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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.
| 1. |
Assuming the ideal diode, draw the output waveform for the circuit given in Fig.Explain the waveform. |
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Answer» When the input voltage is equal to or less then 5V, diode will be reverse biased. It will offer high resistance in comparison to resistance (R) in series. Now diode appears in open circuit. The input wave from is then passed on to the output terminals. The result with sine wave input is to clip off all positive-going portion above `+5V` volt. If input voltage is more than `+5V`, diode will be conducting as if forward biased offering low resistance incomparison to R. But there will be no voltage in output beyond 5volt as the voltage beyond `+5V` will appear across R. When input voltage is negative, there will be opposition to 5 V battery in p-n junction circuit. Due to it, reverse bias voltage of p-n junction decreases and a voltage appears across output. When input voltage becomes more than -5V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals. The output wave form will be as shown in Fig. |
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| 2. |
Draw the output waveform across the resistor (Fig.) |
| Answer» A p-n junction offers low resistance when forward biased and high resistance, when reverse biased. Therefore, there will be output voltage for the positive half cycle of a.c. voltage used as input. Which is shown in Fig. | |
| 3. |
Calculate the value of output voltage`V_(0)` and I if the Si diode and the Ge diode conduct at 0.7 V and 0.3V respectively, in the circuit given in Fig. If now Ge diode connections are reversed, what will be the new values of `V_(0)` and I. |
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Answer» Refer to given circuit, current, `I=(12-0.3)/(5kOmega)=(11.7V)/(5xx10^(3)Omega)=2.34mA` Output voltage, `V_(0)=RI=(5xx10^(3))xx(2.34xx10^(-3)) =11.7V` When the connections of Ge diode are reverse, then current will be through silicon. In this case, `I=((12-0.7)V)/(5kOmega)=(11.3V)/(5xx10^(3)Omega)=2.26mA` `V_(0)=IR=(2.26xx10^(-3))xx(5xx10^(3))=11.3V` |
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| 4. |
Why does the width of depletion layer of a p-n junction increase in reverse biasing? |
| Answer» When p-n junction is reversed biased, the positive terminal of the external battery is connected to n-side of p-n junction and its negative terminal to p-side of p-n junction. The positive terminal of the external battery attracts the majority carrier electrons from the n-region and its negative terminal attract the majority carrier holes from p-region. Due to it, the majority charge carriers move away from the junction. This increases the width of the depletion layer. | |
| 5. |
How does the width of the depletion layer of a p-n junction diode change with decrease in reverse bias? |
| Answer» The width of depletion layer of a p-n junction decreases with the decrease in reverse bias. | |
| 6. |
The V-I characteristics of silicon diode is shown in Fig. Calculate the diode resistance in `` forward bias at `V=+0.9V ` reverse bias `V=-3.0V`. |
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Answer» In forward characteristic for voltage `V=+0.9V` `DeltaV=DE=1.0-0.80=0.2V, DeltaI=CE=10-6=4mA` `:.` Forward resistance, `R_(f)=(DeltaV)/(DeltaI)=(0.2V)/(4mA)=50Omega` in reverse characteristics for voltage `V=-3V ,DeltaV=-2-(-4)=2V, DeltaI=3-2=1muA` `:.` Reverse resistance, `R_(r)=(DeltaV)/(DeltaI)=(2V)/(1muA)=2xx10^(6)Omega` |
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| 7. |
What do the acronyms LASER and LED stands for? Name the factor which determines (i) frequency and (ii) intensity of light emitted by LED. |
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Answer» LASER stands for light amplification by stimulated emission of Radiation. LED stands for light emitting diode. (i) The frequency of light emitted by an LED is related to the band gap of the semiconductor used in LED, i.e., a type of material used in making the LED. (ii) The intensity of light emitted by LED depends upon the doping level of the semiconductor used. |
