Title 14. Semiconductors and Electronic Devices - Explanations.pdf ✅

1 (a) Sodium has bcc structure. The distance between body centre and a corner = √3 a 2 = √3 × 4.225 2 = 3.66 Å 2 (c) In reverse biasing negative terminal of the battery is connected to N-side 3 (d) Boron is a trivalent impurity having three valence electrons. When it is introduced to pure silicon, then such type of semiconductors are called p- type or acceptor type semiconductors. 4 (d) In positive half cycle one diode is in forward biasing and other is in reverse biasing while in negative half cycle their polarity reverses, and direction of current is opposite through R for positive and negative half cycles so out put is not rectified. Since R1 and R2 are different hence the peaks during positive half and negative half of the input signal will be different 5 (d) The output Y is a combination of AND + NOT gate. Hence, the truth table is for NAND gate. 6 (a) For first case, the Boolean expression is, Y = A̅ ̅̅̅ ∙ ̅̅B̅̅ = A + B hence for OR gate and for second case, the Boolean expression is Y = A̅ ̅̅̅ ∙ ̅̅B̅̅ = A ∙ B, hence for AND gate. 7 (a) V − d curve near the junction will be as shown by curve (a). 8 (d) 108 electrons enter the emitter in 10−8 s i. e. , iE = 108 × 1.6 × 10−19 10−8 A = 172.8 × 10−11A ∴ 1% of iE is lost in base i. e. , iB = iE 100 ⇒ 99% iE i. e. , 99 100 iE enters the collector ⇒ IC = 0.99 iE Current amplification factor β = iC iB = 0.99iE 0.01iE = 99 9 (c) Hence, (10111)B = 1 × 2 4 + 0 × 2 3 + 1 × 2 2 + 1 × 2 1 + 1 × 2 0 = 16 + 0 + 4 + 2 + 1 = 23 10 (d) The conditions for a circuit to oscillate are (i) the feedback is positive (ii) the fraction of the output voltage feedback ie, β = 1 A ie, the reciprocal of the voltage gain without feedback. 12 (b) Graph between potential and distance in a p-n junction diode is given by ∴ potential at p is less than that at n. 13 (c) The figure shown in the circuit is a half wave rectifier. During the positive half of the input cycle, when junction diode conducts, output across RL will vary in accordance with AC input. During the negative half cycle, junction diode will get reverse biased and hence no output will be obtained across RL. 14 (c) Energy gap, Eg = hc λ λ = hc Eg = 12400 0.72 = 17222 Å 15 (a) At high reverse voltage, the minority charge carriers acquire very high velocities. These by collision break down the covalent bonds, generating more carriers. This mechanism is called Avalanche breakdown 17 (b) μ = rpgm = 50 From ip = KVp 3/2 ⇒ ∆Vp ∆ip = rp = 2ip −1/3 3K2/3 ⇒ gm = μ rp = 3μK 2/3 /ip 1/3 2 = 3 2 μK 2/3 [K 1/3 (Vp + μVg) 1/2 ] = 3 2 μK(Vp + μVg) 1/2 = 75 K (ip/K) 1/3 Because ip was in mA, gm is substituted as 5 m℧ ⇒ 5 = 75K 2/3 ip 1/3 = 75 K 2/3(8) 1/3 ⇒ K = ( 1 30) 3/2 Cut off grid voltage VG = − VP μ = − 300 50 = −6V 18 (d) In forward biasing both electrons and protons move towards the junction and hence the width of depletion region decreases. 20 (c) Distance Potential p n a 3 a a a
A p-type material is electrically neutral. 21 (b) It is used to convert ac into dc (rectifier) 22 (a) At Vg = −3V, Vp = 300 V and Ip = 5mA At Vg = −1V, for constant plate current i. e. ,Ip = 5mA From Ip = 0.125 Vp − 7.5 ⇒ 5 = 0.125 Vp − 7.5 ⇒ Vp = 100V ∴ change in plate voltage ∆Vp = 300 − 100 = 200V Change in grid voltage ∆Vg = −1 − (−3) = 2V So, μ = ∆Vp ∆Vg = 200 2 = 100 23 (d) Radiowaves of constant amplitude can be produced by using oscillator with proper feedback. 