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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The `V-I` characteristic of a silicon diode is shown in the Fig. Calculate the resistance of the diode at (a) `I_(D) = 15 mA` and (b) `V_(D) = -10 V`. . |
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Answer» Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law. (a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V `r_(fb) = DeltaV//DeltaI = 0.1V//10 mA = 10 Omega` (b) From the curve at `V = –10 V, I = –1 muA,` Therefore, `r_(rb) = 10 V//1muA= 1.0 × 10^(7) Omega` |
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| 2. |
Region which have no free electron and holes in P-N junction isA. x-regionB. p-regionC. depletion regionD. none of these |
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Answer» Correct Answer - C The depletion region created at the junction is devoid of free charge carriers. |
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| 3. |
In an n-p-n circuit transistor, the collector current is 10 mA. If 80% electrons emitted reach the collector, thenA. the emitter current will be 7.5 mAB. the emitter current will be 12.5 mAC. the base current will be 3.5 mAD. the base current will be 1.5 mA |
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Answer» Correct Answer - B Here, `I_C`=80% of `I_E=80/100I_E` or `I_E=I_C/0.8=10/0.8=12.5 mA` `I_B=I_E-IC=12.5-10=2.5 mA` |
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| 4. |
If a change of `100 mu A` in the base current of an `n-p-n` transistor in `CE` causes a change of `10 mA` in the collector current, the `ac` current gain of the transistor isA. 50B. 100C. 200D. 150 |
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Answer» Correct Answer - B Here, `DeltaI_B=100 mu A = 100 xx 10^(-6) A` `DeltaI_C=10 mA = 10 xx 10^(-3)A` `therefore` ac current gain , `beta_(ac)=(DeltaI_C)/(DeltaI_B)=(10xx10^(-3))/(100xx10^(-6))=100` |
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| 5. |
If `alpha and beta` are the current gain in the CB and CE configurations respectively of the transistor circuit, then `(beta-alpha)/(alphabeta)=`A. zeroB. 1C. 2D. 5 |
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Answer» Correct Answer - B Since `beta=alpha/(1-alpha)`...(i) `therefore alphabeta=alpha^2/(1-alpha)-alpha"or" beta-alpha=alpha^2/(1-alpha)`..(ii) `therefore (beta-alpha)/(alphabeta)=(alpha^2/(1-alpha))((1-alpha)/alpha^2)=1` |
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| 6. |
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. 0.02B. 7C. 33D. 4.9 |
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Answer» Correct Answer - A As, `I_E=(n_Exxe)/t` and `I_C=(n_Cxxe)/t=((98//100)n_Exxe)/t=98/100xxI_E` Current transfer ratio, `alpha=I_C/I_E=98/100=0.98` Current amplification factor, `beta=alpha/(1-alpha)=0.98/(1-0.98)=49 " " therefore " " alpha/beta=0.98/49=0.02` |
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| 7. |
The transfer characteristics of a base biased transistor has the operation regions, namely, cutoff, active region and saturation region. For using the transistor as an amplifier it has to operate in theA. active regionB. cutoff regionC. saturation regionD. cutoff and saturation |
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Answer» Correct Answer - A When the transistor is used in the cutoff or saturation region it acts as a switch. On the other hand for using the transistor as an amplifier, it has to operate in the active region. |
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| 8. |
When the voltage drop across a `p.n` junction diode is increased from `0.65V` to `0.70V`, the change in the diode current is `5mA`. What is the dynamic resistance of the diode?A. `5 Omega`B. `10 Omega`C. `20 Omega`D. `25 Omega` |