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| 8. |
In a transistor base is made thin and doped with little impurity atoms. Why? |
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Answer» The base region in a transistor is made very thin so that there is a better conduction of majority carriers from emitter to collector through base. Due to it, the base current is quite weak, the collector current is nealy equal to the emitter current. As result of it, the transistor can give good power gain and voltage gain. The base region in a transistor is doped lightly so that the number density of majority carriers (electrons in p-n and holes in n-p-n transistor) is low. When emitter is forward biased, the majority carriers move from emitter to collector through base. Since base is thin and lightly doped, only a small (about5%) electron-hole combination will take place giving weak base current and remaining majority carriers will be colleced by collector giving collector current nearly equal to emitter current. |
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| 9. |
For the circuit shown in Fig., find the current flowing through the `1Omega` resistor. Assume that the two diode are ideal diode. |
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Answer» Here, diode `D_(1)` is forward biased, it offers infinite resistance. The equivalent circuit of the given circuit is shown in Fig. Current through `1 Omega` resistor, `I=6/(2+1)=2A` |
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| 10. |
Why is a photodide operated in reverse bias mode? Fig. shown reverse bias current, under different illuminating intensities `I_(1)I_(2),I_(3)` and `I_(4)` for a given photodiode. Average the intensities `I_(1), I_(2),I_(3)` and `I_(4)` in decreasing order of wavelength. |
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Answer» The photodiodes are use4d in reverse bias condition because the change in reverse current through the photodiode due to change in light flux or light intensity can be measured easily, as the reverse saturation current is directly proportional to the light flux or light intensity. But it is not so when photodiode is forward biased. As the reverse saturation current through a photodiode increases with the increase in light intensity or light flux, so `I_(4)gtI_(3)gtI_(2)gtI_(1)`. |
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| 11. |
In V-I characteristics of a p-n junction diode, why is the current under reverse bias almost independence of the applied potential upto a critical voltage? |
| Answer» When a p-n junction is reverse biased, a very small current (of the order of few `muA`) flows due to drifting of minority charge carriers whose number density remain constant upto the critical voltage. As result of it, the current under a reverse bias is almost independent of the applied potential upto critical voltage. | |
| 12. |
The current in the forward bias is known to be more (in `mA`) than the current in the reverse bias (in ` mu A ` ). What is the reason then to operate the photodiodes in reverse bias? |
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Answer» In case of n-type semiconductor, let n be the majority carrier (i.e., electrons) density and p be the minority carrier (holes) density. Where `n gt p`. On illumination of semiconductor, there will be production of equal number of electrons and holes. Let `Deltan, Deltap` be the increase in majority carrier density and minority carrier density due to illumination of semiconductor, where `Deltan=Deltap`. Hence, fractional change in majority carrier `(=Deltan//n)`, fractional change in minority carrier `(=Deltap//p)`. Since `n gt gt p`, so `(Deltan)/n lt (Deltap)/p` it means, due to photo-effects the fractional change due to minority carriers dominates. As a result of it, the fractional change in the reverse bias current is more easity measurable than the fractional change in the forward bias current. It is due to this reason, photodiodes are preferably used in the reverse bias condition for measuring light intensity. |
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| 13. |
