Tiêu đề 14. Semiconductors and Electronic Devices - Questions.pdf ✅

14. Semiconductors and Electronic Devices 1(1.)The distance between the body centred atom and a corner atom in sodium (a = 4.225 Å) is (a.) 3.66 Å (b.) 3.17 Å (c.) 2.99 Å (d.) 2.54 Å 2(2.)In order to forward bias a PN junction, the negative terminal of battery is connected to (a.) P-side (b.) Either P-side or N-side (c.) N-side (d.) None of these 3(3.)When boron is added as an impurity to silicon, the resulting material is (a.) n-type semiconductor (b.) n-type conductor (c.) p-type conductor (d.) p-type semiconductor 4(4.)In the circuit given below, V(t) is the sinusoidal voltage source, voltage drop VAB(t) across the resistance R is (a.) Is half wave rectified (b.) Is full wave rectified (c.) Has the same peak value in the positive and negative half cycles (d.) Has different peak values during positive and negative half cycle 5(5.)The truth table given below is for (A and B are the inputs, Y is the output) A B Y 0 0 1 1 0 1 0 1 1 1 1 0 (a.) NOR (b.) AND (c.) XOR (d.) NAND 6(6.)The combination of ‘NAND’ gates shown here under figure, are equivalent to (a.) An OR gate and an AND gate respectively (b.) An AND gate and a NOT gate respectively (c.) An AND gate and an OR gate respectively (d.) An OR gate and a NOT gate respectively 7(7.)The correct curve between potential (V) and distance (d) near p − n junction is (a.) (b.) (c.) (d.) 8(8.)In a NPN transistor, 108 electrons enter the emitter in 10−8 s. If 1% electrons are lost in the base, the fraction of current that enters the collector and current amplification factor are respectively (a.) 0.8 and 49 (b.) 0.9 and 90 (c.) 0.7 and 50 (d.) 0.99 and 99 9(9.)The binary number 10111 is equivalent to the decimal number (a.) 19 (b.) 31 (c.) 23 (d.) 22 R1=100  R2=150  D1 D2 V(t) R VAB
10(10.)When A is the internal stage gain of an amplifier and β is the feedback ratio, then the amplifier becomes as oscillator if (a.) β is negative and magnitude of β = A/2 (b.) β is negative and magnitude of β = 1/A (c.) β is negative and magnitude of β = A (d.) β is positive and magnitude of β = 1/A 11(11.)Intrinsic germanium and silicon at absolute zero temperature behave like (a.) Superconductor (b.) Good semiconductor (c.) Ideal insulator (d.) Conductor 12(12.)In an unbiased p-n junction (a.) Potential at p is more than that at n (b.) Potential at p is less than that at n (c.) Potential at p is equal to that at n (d.) Potential at p is +ve and that at n is −ve 13(13.)In the following combinations of losgic gates, the outputs of A, B and C are respectively (a.) 0, 1, 1 (b.) 0, 1, 0 (c.) 1, 1, 0 (d.) 1, 0, 1 14(14.)The forbidden energy gap in Ge is 0.72 eV, given, hc = 12400 eV-Å. The maximum wavelength of radiation that will generate electron hole pair is (a.) 172220 Å (b.) 172.2 Å (c.) 17222 Å (d.) 1722 Å 15(15.)Avalanche breakdown is due to (a.) Collision of minority charge carrier (b.) Increase in depletion layer thickness (c.) Decrease in depletion layer thickness (d.) None of these 16(16.)What is the number of code combination in a 4 bit byte? (a.) 256 (b.) 4 (c.) 16 (d.) 32 17(17.)The plate current ip in a triode valve is given ip = K(Vp + μVg) 3/2 where ip is in milliampere and Vp and Vg are in volt. If rp = 104 ohm, and gm = 5 × 10−3mho, then for ip = 8 mA and Vp = 300 volt, what is the value of K and grid cut off voltage (a.) −6V, (30) 3/2 (b.) −6V, (1/30) 3/2 (c.) +6V, (1/30) 3/2 (d.) +6V, (1/30) 3/2 18(18.)Which of the following statements is not correct when a junction diode is in forward bias? (a.) The width of depletion region decreases. (b.) Free electrons on n-side will move towards the junction. (c.) Holes on p-side move towards the junction. (d.) Electrons on n-side and holes on p-side will move away from junction. 19(19.)Coating of strontium oxide on tungsten cathode in a valve is good for thermionic emission because (a.) Work function decreases (b.) Work function increases (c.) Conductivity of cathode increases (d.) Cathode can be heated to high temperature 20(20.)A p-type material is electrically ....... (a.) Positive (b.) Negative (c.) Neutral (d.) Depends on the concentration of p impurities 21(21.)The PN junction diode is used as (a.) An amplifier (b.) A rectifier (c.) An oscillator (d.) A modulator 22(22.)On applying a potential of −1 volt at the grid of a triode, the following relation between plate voltage Vp (volt) and plate current Ip (in mA) is found Ip = 0.125 Vp − 7.5. If on applying −3 volt potential at grid and 300 V potential at plate, the plate current is found to be 5mA, then amplification factor of the triode is (a.) 100 (b.) 50 (c.) 30 (d.) 20 23(23.)Radiowaves of constant amplitude can be generated with (a.) FET (b.) Filter (c.) Rectifier (d.) Oscillator 24(24.)For an insulator the forbidden energy gap is (a.) Zero (b.) 1 eV (c.) 5 eV (d.) 2 eV 25(25.)In the figure, potential difference between A and B is