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Answer» Correct Answer - B Dynamic resistance is `r_d=(DeltaV)/(DeltaI)` Here, `DeltaV` = 0.7 - 0.65 V = 0.05 V , `DeltaI=5mA = 5xx 10^(-3) A` `therefore r_d=0.05/(5xx10^(3))=10 Omega` |
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| 9. |
A potential barrier of 0.3 V exists across a p-n junction. If the depletion region is 1 `mu`m wide, what is the intensity of electric field in this region?A. `2xx10^5 V m^(-1)`B. `3xx10^5 V m^(-1)`C. `4xx10^5 V m^(-1)`D. `5xx10^5 V m^(-1)` |
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Answer» Correct Answer - B Electric field `E=V/d=0.3/(1xx10^(-6))=3xx10^5 V m^(-1)` |
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| 10. |
In Fig . `V_(0)` is the potential barrier across a p-n junction, when no battery is connected across the junction A. 1 and 3 both correspond to forward bias of junctionB. 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junctionC. 1 corresponds to forward bias and 3 corresponds to reverse bias of junction.D. 3 and 1 both correspond to reverse bias of junction. |
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Answer» Correct Answer - B Height of potential barrier decreases when p-n junction is forward biased and it increases when junctionis reverse biased. |
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| 11. |
A potential barrier of `0.50 V` exists across a `P-N` junction. If the depletion region is `5.0xx10^(-7)m`, wide the intensity of the electric field in this region isA. `10^6` V/mB. `10^7` V/mC. `10^5` V/mD. `10^4` V/m |
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Answer» Correct Answer - A Electric field `E=V/d=(0.50 V)/(5.0xx10^(-7)m)=1.0xx10^6 V m^(-1)` |
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| 12. |
The equivalent resistance between the points A and B, if `V_A gt V_B` isA. `10 Omega`B. `20 Omega`C. `30 Omega`D. `15 Omega` |
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Answer» Correct Answer - A When `V_A gt V_B` the diode is forward biased and offers no resistance . `therefore R=(20xx20)/(20+20)=10 Omega` |
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| 13. |
A 220 V ac supply is connected between points A and B as shown 220 V in figure. What will be the potential AC difference V across the capacitor? A. 220 VB. 110 VC. 0 VD. `200sqrt2V` |
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Answer» Correct Answer - D Potential difference across capacitor, V= peak voltage `=V_(rms)sqrt2=220sqrt2V` |
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| 14. |
The equivalent resistance of the circuit shown in figure between the points A and B if `V_A lt V_B` is A. `10 Omega`B. `20 Omega`C. `5 Omega`D. `40 Omega` |
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Answer» Correct Answer - B When `V_A lt V_B`, the diode is reverse biased and offers infinite resistance. No current flows through the upper branch `therefore R=20 Omega` |
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| 15. |
A Zener diode is specified having a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can handle.A. 40 mAB. 60 mAC. 50 mAD. 45 mA |
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Answer» Correct Answer - A The maximum permissible current is `I_(Z_("max"))=P/V_z=(364xx10^(-3))/9.1=40 mA` |
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| 16. |
In a half wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would beA. 25 HzB. 50 HzC. 70.7 HzD. 100 Hz |
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Answer» Correct Answer - B As the output voltage obtained in a half wave rectifier circuit has a single variation in one cycle of ac voltage, hence the fundamental frequency in the ripple of output voltage would be = 50 Hz. |
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| 17. |
A p-n photodiode is fabricated from a semiconductor with band - gap of 2.8 eV . Can it detect a wavelength of 6000nm? |