The energy of a photon of sodium light wavelength `5890 Å` equals the energy gap of a semiconducting material. Find the minimum energy E required to create a hole-electron combination. the value of `E//kT` at a temperature of `27^(@)C`, where `k=8.62xx10^(-5) eV//K, h=6.63xx10^(-34)Js`. |
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Answer» Energy required to create a hole-electron combination is equal to energy E of photon which can push an electron from valence band to conduction band. Thus `E=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(5890xx10^(-10)xx1.6xx10^(-19))` `=2.1eV` Here `T=27^(@)C=273+27=300K` `:. E/(KT)=(2.1)/(8.62xx10^(-5)xx300)=81.2` |
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| 14. |
The number density of free electrons in the semiconductor is `10^(18)m^(-3)`. It is doped with a pentavalent impurity atoms of number density `10^(24)m^(-3)`, the number density of free electrons `m^(-3)` increases by a factor ofA. `4//3`B. 6C. `10^(6)`D. `10^(24)` |
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Answer» Correct Answer - C The number density of free electrons in the doped semiconductor `=10^(18)+10^(24)~~10^(24) m^(-3)`. Increase in number density by a factor `=10^(24)//10^(18)=10^(6)` |
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| 15. |
The ratio of number density of free electrons to holes for two different materials, A and B, are (i) equal to one and (ii) less than one respectively. Name the type of semiconductor to which A and B belong. Draw energy level diagram for A and B. |
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Answer» (i) The material A is a pure semiconductor and (ii) the material B is p-type semiconductor. For their energy level diagram, refer to Fig and. |
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| 16. |
How does the energy gap of an intrinsic semiconductor vary, when doped with a trivalent impurity? |
| Answer» When an intrinsic semiconductor is doped with the impurity atoms of valence three like indium or boron, some allowed energy levels are produced, situated in the energy gap slightly above the valence band. These levels are called acceptor energy levels. | |
| 17. |
How does the energy gap in an instrinsic semiconductor vary, when doped with a pentavalent impurity and a trivalent impurity. Draw their energy band diagrams. |
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Answer» When a semiconductor is doped with a pentavalent impurity atoms [like As or P], some additional energy levels are created in the energy gap just below the bottom of conduction band. See Fig. When a semiconductor is doped with a trivalent impurity atoms (like B, A1ect.), some additional energy levels are created in the energy gap just above the upper level of valence band. See Fig. |
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| 18. |
What happens during regualtion action of a Zener diode?A. The current in and voltage across the Zener remains fixedB. The current through the series Resistance `(R_(s))` changesC. The Zener resistance is constantD. The resistance offered by the Zener changs |
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Answer» Correct Answer - B::D During regulation action of a zener diode, the current through the series resistance charges and resistance offered by the zener changes. The current through the zener changes but the voltage across the zener remains constant. |
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| 19. |
In an n-p-n transistor circuit , the collector currents is 10mA . If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?A. The emitter current will be `8mA`B. The emitter current will be `10.53 mA`C. The base current will be `0.53mA`D. The emitter current will be `2mA` |
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Answer» Correct Answer - B::C Here, `I_c=10mA, I_c=95/100I_(e)` `:. 10mA=95/100I_(e) or I_(e)=(10xx100)/95=10.53mA` `I_b=I_(e)-I_c=10.53-10=0.53mA` |
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| 20. |