24 (c) For an insulator Eg> 3 eV, that is why electron transition from valence band to conduction band is not possible. For semiconductor Eg. is 0.2 eV to 0.3 eV while for metals Eg> 3 eV. 25 (c) Here p − n junction is forward biased. If p − n junction ideal, its resistance is zero. The effective resistance across A and B = 10 × 10 10 + 10 = 5kΩ. Current in the circuit I = V R = 30 15×103 = 2 103 A Current in arm AB = I = 2 103 Potential difference across A and B = 2 103 × 5 × 103 = 10 V. 26 (a) hfe = ( Δic Δib ) Vce = 8.2 8.3 − 8.2 = 82 27 (d) β = Ic Ib > 1 or Ic > Ib. 28 (c) 6C = 1S 2 , 2S 2 2P 2 14Si = 1S 2 , 2S 2 2p 6 , 3S 2 3P 2 31 (b) In superconductors, the current flows on the surface so that it exactly cancels the magnetic field inside the metal. Therefore statement (b) is not correct 32 (c) We know that resistance of conductor is directly proportional to temperature (ie, R ∝ ∆t), while resistance of semiconductor is inversely proportional to temperature (ie, R ∝ 1 ∆t ). Therefore, it is clear that resistance of conductor decreases with decrease in temperature of vice- versa, while in case of semiconductor, resistance increase with decrease in temperature of vice- versa. Since, copper is pure conductor and germanium is a semiconductor hence, due to decrease in temperature, resistance of conductor decreases while that of semiconductor increases. 33 (b) In figure 2, 4 and 5 P-crystals are more positive as compared to N-crystals 34 (a) The potential of P-side is more negative that of N-side, hence diode is in reverse biasing. In reverse biasing it acts as open circuit, hence no current flows 35 (c) In space charge limited region, the plate current is given by Child’s law ip = KVp 3/2 Thus, ip2 ip1 = ( Vp2 Vp1 ) 3/2 = ( 600 150) 3/2 = (4) 3/2 = 8 Or ip2 = ip1 × 8 = 10 × 8 mA = 80 mA 36 (d) Conductor has positive temperature coefficient of resistance but semiconductor has negative temperature coefficient of resistance 37 (c) Rp = Vp ip = 50 150 × 10−3 = 333.3 Ω 38 (a) In n-p-n transistor as CE amplifier ie = ib + ic Therefore, the emitter current is more than the collector current. 40 (b) If V is the voltage across the junction and I is the circuit current, then V + IR = E or I = E R − V R = − V R + E R Slope of load line = − 1 R = 1 1000 = 10−3AV −1 . 41 (c) When a p − n junction is formed, n-side attains positive potential and p-side attains negative. When ends of p and n of a p − n junction are joined by a wire, there will be a steady conventional current from n-side to p-side through the wire and p-side to n-side through the p − n junction. 42 (b) According to Pauli’s exclusion principle, the electronic configuration of number of subshells existing in a shell and number of electrons entering each subshell is found. Hence, on the
basis of Pauli’s exclusion principle, the manifestation of band structure in solids can be explained. 43 (d) Rectifier is n device which converts AC into DC. 44 (a) A = μRL rp+RL = 14×12 10+12 = 84 11 . Peak value of output signal V0 = 84 11 × 2√2V ⇒ Vrms = V0 √2 = 84 × 2 11 V ⇒ r. m. s. value of current through the load = 84 × 2 11 × 12 × 103 A = 1.27 mA 45 (b) By using gm = ∆ip ∆vg ⇒ 3 × 10−4 = ∆ip −1−(−3) ⇒ ∆ip = 6 × 10−4A = 0.6 mA 46 (c) The current through the battery is I = 10V 5Ω + 5Ω = 10V 10Ω = 1A 47 (b) v = 1 2π√LC = 1 2π 1 √ 10 π 2 × 10−3 × 0.04 × 10−6 = 25 kHZ 48 (b) Consider the case when Ge and Si diodes are connected as show in the given figure. Equivalent voltage drop across the combination Ge and Si diode = 0.3 V ⇒ Current i = 12−0.3 5 kΩ = 2.34 mA ∴ Out put voltage V0 = Ri = 5 kΩ × 2.34 mA = 11.7 V Now consider the case when diode connection are reversed. In this case voltage drop across the diode’s combination = 0.7 V ⇒ Current i = 12−0.7 5 kΩ = 2.26 mA ∴ V0 = iR = 2.26 mA × 5 kΩ = 11.3 V Hence charge in the value of V0 = 11.7 − 11.3 = 0.4 V 49 (c) The resistance of semiconductor decreases with the increase in temperature 50 (b) The ripple factor for full wave rectifier is 0.482 which is 48.2%. 