AC
Unit-2Multistage Amplifiers Q1) For the circuit shown in below figure find input impendence RI if hie = hre = hoe = 0 and hFe is the same of each of transistor.
A1)Given hie = hre = hoe = 0 and hFe1 = hFe2 = hFe3 = ….. = hFeN = hFe Analysis of QN transistor [CC] : RLN’ = Re Check for approximation : hoeRLN’ 0.1As hoe is zero so all transistor are in valid approximation Now, Ain = 1 + hFe , RiN = hie + (1 + hFe)RLNOr RiN = 0 + (1 + hFe)Re Analysis of QN-1 transistor [CC] : = (1 + hFe)Re ; 1 + hFe , = hie + (1 + hFe)Re Analysis of QN-2 transistor [CC]: Re 1 + hFe , = hie + (1 + hFe)ReHence, QN → (1 + hFe)Re QN-1 → Re QN-2 → ReSimilarly, Q[N-(n-1)] → ReRi1 = Re (Ans)Also, AI = AI1 × AI2 × ……. × AINHence, AI = (1 + hFe)N Q2) Determine VONS for the cascade amplifier shown in figure Use hie = 1 KΩ , hFe = 100 , hre = 0 , hoe = 0
A2)Given hie = 1 KΩ , hFe = 100 , hre = 0 , hoe = 0RC1 = RC2 = 5 KΩQ1 → CEQ2 → CEThe equivalent circuit can be drawn as –
Analysis of Q2 [CE] : Ri2 = 5KΩ hoeRL2 = 0 (approximate analysis)AI2 = = -hfe = -100Ri2 = hie = 1KΩAV2 = = AV2 = Analysis of Q1 [CE] : RL1 = RC1 || Ri2 = 5 K || 1 K = 0.8333 KhoeRL1 = 0 (approximate analysis)AI1 = hFe = -100Ri1 = hie = 1KΩAV1 = = AV = AV1 × AV2AV = (-500) × (-83.33) = 41650 (Ans) Q3) Find the voltage gain AVS of the amplifier shown. Assume hie = 1000Ω , hre = 10-4 , hFe = 50 , hoe = 10-4 A/V
A3)– Given : hie = 1000Ω , hre = 10-4 , hFe = 50 , hoe = 10-4 A/V , RL = 5 KΩQ1 → CE , Q2 → CCFor analysis of amplifier we consider DC aupply voltage as ground and modified circuit is given as –
Analysis of Q2 [CC] : hoeRL’ = 10-4 × 5 × 103 = 0.5 > 0.1 approximate analysis is not valid, therefore exact analysis.AI2 = = = = 34Ri2 = hie + AI2RL’ = 1 + 34 × 5 = 171 KΩAV2 = = 0.994Analysis of Q1 [CE] : RL1 = 10 || Ri2 = 9.447 KΩhoeRL1 = 10-4 × 9.447 × 103 = 0.9447 > 0.1 so exact analysisAI1 = = -25.71Ri1 = hie + hreAI1RL1 = 1 + 10-4 × (-25.71) × 9.447 = 975.71 ΩAV1 = = AV = But Vi2 = VO1AV = AV1 × AV2AV = (-249.11) × (0.994) = -247.615 AVS-247.615 × 0.66101 = -163.675 (Ans.)AI = 25.71AI2 = IO = Ie2 , Ib1 = IiAI = (-1) × (25.71) ×
-0.05524AI = -25.71 × 34 × 0.05524 = -48.288 (Ans) Q4) For the circuit shown, compute AI , AV , AVS , RI and RO’
A4) Given: Re = 0.1 KΩ , RL = 3 KΩ RS = 10 KΩ Q1 → CC , Q2 → CEAC analysis –
Analysis of Q2 [CE] : hoe(Re + RL) = (0.1 + 3) = 0.0775 < 0.1 approximate analysisAI2 = = -hFe = -50Ri2 = hie + (1 + hFe)Re = 1.1 + 51 × 0.1 = 6.2 KΩAV2 = = -24.193Analysis of Q1 [CE] : hoe(Re + RL’) RL’ = Ri2 || 8 K= 3.492 K(0.5 + 3.492) = 0.0998 < 0.1So approximate analysisAI1 = = -50Ri1 = hie + (1 + hFe)Re = 1.1 + 51 × 0.5 = 26.6 KΩ (Ans)AV1 = = Hence, AV = AV1 × AV2 = 158.8 (Ans)AVS
= AVS = 158.8 × 0.7261 = 115.41 (Ans)AI = AI1 = 50AI2 = AI = (-1) × (50) ×
-0.5633AI = 50 × 50 × 0.5633 = 1408.25 (Ans)RO1 = ∞ , RS2 = 8 || ∞ = 8 KΩRO2 =∞ , RO’ = ∞ || 3 K = 3 K (Ans) Q5) Calculate AI , AV , AVS and Ri
A5) Given: RS = 1 KΩ , RL = 5 KΩAmplifier configuration: cascade amplifier [CE – CB ]Analysis of Q2 [CB] :hoeRL 0.1 2.45 × 10-3 < 0.1 which is less than 0.1Approximate analysis:AI2 = = -hFb = 0.98Ri2 = hib = 21.6 ΩAV2 = = 226.85Analysis of Q1 [CE] : RL1 = Ri2 = 21.6 ΩChecking approximation:hoeRL1 0.1 = 5.4 × 10-4 < 0.1 , which is less than 0.1So approximate analysisAI1 = = -50Ri1 = hie = 1.1 KΩ (Ans)AV1 = = Hence, AV = AV1 × AV2 = -222.723 (Ans)AVS = 0.5238 × (-222.732) = -116.66 (Ans) [ since can be given as ]AVS = 158.8 × 0.7261 = 115.41 (Ans)AI = AI1 × AI2 = -50 × 0.98 = -49 (Ans) Q6) For the circuit shown in figure. Calculate the values of Ri , AI , AV and Ro. Assume the following h-parameters for both the transistors. Hie = 1.1 KΩ , hFe = 50 , hre = 2.5 × 10-4 , hoe = 25 μA/V
