2017 WORKBOOK Detailed Explanations of Try Yourself Questions Electrical Engineering Analog Circuits www.EngineeringBooksPdf.com Diode Applications T3 T1 :: Solution Solution (a) In this question we need to determine which diode is on and which diode is OFF, clearly diode D3 is OFF because if it is on then current from current source will flow from n to p terminal of the diode D3 and this is not possible, hence D3 is OFF Applying the same concept, we can say diode D2 is also OFF Diode D1 is on because it is forced by the battery of 10 V T2 : Solution (c) Assume D1on, D2off, D3on 10 V 0.6 V 10 kΩ 10 kΩ 0.6 V D2 off 10 kΩ – 10 V – 20 V ID1 = 10 0.6 (20) = 1.47 mA 20 k ID2 = ID3 = 0.6(10) = 0.94 mA 10k www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright Workbook (c) Since all the voltage are positive all the diodes will try to be forward biased but only the diode with highest voltage will be switched on rest will be in off state T4 : Solution (b) + – Vin V0 kΩ kΩ Vin kΩ – V0 kΩ kΩ kΩ + For negative cycle of input For positive cycle of input Thus the Vin will appear across the series combination of the two kΩ resistors and we are taking output across kΩ resistance only hence the output will be reduced by 50 % and the above circuit will work as a full wave rectifier with an attenuation of 1/2 T5 : Solution (b) Since there is a D.C level shift in the output waveform thus the circuit must be a clamper circuit and when the diode is conducting then the voltage at the output must be V as seen from the output waveform hence option (b) © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com Bipolar Junction Transistor T1 : Solution (c) I 0.6 × 10−3 (β + 1) = CEO = = 200 ICBO × 10−6 ∵ ∴ β = 199 T2 : Solution (c) – + (R.B) (R.B) 9V 6V (F.B) (R.B) – – 6V 1V (F.B) (Active) – – – + 1V 1V (F.B) (R.B) 9V 2V + + – + + + (F.B) (Saturation) – (Active) + (Cut off) T3 : Solution (d) ∵ IB = then only emitter to collector current will flow ∴ ICEO = (β + 1)ICBO = 101 × 15 × 10–6 = 1515 μA = 1.515 mA www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright Workbook T4 : Solution (c) If base length > length of diffusion then the carriers will not enter the collector T5 : Solution IC = βIC + (β + 1)ICO Now, I 0.6 × 10 −3 = 200 β + = CEO = ICBO × 10 −6 ∴ β = 199 ∴ I C = 199(10 μA) + (1 + 199) × × 10 –6 = 2.59 × 10 –3 Amp © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com BJT Biasing T1 : Solution VCC – IC1R2 – VCE1 = – 1.5 mA × R2 – = R2 = kΩ IC1 1.5mA = = 0.01 mA β1 150 IB1 = IB2 will be equal to IB1 as there is no change in R1 IC2 = β2 IB2 = 200 × 0.01 mA = mA VCE2 = VCC – IC2R2 = – mA × 2kΩ = V The new operating point is Q(2 V, mA) T2 : Solution Assume Q is in active region kΩ 50 kΩ 10V 5V kΩ I B active IBactive = = 0.7 50k + 101× 2k 4.3 = 17 μA 252 www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright Workbook ICactive = β IBactive = 100 × 17 μA = 1.7 mA KVL in output loop, 10 V ICsat kΩ VCsat = 2.5 V VCEsat = 0.2 V VE = kΩ × 1.7 mA 2Ω VCsat = 3.6 V VCEsat = 0.2 V VE = kΩ × 1.7 mA = 3.4 V 10 3.6 = 2.13 mA 3kΩ > ICactive ICsat = ICsat (Active region) T3 : Solution