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Answer» Energy band gap of the given photodiode, `E_(g) = 2.8 eV` Wavelength, `lamda = 6000 nm = 6000 × 10^(−9) m` The energy of a signal is given by the relation: `E=overset((hc)/(lamda)` Where, h = Planck’s constant `= 6.626 xx 10−^(34) "Js"` c = Speed of light `= 3 × 10^(8) " m/s"` `=(6.626xx10^(-34)xx3xx10^(8))/(600xx10^(-9))` E `= 3.313 xx 10^(−20) J` But `1.6 xx 10^(−19) J = 1 eV` `thereforeE = 3.313 xx 10^(−20) J` `=(3.313xx10^(-20))/(1.6xx10^(-19))=0.207 " eV"` The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal. |
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| 18. |
In an intrinsic semiconductor the energy gap `E_(g) is 1.2 eV`. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at `600K` and `300K`? Assume that temperature dependence intrinstic concentration `n_(i)` is given by `n_(i)=n_(0) exp ((-E_(g))/(2k_T))`, where `n_(0)` is a constant and `k_=8.62xx10^(-5)eV//K`. |
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Answer» Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV The temperature dependence of the intrinsic carrier-concentration is written as: `n_(i)=n_(0)exp[-(E_(g))/(2K_(B)T)]` Where, `k_(B)` = Boltzmann constant `= 8.62 xx 10^(−5) eV//K` T = Temperature `n_(0)` = Constant Initial temperature, `T_(1) = 300 K` The intrinsic carrier-concentration at this temperature can be written as: `n_(i_(1))=n_(0)exp[-(E_(g))/(2k_(B)xx300)]_(...(1))` Final temperature, `T_(2) = 600 K` The intrinsic carrier-concentration at this temperature can be written as: `n_(i2)=n_(0)exp[-(E_(g))/(2k_(B)xx600)]_(...(2))` The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures. `(n_(i2))/(n_(i1))=(n_(0)exp[-(E_(g))/(2k_(B)xx600)])/(n_(0)exp[-(E_(g))/(2k_(B)xx300)])` `=exp""E_(g)/(2k_(B))[(1)/(300)-(1)/(600)]=exp[(1.2)/(2xx8.62xx10^(-5))xx(2-1)/(600)]` `=exp[11.6]=1.09xx10^(5)` Therefore, the ratio between the conductivities is `1.09 xx 10^(5)`. |
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| 19. |
If a small amount of antimony is added to germanium crystalA. its resistance is increasedB. it becomes a p-type semiconductorC. there will be more free electrons than holes in the semiconductorD. none of these |
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Answer» Correct Answer - C Adding fifteenth group element to germanium makes it an n-type semiconductor. Antimony is a fifteenth group element and so germanium becomes n-type semiconductor. |
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| 20. |
The ac current gain of a transistor is 120. What is the change in the collector current in the transistor whose base current changes by `100 muA`?A. 6 mAB. 12 mAC. 3 mAD. 24 mA |
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Answer» Correct Answer - B Here, `I_B=100 muA=100xx10^(-6) A,beta_(ac)=120` Using, `beta_(ac)=(DeltaI_C)/(DeltaI_B) " " therefore 120 =(DeltaI_C)/(100xx10^(-6))` Change in collector current , `DeltaI_C=120xx100xx10^(-6)=12xx10^(-3)=12 mA` |
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| 21. |
The symbolic representation of four logic gates are given in Fig.The logic symbol for OR, NOT and NAND gates are respectively A. (iv), (i), (iii)B. (iv), (ii), (i)C. (i), (ii), (iv)D. (ii), (iv), (i) |
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Answer» Correct Answer - B The given circuit represents an OR gate when either A or B or both inputs are high, the output C is high. OR gate, NOT gate and NAND gates are (iv), (ii) and (i) respectively |
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| 22. |
Two amplifiers are connected one after the other in series ( cascaded) . The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal. |
