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