Find the number density of impurity atoms that must be added to a pure silicon crystal inorder to convert it to have resistivity (i) `10^(-1)Omegam` n-type silicon (ii) `10^(-1)Omegam` p-type silicon. Give for silicon: `mu_(e)=0.135m^(2)V^(-1)s^(-1)` and `mu_(h)=0.048m^(2)V^(-1)s^(-1)`. |
| Answer» (i) `1/rho=en_(e)mu_(e) or n_(e)=(1)/(rho emu_(e)) (ii) n_(h)=(1)/(rho emu_(h))` | |
| 21. |
If the emitter and base of npn transistor have same doping concentration, explain how will the collector and base currents be affected? |
| Answer» In such silution, base current will increase and collector current will decrease. Most of majority carriers moving from emitter towards base will get neutralised by electron-hole combination in base, resulting in increase of base current and few majority carriers will reach the collector. Due to it there will be decrease of collector current. | |
| 22. |
Predict effect on the electrical properties of a silicon crystal at room temperature if every millionth silver atom is replased by an atom of indium. Given, concentration of silicon atoms `=5xx10^(28) m^(-3)`, Intrinsic carrier concentration `=1.5xx10^(16) m^(-3), mue=0.135 m^(2)//Vs and mu_(h)=0.048m^(2)//Vs`. |
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Answer» Concentration of silicon atoms `=5xx10^(28)m^(-3)`. Doping of indium is 1 atoms in `10^(6)` atoms of Si. Indium has valence three. Each doped indium atom creates one hole in Si- crystal and hence acts as acceptor atom. `:.` Concentration of acceptor atoms, `n_(h)=5xx10^(28)xx10^(-6)m^(-3)=5xx10^(22)m^(-3)` Intrinsic carrier concentration, `n_(i)=1.5xx10^(16)m_(-3)` `:.` Hole concentration is increased by an amount `=n_(h)/n_(i)=(5xx10^(22))/(1.5xx10^(16))=3.33xx10^(6)` New electron concentration, `n_(e)=n_(i)^(2)/n_(h)=((1.5xx10^(16))^(2))/(5xx10^(22))=0.45xx10^(10)m^(-3)` It means the hole concentration has been increased over its intrinsic concentration by the same amount with the electron concentration has been decreased. The aonductivity of doped silicon is given by `sigma=e(n_(e)mu_(e)+n_(h)mu_(h)) =(1.6xx10^(-19))[(0.45xx10^(10))xx0.135+(5xx10^(22))xx0.048]` `=(9.72xx10^(-11)+384)Sm^(-1)=384Sm^(-1)` Resistivity, `rho=1/sigma=1/384=0.0026Omegam` Electrical conductivity of pure Si-crystal, `sigma=en_(i)(mu_(e)+mu_(h)) =(1.6xx10^(-19))xx(1.5xx10^(16))[0.135+0.048]` `=0.4392xx10^(-3)Sm^(-1)` Resistivity, `rho=1/sigma=(1)/(0.4392xx10^(-3))=2272Omegam` Thus we see that the conductivity of silicon doped with indium `(384Sm^(-1))` becomes much greater than its intrinsic conductivity `(0.4392xx10^(-3)Sm^(-1))` and the resistivity `(=0.0026Omega m )` has become much smaller than the instrinsic resistivity `(2272Omegam)`. |
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| 23. |
A doped semiconductor has impurity levels `40 me V` below the conductor band. Is the material n-type or p-type? In a thermal collision, an amount kT of energy is given to the extra electron loosely bound to the impurity ion and this electron is just able to jump into the conduction band. Calculate the temperature T. Given `k=8.62xx10^(-5) eV//K`. |
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Answer» Since energy gap between impurity levels and conduction band `=40meV,` which is much smaller than energy gap of pure semiconductor between valence band and conduction band `(~1eV)`, hence the impurity levels are of donor levels and the impurity added has valence five. Thus, the doped semiconductor is of n-type. As per question, `kT=40meV` or `T=(40meV)/k` `=(40xx10^(-3)eV)/(8.62xx10^(-5)eV//k)=464K` |
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| 24. |
What is the decimal number of binary number `(111001.01)_(2)`? |
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Answer» `(111001.01)_(2) =(111001)_(2)+(0.01)_(2)` `(111001)_(2)=1 xx2^(0)+0 xx2^(1)+0xx 2^(2)=1 xx2^(3)+1 xx2^(4)+1 xx 2^(5)` `1+0+0+8+16+32=(57)_(10)` `(0.01)_(2)=0xx2^(-1)+1xx2^(-2)=0+0.25=(0.25)_(10)` `:. (111001.01)_(2)=(57)_(10)+(0.25)_(10)` `=(57.25)_(10)` |
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| 25. |
Subtract `(10101)_(2)` from `(111001)_(2)`. |
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Answer» `{:("1 1 1 0 0 1"),(" 1 0 1 0 1"),(ulbar("1 0 0 1 0 0")):}` i.e. `(111001)_(2)-(10101)_(2)=(100100)_(2)` |