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Answer» Given, voltage gain of first Amplifier, `A_(V_(1))= 10` Voltage gain of second Amplifier, `A_(V_(2)) = 20` Input voltage `V_(i) = 0.01 V ` Total voltage gain `A_(v) = ( V_(0))/(V_(i)) = A_(V_(1))xx A_(V_(2))` `(V_(0))/(0.01) = 10 xx 20 , V_(0)= 2V `. |
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| 23. |
For a transistor amplifier, the voltage gain (a) remains constant for all frequencies is high at high and low frequencies and constant in the middle frequency range. (c ) is low frequencies and constant at mid frequencies. (d ) None of the above. |
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Answer» Correct Answer - c The voltage gain is low at high and low frequencies and constant at mid- frequencies. |
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| 24. |
`M_(x) and M_(y)` denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q - value for a `beta-` decay is `Q_(1)` and that for a `beta^(+)` decay is `Q_(2)`. If `m_(e)` denotes the mass of an electrons, then which of the following statements is correct?A. `Q_(1) =(M_(x)-M_(y)) c^(2) " and " Q_(2) =[M_(x)-M_(y) -2m_(e)]c^(2)`B. `Q_(1) =(M_(x)-M_(y))c^(2) " and " Q_(2) =(M_(x) -M_(y))c^(2)`C. `Q_(1)=(M_(x) -M_(y)) c^(2) " and " Q_(2) = (M_(x) -M_(y) +2c_(e))c^(2)`D. `Q_(1) =(M_(x)-M_(y) +2m_(e)) c^(2) " and " Q_(2) =(M_(x)-m_(y) +2m_(e))c^(2)` |
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Answer» Correct Answer - a Let the nucleus is `._(z)X^(A).beta^(+)` decay is represented as `._(z)X^(A) to ._(z-1)y^(A) +._(+ 1) e^(0) + v +Q_(2)` ` Q_(2) =[M_(n) (._(z)X^(A)) -m_(n) (._(z-1)y^(A)) -m_(e)]c^(2)` `=[m_(2) (._(z)X^(A)) +zm_(e)-m_(n) (._(z-1)y^(A)) -(Z-1)m_(e)-2m_(e)]c^(2)` , `=[m_(e)(._(z)X^(A))-(._(Z-1)Y^(A))-2m_(e)]c^(2)` `=(m_(x)-M_(y)-2m_(e))c^(2)` `beta-`decay is represented as `=._(z)X^(A) to ._(z+1)Y^(A) + ._(-1)e^(0) +bar(v)+alpha_(1)` `alpha_(1) =[m_(n)(._(z)X^(A))-m_(n)(._(z+1)Y^(A))-m_(e)]c^(2)` `[m_(n)(._(z)X^(A)) +zm_(e) -m_(n) (._(z+1)Y^(A)) - (z+1) me ]c^(2)` `=[m (._(z)X^(A)) -m (._(z-1)Y^(A))]c^(2)` `=(M_(x) -M_(y)) c^(2)` |
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| 25. |
What is the voltage gain in a common emitter amplifier, where input resistance is `3Omega` and load resistance `24Omega` and `beta=61` ?A. 8.4B. 488C. 240D. 0 |
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Answer» Correct Answer - B Voltage gain, `A_V=betaR_o/R_i=(61xx24)/3=488` |
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| 26. |
In a transistor connected in a common emitter mode `R_C = 4kOmega, R_1 = 1kOmega, I_C = 1mA` and `I_B = 20 muA`. Find the voltage gain.A. 100B. 200C. 300D. 400 |
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Answer» Correct Answer - B Voltage gain, `A_V=(betaR_C)/R_I=I_C/I_BxxR_C/R_I` `=(1xx10^(-3)xx4xx10^3)/(20xx10^(-6)xx1xx10^3)=200` |
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| 27. |
A common emitter amplifier gives an output of 3 V for an input of 0.01 V. If `beta` of the resistance is 100 and the input resistance is `1 kOmega`. then the collector resistance isA. `3 kOmega`B. `30 kOmega`C. `1 kOmega`D. `5 kOmega` |
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Answer» Correct Answer - A `A_V=V_o/V_i=betaR_o/R_i` or `R_o=(V_oR_i)/(V_ibeta)=(3xx1xx10^3)/(0.01xx100)=3 kOmega` |
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| 28. |
A transistor connected in common emitter mode, the voltage drop across the collector is 2 V and `beta` is 50, the base current if `R_C` is `2 kOmega` isA. `40 muA`B. `20 muA`C. `30 muA`D. `15 muA` |
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Answer» Correct Answer - B Using, `I_C=(V_(CE))/R_C=2/(2xx10^3)=10^(-3)`=1 mA `therefore beta=I_C/I_B,I_B=I_C/beta=10^(-3)/50A=20 muA` |
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| 29. |
The input resistance of a common emitter transistor amplifier, if the output resistance is `500 kOmega`, the current gain `alpha=0.98` and the power gain is `6.0625 xx 10^(6)` isA. `198 Omega`B. `300 Omega`C. `100 Omega`D. `400 Omega` |