Since I1 = 0.2 mA and I2 = 0.3 mA So n1 = and n2 = 3, because we need to find minimum number of BJT required © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com Small Signal Analysis of BJT T1 : Solution (d) 270 Ω Vi RL V Th gm = mS ; ro = 250 kΩ rπ = βre = β 100 = = 50 kΩ gm mS ro 270 Ω E C + IE Vi – rπ + Vπ IB – g m vπ + VTH RL – B Vπ = v i × rπ 50 k = v = – 0.994 vi rπ + 270 50 k + 270 i VTh + rogmvπ – vπ = VTh = –rogmvπ – vπ = –(1 + gmro) (–0.994vi) = – (1 + 2mS × 250 kΩ) × 0.994vi VTh = 497.9 vi www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright Workbook T2 : Solution (d) 12 V 3.9 k 220 k C2 C1 Vi DC circuit: 12 V 3.9 k (1 + β)IB 220 k IB IB = + VBE – 12 0.7 = 0.0163 mA (1 + β)3.9 k + 220 k IE = (1 + β)IB = 1.97 mA 26 mV re = VI = = 13.15 Ω 1.97 mA IE Vi 220 k – Av Vo C B βre 1.578 kΩ βIB 220 kAV Av – 3.9 k = 220 k Vo = – (220 k ⎥⎥ 3.9 k) βIB Av = 3.83 k RC P RL = = –291.41 13.15 re Zi = 220 k Vi = P βre = 0.752 k ⎥⎥ 1.578 k = 0.509 kΩ = 509.4 Ω 1A v Ii © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com FET Biasing T1 : Solution (a) Ad VD = VG ∴ we conclude that each MOSFET is in saturation 10 v ID = kn1 (VGS – VT)2 MOSFET M1 M1 ID = kn1 (VGS1 – VT)2 v1 VGS1 = 10 – = v ⎛W ⎞ 0.5 mA = 36μ × ⎜⎝ ⎟⎠ × (5 – 1)2 L M2 v2 M3 ⎛W ⎞ ⎜⎝ ⎟⎠ = 1.73 L MOSFET M2 ID = kn2 (VGS2 – VT)2 0.5 mA = 36μ × ⎛W ⎞ ⎜ ⎟ ⎝L⎠ (3 – 1)2 ⎛W ⎞ ⎜⎝ ⎟⎠ = 6.94 L MOSFET M3 ID = kn3 (VGS3 – VT)2 0.5 mA = 36μ × ⎛W ⎞ ⎜ ⎟ ⎝L⎠ (2 – 1)2 ⎛W ⎞ ⎜⎝ ⎟⎠ = 27.8 L www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright 11 Workbook T2 : Solution (c) If VTH = 0.4 v PMOS in depletion mode VS = 1.5 V 1.5 V VD = VG = 0.5 V + – + – 0.5 V VSD > VSG + VTH → current saturation VSD < VSG + VTH → Triode region VSD = VS – VG = 1.5 – 0.5 = v VSD = VS – VD = 1.5 – = 1.5 v 1.5 > V + 0.4 current saturation region VS = 0.9 V 0.9 V VD = 0.9 V + – VG = + – 0.9 V PMOS in depletion mode VSD = VS – VD = 0.9 – 0.9 = VSG = VS – VG = 0.9 – = 0.9 < 0.9 + 0.4 triode region T3 : Solution 3v (b) Given: VTH = V So MOSFET is an n channel enhancement MOSFET in both transistors VD = VG vD1 M1 I1 M1 and M2 are in current saturation v2 – VDS1 – VDS2 = M2 VDS1 + VDS2 = V VDS2 = = 1.5 v (VDS1 = VDS2) VGS2 = VDS2 = 1.5 v = V2 VGS1 = VDS1 = 1.5 v I1 = ID1 = ID2 = 1 W 2 μn Cox (VGS − VTH ) = × 20 × (1.5 − 1) μA = 7.5 μ A L © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com Small Signal Analysis of FET T1 : Solution (b) It is common drain amplifier Av = gm R s gm 4kΩ = = 0.95 + g m Rs + gm 4kΩ gm = 4.75 m gm = kn (VGS – VT) ⎛ ID ⎞ = kn ⎜ k + VT − VT ⎟ ⎝ n ⎠ gm = ID kn gm = ID × ⎛W ⎞ μn Cox ⎜ ⎟ ⎝L⎠ W = 47 L www.EngineeringBooksPdf.com Multistage Amplifiers T1 : Solution (b) The input resistance will be VCC Vin β1 = 50 β2 = 100 Rin Vout R′ kΩ R ′ = rπ + (β2 + 1)RE = k + (101)(1k) = 102 kΩ Rin = rπ + (β1 + 1)R ′ = k + (51) (102 k) = 5.203 MΩ © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com Op-Amps and 555 Timer T1 : Solution (b) Output of op-amp R V1 – Vin V + R – R + R It is connected to schmitt trigger (inverting mode) → clockwise But inverting amplifier + inverting schmitt trigger → anticlockwise 12 –6 