52 (d) If we give the following inputs to A and B, then corresponding output is shown in table. A B Y 0 0 1 0 1 0 0 1 1 1 1 1 The above table is similar to OR gate. 55 (a) gm = ∆ip ∆Vg = (20 − 15) × 10−3 (4 − 2) = 2.5 milli mho 56 (d) For Vi < C the diode is reverse biased and hence offer infinite resistance, so circuit would be like as shown in Fig. (2) and Vo = 0. For Vi > 0,the diode is forward biased and circuit would be as shown in Fig. (3) and Vo = Vi . Hence, the optical (d) is correct. 57 (c) The probability of occupation of the highest electron state in valence band at room temperature becomes half according to Fermi distribution. 58 (a) If A = 1, B = 1 and Y = 0, the gate can be NOR gate, NAND gate or exclusive NOR gate (ie, XOR gate). 59 (d) Equivalent circuit can be redrawn as follows i = 10 2 = 5 mA = i2 i1 = 0 60 (c) The potential of N-Side is more negative than that of P-Side, hence diode is in forward biasing from circuit resultant potential is 2V ∴ I = 2 100 = 20 mA 61 (b) (37)10 = (100101)2 ⇒ 6 binary digits 62 (b) Child Langmuir law states that the space charge limited current in a diode valve is given by as I ∝ k V 3/2 [where V is voltage] 63 (a) When p-side of junction diode is connected to positive of battery and n-side to the negative, then junction diode is forward biased. 5V -5 V Vi R L Vo Fig. (1) Vo V R L i Fig. (2) Vi V o Fig. (3) 12k 10 V i1 14k 2k i2 i
In this condition, more number of electrons enter in n-side from battery theryby increasing the number of donors on n-side. 64 (d) GaAs (Eg = 1.5 eV) is used for making infrared LED 65 (c) The given figure is the symbol of NAND gate. 66 (b) In a body centred crystal each atom is in contact with eight atoms at the corners of the simple cubic cell, hence coordination number of bcc crystal is 8. 67 (c) A solid is not transparent to visible light if the value of wavelength of light is greater than the bond length (covalent bonds) between the atom/molecules/ions of material. Conductivity depends upon the number of free charge carriers present in the substance at a given temperature or resistance of that material at that temperature. 68 (c) In the given circuit, diode D1 is reverse biased, so it will not conduct. Diodes D2 and D3 are forward biased, so they will conduct. The corresponding equivalent circuit is as shown in the figure The equivalent resistance of the circuit is Req = (5 + 5) × 20 (5 + 5) + 20 = 10 × 20 10 + 20 = 200 30 = 20 3 Ω Current through the battery, I = 10V 20 3 Ω = 1.5A 69 (c) Eg = hv = hc λ = ( 6.63×10−34×3×108 2480×10−9×1.6×10−19) = 0.5 eV 70 (a) A B C D = ̅A̅̅.̅C̅ E = C̅̅̅.̅B̅ Y 0 0 1 1 1 0 0 1 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 0 73 (a) When diode is forward biased, then there is a small voltage drop across it i = V − V ′ R Given, V = 8 volt, V ′ = 0.5 volt, R = 2.2 k Ω = 2.2 × 103Ω ∴= 8−0.5 2.2×103 = 7.5×103 2.2 = 3.4 × 10− 3 A i = 3.4 mA. 74 (a) Voltage gain, AV = 1000 In dB, voltage gain A = 10 log10 1000 dB = (10 × 3) log10 10 dB = ∵ log10 10 = 1 75 (a) Zener diode has a relatively constant voltage across it, regardless of the value of current through the device. This permits the zener diode to be used as a voltage regulator. 76 (a) From circuit Y = A̿̿̿∙̿̿B̿ = A ∙ B This is an output of an AND gate. 78 (c) According to |Av | = μ 1+ rp RL As RL increases Av also increases When RL becomes too high then Av = maximum = μ Hence only option (c) is correct 79 (d) The ratio of collector current (Ic ) to emitter current (Ie ) is known as current gain (α) of a transistor. Therefore, α = ∆Ic ∆Ie ...(i) Also, emitter current is equal to sum of change of base current and collector current. Therefore, ∆Ie = ∆Ib + ∆Ic ...(ii) From Eqs. (i) and (ii), we get