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Answer» Correct Answer - A Voltage gain =`Av=betaR_o/R_i` Also, current gain:`beta=alpha/(1-alpha)=0.98/(1-0.98)=49` `A_V=(49)((500xx10^3)/R_i)` Power gain= current gain x voltage gain Power gain=`6.0625xx10^6=49xx((500xx10^3)/R_i)xx49` or `R_i=198 Omega` |
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| 30. |
The potential difference across the collector of a transistor, used in common emitter mode is 1.5 V, with the collector resistance of `3 kOmega`, the emitter current is [`beta- 50`]A. 0.70mAB. 0.49 mAC. 1.1 mAD. 1.9 mA |
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Answer» Correct Answer - B Here, `V_(CE)`=1.5 V `R_C=3kOmega=3xx10^(3) Omega, beta=50` `therefore I_C=V_(CE)/R_C=1.5/(3xx10^3)=0.5 xx 10^(-3) A` `I_B=I_C/beta=(0.50xx10^(-3))/50=0.01xx10^(-3)A` `I_E=I_C-I_B=(0.50-0.01)xx10^(-3)` `=0.49xx10^(-3) A =0.49 mA` |
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| 31. |
For a common emitter transistor amplifier, the audio signal voltage across the collector resistance of `2 kOmega` is 2 V. Suppose the current amplification factor of the transistor is 100, the base current if base resistance is `1 k Omega` isA. `10 muA`B. `20 muA`C. `5 muA`D. `2 muA` |
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Answer» Correct Answer - A Here, `R_C=2 kOmega =2000 Omega, R_B=1 kOmega=1000 Omega, beta=100, V_o=2V` `therefore beta=I_C/I_B=(V_o//R_C)/I_B` `therefore` Base current , `I_B=V_o/(betaR_C)=2/(100xx2000)=10^(-5) A=10 muA` |
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| 32. |
For a CE transistor amplifier, the aurdio signal voltage across the collector resistance of`2.0 kOmega` is 2.0 V.Suppose the current amplification factor of the transistor is 100, what should be the value of`R_(B)` in series with `V_(B B)` supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance (Refer to Figure). |
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Answer» The output ac voltage is2.0 V .So, the ac collector current `i_(C ) = 2.0//2000 = 1.0 mA.` The signal current through the base is , therefore given by `i_(B ) = i_(C ) //beta = 1.0 mA// 100 = 0.010 mA`. The dc base current has to be `10 xx 0.010 = 0.10 mA` From `V_(B B ) = V_(BE ) + I_(B) R_(B) R_(B) = (V_( B B) - V_(B E) )//I_(B) `. Assuming `V_(BE) = 0.6 V` . ` R_(B) = ( 2.0 - 0.6 ) //0.10 = 14kOmega ` |
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| 33. |
Boolean algebra is essentially based onA. numberB. truthC. logicD. symbol |
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Answer» Correct Answer - C The Boolean algebra is based on logic. |
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| 34. |
In boolean algebra, if `A=1` and `B=0` then the value of `A+barB` isA. AB. A.BC. A+BD. Both (a) and (c ) |
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Answer» Correct Answer - D Here, A=1, B=0 , then `A+barB=1+bar0=1+1=1` `therefore ` A+B=1+0=1, But B=0 and A.B=1.0=0 |
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| 35. |
In figure , assuming the diodes to be ideal , A. `D_1` is forward biased and `D_2` is reverse biased and hence current flows from A to B.B. `D_2` is forward biased and `D_1` is reverse biased and hence no current flows from B to A and vice versa.C. `D_1 and D_2` are both forward biased and hence current flows from A to B.D. `D_1 and D_2` are both reverse biased and hence no current flows from A to B and vice versa |
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Answer» Correct Answer - B Diode `D_1` is reverse biased as p side is connected to negative potential and n side to ground. Diode `D_2` is forward biased as p side is grounded and n side is at negative potential. |
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| 36. |