www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright 15 Workbook T2 : Solution (b) Rif = Ri Ri = + Aβ A β Ab >> + Vf – I f = 10 kΩ kΩ – vin v0 + + 20 kΩ Voltage shunt: β= Vf = –1 V0 kΩ – I β= f = − V0 10k vin β = 10k Rif = Ri 10k = A β 105 × = + v0 + 10 kΩ – Vf 10 kΩ + 20 kΩ If 10 × 10 × 106 105 10k R i f = kΩ T3 : Solution (b) Redrawing the circuit by replacing amplifier with its block diagram from the given properties Ri = ∞ ; R0 = ; voltage gain = AV iin = V − [1 − Av ] V − Av Vin iin = in = in Rf Rf Rin = © Copyright Rf Vin − V0 Rf iin vin + + v0 = AV vin AV vin Vin Rf = iin − Av www.madeeasypublications.org www.EngineeringBooksPdf.com Electrical Engineering • Analog Circuits 16 T4 : Solution (b) i3 10 kΩ i3 10 kΩ vin i2 – C + – + B 10 kΩ D A 10 kΩ i1 Rf – E + io RL From the circuit, VE = io RL VE = VA i1 = i2 = i3 If we apply KVL between node B and C, ∴ VB = VC i1 = i2 = i3 = ∴ ∵ (Virtual short concept) v in 20 kΩ VC – VD = i3 × 10 kΩ = and (Virtual short concept) v in v in ⇒ VD – VE = –vin VA – VB = i1 × 10 kΩ = VB = VC io = −v in Rf T5 : Solution (a) The duty cycle of the above astable multivibrator (designed using 555 timer) is ∵ R A + RB Ton = R A + 2RB T Thus duty cycle > 50% www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright 10 Negative Feedback Amplifiers and Oscillators T1 : Solution (a) The overall forward gain is 1000 and close loop gain is 100 Thus, β = 0.009 Now, when gain of each stage increase by 10% then overall forward gain will be 1331 and using the previous value of β the close loop will be 102.55 ⇒ Close loop Voltage gain increase by 2.55% T2 : Solution (b) The feedback element is Rf it samples voltage and mix current so shunt-shunt feedback T3 : Solution (a) The output can be ±12 V only, when output is 12 V then R1 R2 C – Vo + kΩ P 10 kΩ 10 kΩ So, Vp = V when output is –12 V then â Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com Electrical Engineering ã Analog Circuits 18 R1 R2 C – Vo + kΩ P 10 kΩ 10 kΩ So, Vp = –10 V T4 : Solution (a) Since their are capacitors the maximum phase shift that can be provided will be 270° but due to the presence of the RC circuit the phase shift is equal to 60° for the individual RC circuit, making the phase shift of the feedback network equal to 180° Thus the amplifier should be an inverting amplifier so that it can be a positive feedback circuit and because the amplifier is a practical amplifier thus ⎥ Aβ⎥ > for the circuit to work www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright ... = AV vin AV vin Vin Rf = iin − Av www.madeeasypublications.org www.EngineeringBooksPdf.com Electrical Engineering • Analog Circuits 16 T4 : Solution (b) i3 10 kΩ i3 10 kΩ vin i2 – C + – + B 10... when output is –12 V then © Copyright www.madeeasypublications.org www.EngineeringBooksPdf.com Electrical Engineering • Analog Circuits 18 R1 R2 C – Vo + kΩ P 10 kΩ 10 kΩ So, Vp = –10 V T4 : Solution... flow ∴ ICEO = (β + 1)ICBO = 101 × 15 × 10–6 = 1515 μA = 1.515 mA www.madeeasypublications.org www.EngineeringBooksPdf.com © Copyright Workbook T4 : Solution (c) If base length > length of diffusion