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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.
| 101. |
Which of the following statements is not true ?A. the resistance of intrinsic semiconductors decreases with increase of temperatureB. doping pure `Si` with trivalent impurities gives p-type semiconductorsC. the majority charge carries in n-type semiconductors are holesD. a `p-n` junction can act as a semiconductor diode |
| Answer» Correct Answer - C | |
| 102. |
The majority carries in a p-type semiconductor are….A. ElectronsB. HolesC. BothD. Impurities |
| Answer» Correct Answer - B | |
| 103. |
The element that can be used as acceptor impurity to dope silicon isA. AntimonyB. ArsenicC. BoronD. phosphorous |
| Answer» Correct Answer - C | |
| 104. |
In n-type semiconductors the electron concentration is equal toA. density of donor atomsB. density of acceptor atomsC. density of both type of atomsD. neither density of acceptor atoms nor density of donor atoms |
| Answer» Correct Answer - A | |
| 105. |
On increasing temperature, the conductivity of pure semiconductorsA. decreasesB. increasesC. remains unchangedD. becomes zero |
| Answer» Correct Answer - B | |
| 106. |
The objective of adding impurities in the extrinsic semiconductor isA. to increase the conductivity of the semiconductorB. to increase the density of total current carriesC. to increase the density of either holes or electrons but not bothD. to eliminate the electron-hole pairs produced in intrinsic semiconductor. |
| Answer» Correct Answer - C | |
| 107. |
In extrinsic semiconductorsA. the conduction band and valence band overlapB. thegap between conduction band and valence band is near about `16 eV`C. the gap between conduction band and valence band is near about `1 eV`D. The gap between conduction band and valence band will be `100 eV` and more |
| Answer» Correct Answer - C | |
| 108. |
In the diagram `D` an ideal diode and an alternating voltage of peak value `10 V` is connected as input `V_(1)`. Which of the following diagram represents the correct wavelength of output voltage `V_(theta)` ? .B. C. D. |
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Answer» Correct Answer - D For `V_(1) lt 5V`, the diode is forward-biased, output will be fixed at `5V`. For `V_(1) lt 5V`, the diode is reverse biased. The output will follow `V`. |
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| 109. |
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. Statement `-1` is false, statement `-2` is trueB. statement `1-` is true statement `-2` is true statement `-2` is correct explanation of statement `-1`.C. statement `1-` is true statement `-2` is true statement `-2` is not correct explanation of statement `-1`D. statement `1-` is true statement `-2` is false. |
| Answer» Correct Answer - B | |
| 110. |
For a transistor `x=(1)/(alpha)&y=(1)/(beta)` where `alpha & beta` are current gains in common base and common emitter configuration. ThenA. `x+y =1`B. `x -y =1`C. `2x =1 - y`D. `x+y =0` |
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Answer» Correct Answer - B `alpha -(Delta I_(c))/(Delta I_(e)), beta =(Delta I_(c))/(Delta I_(b))`. |
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| 111. |
Current amplification factor of a common base configuration is `0.88`. Find the value of base current when the emitter current is `1 m A`. |
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Answer» In a common-base arrangement, the current amplification factor `alpha =((Delta I_(C))/(Delta I_(E)))_(V_(CB)) = (I_(C))/(I_(E))` Given `alpha = 0.88, I_(E) = 1 mA` `:.` Collector current `I_(C) = alpha I_(E) = 0.88 xx 1 = 0.88 mA` Now since `I_(E)=I_(B) + I_(C)` `:.` Base current `I_(B) =I_(E) -I_(C) = 1 - 0.88 = 0.12 mA`. |
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| 112. |
A cell of emf. `4.5 V` is connected to a junction diode whose barrier potential is `0.7 V`. If the external resistance in the circuit is `190 Omega`. The current in the circuit isA. `20 mA`B. `2 mA`C. `23 mA`D. `200 mA` |
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Answer» Correct Answer - A `V=4.5-0.7=3.8V, R=190 Omega,i=(V)/(R)`. |