(a.) Zero (b.) 5 V (c.) 10 V (d.) 15 V 26(26.)While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio hfe is (a.) 82 (b.) 83 (c.) 8.2 (d.) 8.3 27(27.)Current gain in common emitter configuration is more than 1 becomes (a.) Ic < Ib (b.) Ic < Ie (c.) Ic > Ie (d.) Ie > Ib 28(28.)C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator where as Si is intrinsic semiconductor. This is because (a.) In case of C the valance band is not completely filled at absolute zero temperature (b.) In case of C the conduction band is partly filled even at absolute zero temperature (c.) The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third (d.) The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit 29(29.)Figure shows a diode connected to an external resistance and an emf. Assuming that the barrier potential developed in diode is 0.5 V, obtain the value of current in the circuit in milli ampere. (a.) 40 mA (b.) 60 mA (c.) 80 mA (d.) 100 mA 30(30.)The logic circuit shown below has the input waveforms A and B as shown. Pick out the correct output waveform. (a.) (b.) (c.) (d.) 31(31.)Identify the incorrect statement regarding a superconducting wire (a.) Transport current flows through its surface (b.) Transport current flows through the entire area of cross-section of the wire (c.) It exhibits zero electrical resistivity and expels applied magnetic field (d.) It is used to produce large magnetic field 32(32.)A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of (a.) Each of them increases (b.) Each of them decreases (c.) Copper decreases and germanium increases (d.) Copper increases and germanium decreases 33(33.)In the given figure, which of the diodes are forward biased 4.5 V 100 Input A Input B 10 kΩ 10 kΩ 10 kΩ 30 V R – 10V – 10 V R – 12V – 5V R +5V +5V +10V
(a.) 1, 2, 3 (b.) 2, 4, 5 (c.) 1, 3, 4 (d.) 2, 3, 4 34(34.)What is the current in the circuit shown below (a.) 0 A (b.) 10−2 A (c.) 1 A (d.) 0.10 A 35(35.)In space charge limited region, the plate current in a diode is 10 mA for plate voltage 150 V. If the plate voltage is increased to 600 V, then the plate current will be (a.) 10 mA (b.) 40 mA (c.) 80 mA (d.) 160 mA 36(36.)Wires P and Q have the same resistance at ordinary (room) temperature. When heated, resistance of P increases and that of Q decreases. We conclude that (a.) P and Q are conductors of different materials (b.) P is N-type semiconductor and Q is P-type semiconductor (c.) P is semiconductor and Q is conductor (d.) P is conductor and Q is semiconductor 37(37.)The plate characteristic curve of a diode in space charge limited region is as shown in the figure. The slope of curve at point P is 5.0 mA/V. The static plate resistance of diode will be (a.) 111.1Ω (b.) 222.2Ω (c.) 333.3Ω (d.) 444.4Ω 38(38.)In a CE, n-p-n transistor circuit, the emitter current is (a.) More than the collector current (b.) Less than the collector current (c.) Less than the base current (d.) Equal to the difference of the collector current and the base current 39(39.)The output of given logic circuit is (a.) A + B + C (b.) (A + B) ∙ (A + C) (c.) A ∙ (B ∙ C) (d.) A ∙ (B + C) 40(40.)A junction diode is connected to a 10 V source and 103Ω rheostate figure. The slope of load line on the characteristic curve of diode will be (a.) 10−2AV −1 (b.) 10−3AV −1 (c.) 10−4AV −1 (d.) 10−5AV −1 41(41.)If the ends p and n of p − n diode junction are joined by a wire (a.) There will not be a steady current in the circuit (b.) There will be a steady current from n-side to p- side (c.) There will be a steady current from p-side to n- side (d.) There will not be a current depending upon the resistance of the connecting wire 42(42.)The manifestation of band structure in solids is due to decreases the majority charge carries (a.) Heisenberg’s uncertainty principle (b.) Pauli’s exclusion principle (c.) Bohr’s correspondence principle (d.) Boltzmann’s law 43(43.)An alternating current can be converted into direct current by a (a.) Dynamo (b.) Motor (c.) Transformer (d.) Rectifier 44(44.)In the grid circuit of a triode a signal E = 2√2 cos ωt is applied. If μ = 14 and rp = 10 kΩ then root mean square current flowing through RL = 12 kΩ will be (a.) 1.27 mA (b.) 10 mA (c.) 1.5 mA (d.) 12.4 mA 45(45.)The grid voltage of any triode valve is changed from −1 volt to −3 volt and the mutual conductance is 3 × 10−4 mho. The change in plate circuit current will be (a.) 0.8 mA – 4V PN 300 – 1V P B IP VP 150 O A 50 Volt (mA) 103 Ω 10 V