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| 26. |
The current gain for common emitter amplifier is 69. If the emitter current is `7.0 mA`, find (i) base current and (ii) collector current. |
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Answer» `beta=I_c/I_b or I_c=betaI_b=69I_b` `I_(e)=I_b+I_c=I_b+69I_b=70I_b` `:.7.0=70I_(b) or I_(b)=7.0/70=0.1 mA` `I_(c)=I_(e)-I_(b)=7.0-0.1=6.9 mA` |
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| 27. |
What is the decimal number of binary number `(111001.01)_(2)`?A. 9.625B. 25.265C. 26.625D. 26.265 |
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Answer» Correct Answer - C `(11010.101)_(2)=(11010)_(2)+(0101)_(2)` `=[0xx2^(0)+1xx2^(1)+0xx2^(2)+1xx2^(3)+1xx2^(4)]+[1xx2^(-1)+0xx2^(-2)+1xx2^(-3)]` `=[0+2+0+8+16]+[1/2+0+1/8]` `=26+5//8=26.625` |
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| 28. |
The decimal equivalent of the binary number `(11010.101)_(2)` isA. `17+9.625`B. `15+9.625`C. `26+0.625`D. `24+0.625` |
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Answer» Correct Answer - A::C `(11010.101)_(2)=0xx2^(0)+1xx2^(1)+0xx2^(2)+1xx2^(3)+1xx2^(4)+1xx2^(-1)+0xx2^(-2)+1xx2(-3)` `=0+2+0+8+16+1/2+0+1/8` `=265/8 =26.625=17+9.625` |
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| 29. |
The current gain for common emitter amplifier is `59`. if the emitter current is 6.0mA, find the base of current and collector current |
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Answer» Here, `beta_(dc)=59,I_(e)=6.0mA,I_(b)=?` and `I_(b)=?` `beta_(dc)=I_c/(I_b)=(I_(e)-I_b)/I_b or I_b=(I_(e))/(1+beta_(dc))=6.0/(1+59)=0.1mA` `I_c=I_(e)-I_b=6.0-0.1=5.9mA` |
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| 30. |
For decimal number 10, the binary representation is............... |
| Answer» Correct Answer - `(1010)_(2)` | |
| 31. |
A digital circuit which either allows a signal to pass through or stops it, is called a.............. |
| Answer» Correct Answer - gate | |
| 32. |
Convert binary number `1010` into decimal number. |
| Answer» `underset(MSB" "LSB)underset(uarr" "uarr)(("1 0 1 0")_(2)) =0 xx 2^(@) +1 xx 2^(1) +0 xx 2^(2) +1 xx 2^(3) =0 +2 +0 +8 =(10)^(10)` | |
| 33. |
Convert binary number `10111` into decimal number. |
| Answer» Correct Answer - `(23)_(10)` | |
| 34. |
Two Zener diodes having specification `12V, 1/4W` are connected in series. If breakdown voltage of each diode is 5V, then what is the breakdown voltage in the series combination of the diodes?A. 2.5VB. 5VC. 10VD. 12V |
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Answer» Correct Answer - C The two zener diodes in series will double the breakdown voltage i.e. it will be `10V`. |
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| 35. |
The base current of a transistor is 105 muA and collector current is `2.05mA`. Determine the value of `beta, I_(e)` and `alpha` A change of `27muA` in the base current produces a change of 0.65 mA in the collector current. Find `beta_(a.c.)`. |
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Answer» `beta=I_c/I_b=(2.05mA)/(105muA)=(2.05xx10^(-3))/(105xx10^(-6))=19.5` `I_(e)=I_b+I_c=105muA+2.05mA` `=0.105mA+2.05mA=2.155mA` `alpha=I_c/I_(e)=(2.05mA)/(2.155mA)=0.95` `beta_(a.c.)=(DeltaI_c)/(DeltaI_b)=(0.65mA)/(27muA)=(0.65xx10^(-3))/(27xx10^(-6)) =24.07` |
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| 36. |
In a good conductor of electricity the type of bonding that exists is :A. ionicB. vander waalsC. covalentD. metallic |
| Answer» Correct Answer - D | |
| 37. |
Explain, why the input resistance of a transistor is low and output resistance is high. |
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Answer» While using a transister, the emitter-base junction is always forward biased and collectorbase junction is always reverse-biased. Due to it, a small change in emitter current. This means that a small signal voltage variation at the input of the transistor produces a large emitter current variation. This shown that the input resistance of a transistor is low. Since collector is reverse-biased, it collects all the charge carriers which diffuse into it, through base. Due to it a very large change in collector voltage shows only a small change in the collector current. This shows that the output resistance of the transistor is high. |