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| 113. |
In the above problem the current amplification factor `beta` isA. `83`B. `100`C. `133`D. `203` |
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Answer» Correct Answer - C `B=(I_(C))/(I_(B))=(3.33 xx10^(-3))/(25xx10^(-6))=1.33 xx10^(2)=133`. |
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| 114. |
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than `1240 nm` is incident on it. The forbidden band energy for the semi conductor is (in eV).A. `0.5`B. `0.97`C. `0.7`D. `1.1` |
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Answer» Correct Answer - B `E=(hc)/(lamda)=(12000)/(12400 Å)(e_(v)), 2.n_(e) gt n_(h)`. |
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| 115. |
In a `p-n` junction diode the thickness of deplection layer is `2 xx 10^(-6) m` and barrier potential is `0.3 V`. The intensity of the electrical field at the junction is.A. `0.6 xx 10^(-6) Vm^(-1)` from `n` to `p` sideB. `0.6 xx 10^(-6) Vm^(-1)` from `p` to `n` sideC. `1.5 xx 10^(5) Vm^(1)` from `n` to `p` sideD. `1.5 xx 10^(5) Vm^(-1)` from `p` to `n` side. |
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Answer» Correct Answer - C `E=(v)/(d)=(0.3)/(2 xx 10^(-6)) =1.5 xx 10^(5) Vm^(-1)` Its direction from `n` to `p` side. |
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| 116. |
In the following, reverse biased diode is.A. B. C. D. |
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Answer» Correct Answer - D For reverse biasing of an ideal diode, the potential of n-side should be higher then potential of p-side. Only option (4) is satisfying the criterion for reverse biasting. |
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| 117. |
In a single state transistor amplifier, when the signal changes by `0.02 V` the base current by `10 mu A` and collector current by `1 mA`. If collector load `R_(C) = 2k Omega` and `R_(L) = 10 k Omega`, Calculate : (i) Current Gain (ii) Input Impedance, (iii) Effective `AC` load, (iv) Voltage gain and (v) Power gain.A. `50,2k Omega,1.66 k Omega,83,8300`B. `100,1 k Omega,1.66 Omega,83,8300`C. `100, 2k Omega,1.66 k Omega,83,830`D. `100,2 k Omega,1.66 k Omega,83.8300` |
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Answer» Correct Answer - D (i) Current Gain `beta -(delta i_(c))/(delta i_(b))` (ii) Input impedance `R_(i)=(delta V_(BE))/(delta i_(b))` (iii) Effective (ac) load `R_(AC)=(R_(C)R_(L))/(R_(C)+R_(L))` (iv) Voltage gain `A_(v)=beta xx (R_(AC))/(R_(i n))` (v) Power gain, `A_(p) = A_(v) xx beta`. |
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| 118. |
When base -emitter voltage of a transistor connected in the common-emitter mode is changed by `20 mV` the collector current is changed by `25 mA`. Find the transconductance.A. `1.25 Omega^(-1)`B. `2.50 Omega^(-1)`C. `0.5 Omega^(-1)`D. `5.5 Omega^(-1)` |
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Answer» Correct Answer - A `g_(m) = (Delta I_(c))/(Delta V_(BE)) = (25 xx 10^(-3))/(20 xx 10^(-3))=1.25 Omega^(-1)`. |
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| 119. |
In a `P-N-P` transistor, the collector current is `10 mA`. If `90%` of the holes reach the collector, then emitter current will be :A. `13 mA`B. `12 mA`C. `11 mA`D. `10 mA` |
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Answer» Correct Answer - C `I_(C)=(90)/(100)I_(E) rArr I_(E)=(10)/(9) I_(C)`. |
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| 120. |
A change of `8mA` in the emitter current brings a change if `7.9 mA` in the collector current. The change in base current required to have the same change in the collector isA. `0.01 mA`B. `1 A`C. `10 mA`D. `0.1 mA` |
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Answer» Correct Answer - D `Delta I_(B) = Delta I_(E) - Delta I_(C)`. |
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| 121. |
If `A = B = 1`, then in terms of Boolean algebra the value of `A.B + A` is not equal to.A. `B.A + B`B. `B + A`C. `B`D. `overline A.B` |
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Answer» Correct Answer - D If `A =B=1`, then `A.B +A =1.1 +1=1` then `overline(A).B= overline(1).1=0.1 =0`. |