In half - wave rectification, what is the output frequency, if the input frequency is 50 Hz ? What is the output frequency of a full - wave rectifier for the same input frequency ? |
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Answer» Input frequency = 50 Hz For a half-wave rectifier, the output frequency is equal to the input frequency. `therefore"Output frequency" = 50 Hz` For a full-wave rectifier, the output frequency is twice the input frequency. `therefore"Output frequency" = 2 xx 50 = 100 Hz` |
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| 37. |
The power gain for common base amplifier is 800 and the voltage amplification factor is 840. The collector current when base current is 1.2 mA isA. 24 mAB. 12 mAC. 6 mAD. 3 mA |
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Answer» Correct Answer - A Current gain, `alpha=("Power gain")/("Voltage gain")=800/840 =20/21 ` Now, `beta=alpha/(1-alpha)=((20//21))/(1-(20//21))=20` As, `beta=I_C/I_B` `therefore I_C=betaI_B=20xx1.2=24 ` mA (Given `I_B` =1.2 mA) |
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| 38. |
When a forward bias is applied to a p -n junction. ItA. raises the potential barrier.B. reduces the majority carrier current to zero.C. lowers the potential barrier.D. None of the above. |
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Answer» The correct statement is (c). When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced. |
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| 39. |
At absolute zero , Si acts asA. metalB. semiconductorC. insulatorD. none of these |
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Answer» Correct Answer - C At absolute zero. Si acts as an insulator due to the absence of free electrons in the conduction band. |
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| 40. |
Write the expressions for the efficiency of a full wave rectifier and a half wave rectifier. |
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Answer» (i) Efficiency of half wave rectifier `( eta)= ( 0.406R_(L))/( r_(f)+ R_(L))` (ii) Efficiency of full wave rectifier `( eta )= ( 0.812R_(L))/( r_(f)+ R_(L))` |
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| 41. |
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier. |
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Answer» Internal resistance `( r_(f)) = 20 Omega ` `R_(L) = 2k Omega = 2000 Omega ` `eta = ( 0.406R_(L))/(r_(f) + R_(L)) = ( 0.406 xx 2000)/( 20+ 2000) xx 100 = ( 812 xx 100)/( 2020)` `eta = 40.2 %` |
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| 42. |
In which bias can be a Zener diode be used as voltage regulator ? |
| Answer» In reverse bias Zener diode can be used as voltage regulator . | |
| 43. |
Which gates are called universal gates? |
| Answer» NAND gate and NOR gates are called universal gates. | |
| 44. |
What is the maximum percentage of rectification in half wave and full wave rectifiers? |
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Answer» (i) The percentage of rectification in half-wave rectifer is `40.6%`. (ii) The percentage of rectification in full-wave rectifier is`81.2 %` |
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| 45. |
What is Zener voltage `( V_(2))` and how will a Zener diode be connected in circuits generally ? |
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Answer» (i) When Zener diode is in reverse biased, at a particular voltage current increases suddenly is called Zener (or ) break down voltage. (ii) Zener diode always connected in reverse bias. |
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| 46. |
Which of the junction diodes shown below are forward biased ?A. B. C. D. |
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Answer» Correct Answer - A The p-n junction diode is forward biased when p is at high potential with respect to n. Hence option (a) is correct |