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| 38. |
The number of silicon atoms per `m^(3) is 5xx10^(28)`. This is doped simultaneously with `5xx10^(22)` atoms per `m^(3)` of Arsenic and `5xx10^(20) per m^(3)` atoms of indium. Calculate the number of electrons and holes. Given that `n_(i)=1.5xx10^(16)m^(-3)`. Is the material n-type or p-type? |
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Answer» `n_(e)=5xx10^(22)-5xx10^(20)=(5-0.05)xx10^(22)=4.95xx10^(22)m^(-3)` `n_(h)=n_(i)^(2)/n_(e)=((1.5xx10^(16))^(2))/(4.95xx10^(22))=4.54xx10^(9)m^(-3)` As `n_(e) gt n_(h)`, so the material is n-type semiconductor. |
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| 39. |
The Boolean expression `P+bar(P)Q`, where P and Q are the inputs of the logic circuit, representsA. AND gateB. NAND gateC. NOT gateD. OR gate |
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Answer» Correct Answer - D We draw a truth table for the expression `P+bar(P).Q` as given below `|{:(P,Q,bar(P),bar(P).Q,P+bar(P.Q)),(0,0,1,0,0),(0,1,1,1,1),(1,0,0,0,1),(1,1,0,0,1):}:|` We note that output is `1` when either of the input is `1`. Thus the logic circuit works as OR gate. |
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| 40. |
In a common emitter (CE) amplifier having a voltage gain G, the transistor used has tranconductance 0.02 mho and current gain 20, the voltage gain will be:A. ` 5/4 G`B. `2/3G`C. 1.5 GD. `1/3 G` |
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Answer» Correct Answer - B Voltage gain, `A_(V)=g_(m)xxR_(L)` (in magnitude) or `A_(V)propg_(m)` `:. A_(V_(2))/A_(V_(1))=g_(m_(2))/(g_(m_(1))) or A_(V_(2))/G=0.02/0.03=2/3 or A_(V_(2))=(2G)/3` |
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| 41. |
In the circuit shown in Fig. when the input voltage of the base resistance is `10V, V_(be)` is zero and `V_(ce)` is also zero. Find the values of `I_, I_` and `beta`. |
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Answer» Here, `V_(i)=10V, V_(BE)=0, V_(CE)=0, V_(C C)=10V, R_B=400kOmega=400xx10^(3)Omega, R_C=3kOmega=3xx10^(3)Omega` Now `V_(i)-V_(BE)=R_b I_b` `:. 10-0=(400xx10^(3))I_b` or `I_b=10/(400xx10^(3))=25xx10^(-6)A=25muA` Also `V_(C C)-V_(CE)=I_c R_c` or `10-0=I_cxx(3xx10^(3))` or `I_c=10/(3xx10^(3))=3.33xx10^(-3) A=3.33mA` `beta=I_c/I_b=(3.33xx10^(-3))/(25xx10^(-6))=133` |
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| 42. |
If the base region of a transistor is made large, as compared to a usual transistor, how does it affect (i) the collector current and (ii) current gain of this transistor. |
| Answer» If the base region of a transistor is made large, then most of the majority charge carriers coming from emitter would get neutralised in the base by the electron-hole combination. As a result of it, only few majority carriers will be reaching the collector. Due to it (i) the collector current will reduce and heance (ii) the current gain will reduce. | |
| 43. |
In the common emitter circuit as shown in Fig, the value of the current gain is 100. Find `I_(B),V_(CE),V_(BE),V_(BC)` when `I_(C)=1.5mA`. The transistor is in active, cutoff or saturation state. |
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Answer» Here, `beta=100,I_(C)=1.5xx10^(-3)A, V_(C C)=24V`, `R_(B) = 220 xx 10^(3) Omega, R_(C) = 4.7 xx 10^(3) Omega` `I_(B)=I_C/beta=(1.5xx10^(-3)A)/100=15xx10^(6)A` In a close circuit GDCEFG `V_(CE)=V_(C C)-I_CR_C` `=24-(1.5xx10^(-3))xx(4.7xx10^(3))=16.95V` In a close circuit GDABEFG `V_(BE)=V_(C C)-I_B R_B` `=24-(15xx10^(-6))xx(220xx10^(3))=20.7V` In a closed circuit ABCDA `V_(BC)=I_C R_C-I_B R_B` `=(1.5xx10^(-3))xx(4.7xx10^(3))-(15xx10^(-6))xx(220xx10^(3))` `=3.75V` Since `V_(BE)gtV_(CE)`, so both the junction of transistor are forward biased. Therefore, the transistor is in a saturation state. |
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| 44. |