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| 122. |
Binary number `1001001` is equivalent to the decimal number :A. `37`B. `73`C. `41`D. `32` |
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Answer» Correct Answer - B `{:(overset(6)5,overset(5)4,overset(4)0,overset(3)1,overset(2)0,overset(1)0,overset(0)1):}_((2)) = 1 xx 2^(6) +0 xx 2^(5) +0 xx 2^(4) +1 xx 2^(3)+0 xx 2^(2) +0 xx 2^(1) +1 xx 2^@ = 73`. |
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| 123. |
In the Binary number system `1 + 1=`A. `2`B. `1`C. `10`D. `100` |
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Answer» Correct Answer - C `1 + 1 = 10` in binary number system. |
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| 124. |
In intrinsic semiconductor at room temperature the no. of electrons and holes areA. equalB. zeroC. unequalD. infinite |
| Answer» Correct Answer - A | |
| 125. |
Carbon , silicon and germanium have four valence electrons each . At room temperature which one of the following statements is most appropriate ?A. The number of free conduction electrons is negligibly small in all the threeB. The number of free electrons for conduction is significant in all the three.C. The number of free electrons for conduction is significant only in `Si` and `Ge` but small in `C`.D. The number of free conductor electrons is significant in `C` but small in `Si` and `Ge`. |
| Answer» Correct Answer - C | |
| 126. |
A piece of copper and another of germanium are cooled from room temperature to `80 K`. The resistance ofA. each of these decreasesB. copper strip decreases and that of germanium decreasesC. copper strip decreases and that of germanium increasesD. each of these increases |
| Answer» Correct Answer - C | |
| 127. |
With increase in temperature in an intrinsic semiconductor the ration of conduction electrons and holes isA. `1 : 1`B. `1 : 2`C. `2 : 1`D. `1 : 3` |
| Answer» Correct Answer - A | |
| 128. |
A pure silicon crystal of length `l(0.1m)` and area `A(10^(-4) m^(2))` has the mobility of electron `(mu_(e))` and holes `(mu_(h))` as `0.135 m^(2)//Vs` and `0.48 m^(2)//Vs`, respectively, If the voltage applied across it is `2V` and the intrinsic charge concen-tration it is `2V` and the intrinsic charge concen-tration is `n_(i) = 1.5 xx 10^(6) m^(-3)`, then the total current flowing through the crystal is.A. `8.78 xx 10^(17) A`B. `6.25 xx 10^(-17) A`C. `7.89 xx 10^(-17) A`D. `2.456 xx 10^(-17) A` |
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Answer» Correct Answer - A `I=e(n_(e)mu_(e)+n_(h)mu_(h))(AV)/(l)` Putting the values, we get `I = 8.78 xx 10^(-17) A`. |
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| 129. |
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.A. `2.4 xx 10^(-11) J`B. `3.4xx 10^(-11) J`C. `5.4 xx 10^(-11) J`D. `6.4 xx 10^(-11) J` |
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Answer» Correct Answer - D `n_(e)=n_(h)=n_(i)` `:. sigma = n_(i)e(mu_(e)+mu_(h)) = 1.6 xx 1.6 xx 0.6 xx10^(-13)` Current produced in germanium plate. `I=JA = sigma E.A = sigma((V)/(d))A` Here generated in the plate, `H = V xx I xx t` `= 5xx1.28 xx10^(-13)xx100 =6.4 xx10^(-11)` joule. |
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| 130. |
Assume that the number of hole-electron pair in an intrinsic semiconductor is proportional to `e^(- Delta E//2KT)`. Here `Delta E` = energy gap and `k=8.62 xx 10^(-5) eV//"kelvin"` The energy gap for silicon is `1.1 eV`. The ratio of electron hole pairs at`300 K` and `400 K` is :A. `e^(-5.31)`B. `e^(-5)`C. `e`D. `e^(2)` |
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Answer» Correct Answer - A ` (N_(1))/(N_(2))=(e^(-Delta E//2KT_(1)))/(e^(-Delta E//2KT_(2)))` `= e^((Delta E)/(2K)((1)/(T_(2))-(1)/(T_(1)))) =(1.1)/(e^(2xx8.62 xx10^(-5))) ((1)/(400)-(1)/(300))`. |
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| 131. |
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))_(C)`B. `(E_(g))_(C) lt (E_(g))_(Ge)) gt (E_(g))_(Si)`C. `(E_(g))_(C) gt (E_(g))_(Si) gt (E_(g))_(Ge)`D. `(E_(g))_(C) = (E_(g))_(Si) = (E_(g))_(Ge)` |
| Answer» Correct Answer - C | |