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| 47. |
A semiconductor has equal electron and hole concentration of `6xx10^(8)//m^(3)`. On doping with certain impurity, electron concentration increases to `9xx10^(12)//m^(3)`. (i) Identify the new semiconductor obtained after doping. (ii) Calculate the new hole concentration. (iii) How does the energy gap vary with doping?A. `2xx10^4` per `m^3`B. `2xx10^2` per `m^3`C. `4xx10^4` per `m^3`D. `4xx10^2` per `m^3` |
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Answer» Correct Answer - C As, `n_e n_h=n_i^2` Here, `n_i=6xx10^8 per m^3 and n_e = 9xx10^12 per m^3` `therefore n_h=n_i/n_e=((6xx10^8)^2)/(9xx10^12)=4xx10^4 per m^3` |
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| 48. |
In a p-n junction diode, the currentI can be expressed as `I=I_(0)`exp `((eV)/(2K_(B)T)-1)` where`I_(0)`is called the reverse saturation current,V is the voltage across the diode andis positive for forward bias and negative for reverse biase,and I is the current through the diode, `k_(B)` is the Boltzmann constant `( 8.6 xx10^(-5) eV//K)` and T is the absolute temperature . If fora given diode`I_(0)=5 xx10^(-12) A` and `T = 300K`, then (i) What will be the forward current at a forward voltage of 0.6V? (b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V ? (c ) What is the dynamic resistance ? (d) What will be the current if reverse biase voltage changes from 1V to 2V ? |
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Answer» Given `I_(0) =5xx 10^(-12) A,T=300K ` `K_(B) = 8.6xx 10^(-5) eV//K =8.6xx 10^(-5) xx 1.6 xx10^(-19) J//K ` (a) Given, voltage `V= 0.6 V ` `(eV)/(K_(B) T)= ( 1.5 xx 10^(-19) xx0.6)/(8.6 xx 10^(-5) xx 1.6 xx 10^(-19) xx 300)=23.26` The current I through a junction diode is given by `I= l_(0)e((e_(v))/(2K_(B)T)-1)= 5xx 10^(-12) ( e^(23.26)-1)= 5xx 10^(-12)( 1.259xx10^(-10)-1)` Change in current`Delta I = 3.025 -0.693 = 2.9A ` (b) Given voltage `V =0.7 V ` `(eV)/( K_(B)T)= ( 1.6xx10^(-19)xx0.7)/( 8.6 xx 10^(-5)xx1.6 xx10^(-19) xx 300) = 27.14 ` Now, `1=I_(0)( eV)/(K_(B)T)-1 =5xx10^(-12) ( e^(27.14 ) -1)` `=5 xx 10^(-12) ( 6.07 xx 10^(11) -1)` `=5 xx 10^(-12)xx 5.07 xx 10^(11) =0.035 A ` Change in current `Delta I =3.035 -0.693 =2.9A` (c )`DeltaI =2.9 A `,voltage`DeltaV= 0.7 - 0.6 =0.1V ` Dynamic resistance `R_(d) = ( Delta V )/(Delta I) = ( 0.1)/( 2.9 ) =0.0336Omega ` (d) As the voltage changes from 1V to 2V , the current 1 will be almost equal to `I_(0) = 5 xx 10^(-12) A ` It is due to that the diode possesses practically infinite resistance in the reverse bias . |
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| 49. |
Suppose a pure Si crystal has ` 5 xx 10^(28)` atoms `m^(-3)` . It is droped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that `n_(i) = 1.5 xx 10^(16) m^(-3)`. |
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Answer» Note that thermally generated electrons`( n_(i) ~ 10^(16)m^(-3))` are negligibly small as compared to those produced by doping. Therefore, `n_(e ) = N_(D) ` Since `n_( e) n_(h ) = n_(i )^(2) `, The number of holes ,`n_(h) = ( 1.5 xx 10^(16))^(2) // 5 xx 10^(28) xx 16^(-6)` `n_(h) =( 2.25 xx 10^(32)) //( 5xx 10^(22)) ~ 4.5xx 10^(9) m^(-3)` |
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| 50. |
In an n-type silicon, which of the following statement is true `:` (a) Electrons are majority carriers are trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c ) Holes are minority carriers and pentavalent atoms are the dopants. (d ) Holes are majority carriers and trivalent atoms are the dopants. |
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Answer» Correct Answer - c In an n-type semiconductor, it is obtained by doping the Ge or Si with pentavalent atoms .In n - type semiconductors, electrons are majority carriers and holes are minority carriers. |
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