The input resistance of a silicon transistor is `665Omega`. Its base current is changed by `15 muA` which result in the change in collector current by `2mA`. This transistor is used as a common emitter amplifier with a load resistance of 5kOmega calculate current gain, `beta_(a.c.)`, transconductance `g_(m)` and voltage gain `A_(v)` of the amplifier. |
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Answer» Here, `R_(i)=665Omega, R_(0)=5kOmega=5xx10^(3)Omega` `beta_(ac)=(DeltaI_c)/(DeltaI_b)=(2mA)/(15muA)=(2xx10^(-3)A)/(15xx10^(-6)A)=133.3` `R_(i)=(DeltaV_(BE))/(DeltaI_b) or Delta V_(BE)=R_(i)xxDeltaI_b=665xx15xx10^(-6)V` `g_(m)=(DeltaI_c)/(DeltaV_(BE))=(2xx10^(-3))/(665xx15xx10^(-6))=0.2Omega_(-1)` `A_(V)=beta_(a.c.)xxR_(0)/R_(i)=133.3xx(5xx10^(3))/665=10^(3)` |
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| 45. |
A change of `0.2mA` in the base current cause a change of `5mA` in the collector current for a common emitter amplifier. (i) Find the a.c. current gain of the transistor (ii) If the input resistance is `2kOmega` and its voltage gain is 75. Calculate the load resistance used in the circuit. |
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Answer» Here, `DeltaI_(B)=0.2mA=0.2xx10^(-3)A=2xx10^(-4)A` `DeltaI_(C)=5mA=5xx10^(-3)A` (i) `beta_(ac)=(DeltaI)/(DeltaI)=(5xx10^(-3))/(2xx10^(-4))=25` (ii)`A_(V)=betaR_(L)/R_(i)` or `R_(L)=(A_(V)xxR_(i))/beta=(75xx2kOmega)/25=6KOmega` |
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| 46. |
A transistor, connected in common emitter configuration, has input resistance `R_(i)=2kOmega` and load resistance `6kOmega`. If `beta=60` and an input signal `10mV` is applied, calculate the (i) resistance gain, (ii) voltage gain (iii) the power gain and (iv) the value of the output signal. |
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Answer» (i) Resistance gain `=R_(0)/R_(i)=6/2=3` (ii) Voltage gain, `A_(V)=betaR_(0)/R_(i)=60xx3=180` (iii) Power gain `=beta^(2)xxR_(0)/R_(i)=(60)^(2)xx3=10800` |
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| 47. |
Two signals A, and B shown in Fig. are used as two inputs of an AND gate obtain its output wave form. Draw the logic symblo of AND gate. |
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Answer» For AND gate, output `y=A.B` and for NAND gate `y_(1)=bar(A.B)`. For time less than 1, `A=0, B=1, y=0, y_(1)=1` For time 1 to 2, `A=1, B=1, y=1, y_(1)=0` For time 2 to 3, `A=1, B=0, y=0, y_(1)=1` For time 3 to 4, `A=0, B=0, y=0, y_(1)=1` For time 4 to 5, `A=0, B=0, y=0, y_(1)=1` For time 5 to 6, `A=1, B=1, y=1, y_(1)=0` For time 6 to 7, `A=0, B=0, y=0, y_(1)=1` For time 7 to 8, `A=0, B=1, y=0, y_(1)=1` |
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| 48. |
Find the binary (i) addition (ii) subtraction of the following set of numbers , 101010 and 010101. |
| Answer» (i) `{:(" 1 0 1 0 1 0"),("+ 0 1 0 1 0 1"),(ulbar(" 1 1 1 1 1 1")):}` (ii) `{:(" 1 0 1 0 1 0"),("- 0 1 0 1 0 1"),(ulbar(" 0 1 0 1 0 1")):}` | |
| 49. |
Two signal A and B shown in the Fig. are used as two inputs of a NAND gate. Draw its output wave form. Give the logic symbol of NAND gate. |
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Answer» For AND gate, output `y=A.B` and for NAND gate `y_(1)=bar(A.B)`. For time less than 1, `A=0, B=1, y=0, y_(1)=1` For time 1 to 2, `A=1, B=1, y=1, y_(1)=0` For time 2 to 3, `A=1, B=0, y=0, y_(1)=1` For time 3 to 4, `A=0, B=0, y=0, y_(1)=1` For time 4 to 5, `A=0, B=0, y=0, y_(1)=1` For time 5 to 6, `A=1, B=1, y=1, y_(1)=0` For time 6 to 7, `A=0, B=0, y=0, y_(1)=1` For time 7 to 8, `A=0, B=1, y=0, y_(1)=1` |
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| 50. |
In the following circuit, if the input wave form is as shown in the figure, what will be the wave form (i) across R in Fig. and (ii) across the diode in Fig. Assume that the diode is ideal. |
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Answer» (i) During the positive half of wave form, p-n junction is forward biased. Due to it we get output voltage across R. During negative half of waveform, p-n junction is reversed biased. No output across R. (ii) During the positive half of waveform, p-n junction is reverse biased and during negative half of wavefrom, p-n junction is forward biased. |
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