| 132. |
A `p-n` diode is used in a half wave rectifier with a load resistance of `1000 Omega`. If the forward resistance `(r_(f))` of diode is `10 Omega`, calculate the efficiency of this half wave rectifier. |
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Answer» Load resistance `R_(L) = 1000 Omega` Forward resistance of the diode `= r_(f) = 10 Omega` Efficiency of half wave rectifier `[(0.406 R_(L))/(r_(f)+R_(L))] =(0.406 xx 1000)/(1010) = 0.4019` The percentage efficiency of the half wave rectifier `eta = 40.19 %`. |
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| 133. |
A full wave rectifier uses two diodes with a load resistance of `100 Omega`. Each diode is having negligible forward resistance. Find the efficiency of this wave rectifier. |
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Answer» Forward resistance of the diode `r_(f) = 0` , Load resistance, `R_(L) = 100 Omega , eta = ?` efficiency of full wave rectifier `= (0.812 xx 100)/(100) = 0.812` The percentage efficiency of the full wave rectifier `= 81.2 %`. |
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| 134. |
If a `p-n` junction diode, a square input signal of `10 V` is applied as shown. Then the out put signal across `R_(L)` will be . |
| Answer» The juction diode will conduct when it is forward biased. Therefore, the output voltage will be obtained during positive half cycle only. So option is `(3)`. | |
| 135. |
Find the effective resistance between `A` & `B` .A. `(2)/(3R)`B. `(3R)/(2)`C. `(2R)/(2)`D. `R` |
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Answer» Correct Answer - B current does not flow through the diode. |
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| 136. |
Find the effective resistance between `A` and `B` .A. `5//18_(Omega)`B. `9//5_(Omega)`C. `18//5_(Omega)`D. `5//9_(Omega)` |
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Answer» Correct Answer - C Use Wheat stones bridge principle. |
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| 137. |
`4` ideal diodes are connected as shown in the circuit the current through `50 Omega` is. .A. `0.1 A`B. `0.5 A`C. `0.6 A`D. `1 A` |
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Answer» Correct Answer - A `i= (V)/( R) =(5)/(50)`. |
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| 138. |
Two ideal junction diodes `D_(1),D_(2)` are connected as shown in the figure. A `3V` battery is connected between `A` and `B`. The current supplied by the battery if its positive terminal is connected to `A` is .A. `0.1 A`B. `0.3 A`C. `0.9 A`D. `90 A` |
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Answer» Correct Answer - B If positive terminal is connected to `A,D_(1)` in forward bias and `D_(2)` is in reverse bias. So `20 W` is ineffective. `:. i=(V)/(R)=(3)/(10)=0.3A`. |
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| 139. |
Figure shows the transfer characteristics of a base biased `CE` transistor. Which of the following statements are true ? (A) At `V_(1) = 0.14 V` transistors is in active state (B) At `V_(1) = 1V` it can be used as an amplifier (C) At `V_(1) = 0.5 V`, it can be used as a switch turned off (D) At `V_(1) = 2.5 V`, it can be used as switch turned onA. `A,B,C`B. `B,C,D`C. `A,C,D`D. `A,B,D` |
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Answer» Correct Answer - B From the given transfer characteristics of a base biased common emitter transistor, we note that (ii) When `V_(1) = 1V`( which is in between `0.6 V` to `2. V`), the transistor circuit is in active state and it used as an amplifier. (iii) When `V_(1) = 0.5 V` there is no collector current, The transistor is in cut-off state. The transistor circuit can be used as a switch turned off (iv) When `V_(1) = 2.5 V` the collector current becomes maximum and transistor is in saturation and can be used as switch turned on state. |
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| 140. |
The following configuration of gate is equivalent to A. `NAND` gateB. `XOR` gateC. `OR` gateD. `NOR` gate |
