electron before it collides with a copper ioN/Atom, `W=2.4 eV =2.4xx1.6xx10^(-19)J`. If E is the electric field applied, then force on electron due to electric field, `F=eE`. Workdone by electric field on electron, before it collides with a copper ioN/Atom is W= force xx mean free path`=eEx` |
|
| 242. |
A transistor has a current gain of 50. If the collector resistance `5kOmega`. Calculate the output voltage if input voltage is `0.01V`. |
|
Answer» Here, `beta=50, R_(0)=5xx10^(3)Omega`, `R_(i)=10^(3)Omega, V_(i)=0.01V` `A_(V)=V_(0)/V_(i)=beta.R_(0)/R_(i)` or `V_(0)=beta.(R_(0)xxV_(i))/R_(i)=50xx(5xx10^(3)xx0.01)/10^(3)` `2.5V` |
|
| 243. |
The number desnity of donor atoms which have to be added to an intrinsic germanium semiconductor to produce an n-type semiconductor of conductivity `5 ohm^(-1) cm^(-1)` is `axx10^(15)cm^(-3)`. Given that the mobility of electron in n-type Ge is `3900 cm^(2)//Vs`, Neglect the contribution of holes to conductivity. What is the integer value of a? |
|
Answer» Here, `sigma=5Omega^(-1)cm^(-1), mu_(e)=3900 cm^(2)//Vs, n_(e)=?` We know that, `sigma=1/rho=e n_(e) mu_(e)` or `n_(e)=sigma/(emu_(e))=5/((1.6xx10^(-19))xx3900)~=8xx10^(15) cm^(-3)`. Since, one donor atom provides one free electron to the germanium semiconductor, therefore, number bensity of donor atoms `=8xx10^(15) cm^(-3)` |
|
| 244. |
The potential difference across the collector of a transistor, used in common emitter mode is `1.5V`, with the collector resistance of `3 kOmega`. Find (i) the emitter current and (ii) the base current, if d.c. gain of the transistor is 50. |
|
Answer» Here, `V_(CE)=1.5V, R_=3xx10^(3)Omega,` `beta=50.` `I_C=V_(CE)/R_C=1.5/(3xx10^(3))=0.5xx10^(-3)A` `I_B=I_C/beta=(0.5xx10^(-3))/50=0.01xx10^(-3)A` `=0.01mA` `I_(E)=I_B+I_C=0.01xx10^(-3)+0.5xx10^(-3)` `=0.51xx10^(-3)A=0.51mA` |
|
| 245. |
Is it difficult to make an intrinsic semiconductor? Name two factors on which the electrical conductivity of intrinsic semiconductor at a given temperature depends. How does the conductivity of a semiconductor change with the rise in its temperature? |
|
Answer» An instrinsic semiconductor is one which is extremely pure with a purity level of 99.9999%. It is very difficult to achieve this much purity in a semiconductor. Electrical conductivity of an intrinsic semiconductor at a given temperature depends on (i) the number densities of intrinsic charge carriers present in it and (ii) the width of the forbidden energy gap between conduction band and valence band. The conductivity of a semiconductor increases exponentially with temperature. |
|
| 246. |
For a transistor connected in common emitter mode, the voltage drop across the collector is 2V and beta is 50. If `R_(c)` is `2 kOmega`, the base current is `axx10^(-5) A`. What is the value of a ? |
|
Answer» Correct Answer - B Voltage drop across collector `=I_(c)R_(c)` `:. 2=I_(c)xx(2xx10^(3)) or I_(c)=10^(-3) A` `I_(b)=I_c/beta=10^(-3)/50=2xx10^(-5) A` |
|
| 247. |
For a Transistor, the ratio of change in base-emitter voltage to the resulting change in the base current at constant emitter voltage is called........... |
| Answer» Correct Answer - Input dynamic resistance of a transistors | |
| 248. |
In a common emitter a transistor, the ratio of small change in the collector current `(DeltaI_(c))` to the corresponding small change in the collector emitter voltage `(DeltaV_(CE))` at constant base current `(I_(b))` is called........... |
| Answer» Correct Answer - Output admittance | |
| 249. |
The semiconduting devices are more...............than the vacuum tubes. They can withstand.............. |
| Answer» Correct Answer - rugged; rougth handing. | |
| 250. |
In a common emitter transistor amplifier, the input resistance of a transistor is `1000Omega`. On changing its base current by 10muA, the collector current increases by `2mA`. If a load resistance of `5kOmega` is used in the circuit, calculate: (i) the currnet gain (ii)the voltage gain of the amplifier. |
|
Answer» Here, `R_(i)=1000Omega, DeltaI_=10muA=10^(-5)A`, `DeltaI_C=2mA=2xx10^(-3)A` `R_(0)=5kOmega=5xx10^(3)Omega`. (i) Current gain, `beta=(DeltaI_C)/(DeltaI_B)=(2xx10^(-3))/(10^(-5))=200` (ii) Voltage gain, `A_(V)=betaR_(0)/R_(i)=(200xx5xx10^(3))/1000=1000` |
|