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Answer» Correct Answer - B `Y=(A+B).(overline(A.B))=(A+B).(overline(A)+overline(B))` `=A. overline(A) + A. overline(A) + overline(A).B+ B. overline(B)` `Y=(A.overline(B))+(overline(A).B)`. It is `XOR` gate. |
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| 141. |
In the figure shown the potential drop across the series resistor is A. 30 VB. 60 VC. 90 VD. 120 V |
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Answer» Correct Answer - A `V_("in")=V + V_(Z)`. |
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| 142. |
A 220 V AC supply is connected between points A and B . What will be the potential difference V across the capacitor ? A. 200 VB. 110 VC. 0 VD. `220 sqrt(2) V` |
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Answer» Correct Answer - D Capacitor once gets charged upto maximum potential. After that for any other lesser value of `p.d` across `A` and `B` diode is reverse biased and it does not allow charged to flow in opposite direction. From `V_(rms)=(V_(max))/(sqrt(2))` `rArr V_(max)=V_(rms) sqrt(2) =220 sqrt(2)` volts. |
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| 143. |
In the circuit shown(Fig.) if the diode forward voltage drop is `0.3 V`, the voltage difference between `A` and `B` is : .A. `1.3 V`B. `2.3 V`C. `0`D. `0.5 V` |
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Answer» Correct Answer - B `V_(A)-0.2 xx10^(-3) xx5 xx 10^(3) -0.3` `-0.2 xx10^(-3) xx5 xx 10^(3) -V_(B) =0` `rArr V_(A) - V_(B) = 2.3` volt. |
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| 144. |
In an n-p-n transistor `10^(10)` electrons enter the emitter in `10^(-6)`s. If 2% of the electrons are lost in the base, find the current transfer ratio and the current amplification factor.A. `0.98,49`B. `0.49,49`C. `0.98,98`D. `0.49,98` |
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Answer» Correct Answer - A `alpha =(I_(c))/(I_(e)),I_(c)=(98)/(100)I_(e),beta =(alpha)/(1-alpha)`. |
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| 145. |
Diffusion current in a p-n junction is greater than the drift current in magnitudeA. forward biasedB. reverse biasedC. un biasedD. both forward and reverse biased |
| Answer» Correct Answer - A | |
| 146. |
The output of the given circuit in Fig. .A. would be zero at all timesB. would be like a half-wave rectifier with positive cycles in outputC. would be like a half-wave rectifier with negative cycles in outputD. would be like that of a full-wave rectifier. |
| Answer» Correct Answer - C | |
| 147. |
In the figure shown below .A. In both Fig `a` and Fig `b` the diodes are forward biasedB. In both Fig, `a` and Fig `b` the diodes are reverse biasedC. In Fig a the diode is forward biased and in Fig `b`, the diode is reverse biasedD. In Fig a the diode is reverse biased and in Fig `b`, it is forward biased |
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Answer» Correct Answer - C For forward bias, `V_(p) gt V_(n)`, and for reverse bias `V_(p) gt V_(n)`. |
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| 148. |
Germanium diode.A. may be used as rectifier because it offers a relatively low resistance for forward bias and very high resistance for reverse bias.B. may be used as a rectifier because it offers a relatively high resistance for forward bias and very low resistance for reverse bias.C. cannot be used as a rectifierD. may be used as an amplifier |
| Answer» Correct Answer - A | |
| 149. |
The diagram correctly represents the direction of flow of charge carriers in the forward bias of `p-n` junction is.A. B. C. D. |
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Answer» Correct Answer - C In forward bias holes move from `P` to `n`, side electrons move from `n` to `P`. |
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| 150. |
The following data are for intrinsic germanium at `300 K. n_(i) = 2.4 xx 10^(19)//m^(3), mu_(e) = 0.39 m^(2)//Vs, mu_(h) = 0.19 m^(2)//Vs`. Calculate the coductivity of intrinsic germanium.A. `4.3 Sm^(-1)`B. `1.21 Sm^(-1)`C. `2.22 Sm^(-1)`D. `4.22 Sm^(-1)` |
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Answer» Correct Answer - C `sigma_(i)=n_(i)e(mu_(e)+mu_(h))`. |
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