A6) Given: hie = 1.1 KΩ , hFe = 50 , hre = 2.5 × 10-4 , hoe = 25 μA/V , RS = 1 KΩ , RE = 1 ΩAmplifier configuration:Q1 → CC , Q2 → CCAnalysis of second stage :Checking approximation : hoeRc2 0.1Let us first analyse the second stage. For this stage,Analysis of Q2 [CE] : hoeRL2 = 1 × 103 × 25 × 10-6 = 25 × 10-3 [ RC = RE = 1] which is less than 0.1approximate analysis:a) Current gain of second stage (AI2):AI2 1+ hFe = 1 + 50 = 51 ----------(1)b) Input resistance Ri2 = hie + (1 + hFe)RE = 1.1 K + (51 × 1 K) Ri2 = 52.1 KΩ -----------(2)c) Voltage gain (AV2) :AV2 = = 1 - 0.978 -----------(3)Analysis of the first stage :The load resistance for the first stage is the input resistance of the second stage Ri2hoeRi2 = 25 × 10-6 × 52.1 × 10-3 = 1.3As hoeRi2 > 0.1 , we cannot uswe the approximate analysis. Hence we will have to use the exact analysis.a) Current gain (AI1) :We can write the exact equation for the current gain as, AI1 = Substituting the values, we get,AI1 = = 22.41 ----------(4)b) Input resistance (RI1):Ri1 = hie + AI1Ri2 = 1.1 K + (22.41 × 52.1 K) = 1.169 MΩ ----------(5)c) Voltage gain (AV1):AV1 = = 1 - 0.999 Ω ----------(6)d) Output resistance (RO1):RO1 = Ω ------------(7)e) Output resistance of second stage (RO2):RO2 = Substituting the values, we get,RO2 = = 22.38 Ω -------------(8)Combine the result of analysis of stage 1&2:Overall voltage gain (AV): AV = AV1 × AV2 = 0.978 × 0.999 = 0.977 Ans.Overall current gain (AI): AI = AI1 × AI2 = 22.41 × 51 = 1142.9 Ans.Overall input resistance (Ri) = Ri1 = 1.169 MΩ Ans.Overall output resistance (RO) = RO2 = 22.38 Ω Ans. Q7). For the two-stage cascade, shown calculate AI , AV , Ri and RO’ .
A7) Given: RB = R1 = 100 KΩ, RL = 4 KΩ Re = 0.1 KAmplifier configuration: Q1 → CE, Q2 → CC with ReAC analysis: For a.c. analysis of amplifier we consider the d.c. supply as ground and apply Miller’s theorem across feedback resistance (RB = 100 KΩ) then modified circuit is given as:
Analysis of Q2 [CC]:Check for approximation, hoeR1 0.1 hoeRL = × 4 = 0.1 Hence, valid approximation Approximate analysis:Current gain:AI2 = = 1 + hFe = 1 + 50 = 51Input Impendence:Ri2 = hie + (1 + hFe)RL = 1.1 + 51 × 4 = 205.1 KΩVoltage gain:AV2 = = 0.995Analysis of Q1 [CE with Re]: hoe(Re + RL’) RL’ = || 2 || Ri2|AV1| >> 1 for CE, << 1, 1RL1 = 100 || 2 || 205.1 = 1.942 KΩCheck for approximation hoe(RL’ + Re) 0.1 = 0.051 < 0.1Hence, valid approximation Approximate analysis:Current gain:AI1 = = -50Ri1 = hie + (1 + hFe)Re = 1.1 + 51 × 0.1 = 6.2 KΩ Voltage gain :AV1 = = AV = AV1 × AV2 = -15.661 × 0.995 = - 15.583 (Ans)Input Impendence:
Ri = || Ri1 = 6.03 K || 6.2 K = 3.05 KΩ (Ans)Now, RO’ = source resistance for Q2 & RO1 = ∞RB’ = ∞ || || 2K = 1.958 KΩFrom the fig. (a) :RO’ = RO2 || 4 K = 0.060 || 4 = 0.059 KΩ (Ans)Overall current gain :AI =
And AI1 = 50AI2 = AI = (-1) × (51) × From fig. (b) : From fig. (c) : AI = 51 × (- 0.009) × 50 × 0.492 = -11.29 Ans. Q8) For the circuit shown, find AV , AVS , AI = , Ri and RO
A8) Given RC = 100 KΩ , Ri = 470 KΩ , R1 = 40 KΩ , R2 = 100 KΩAmplifier configuration: Q1 → CC , Q2 → CC with ReAC analysis : For a.c. analysis of amplifier we consider the d.c. supply voltage as ground and all capacitor as short circuit and apply Miller’s theorem across 470K feedback resistance then modified circuit is given as :
Let us assume AV = 0.96 [Since in CC-CC configuration AV 1]R2’ = 1K || 40K || 100K || 100K || = 1K || 40K || 50K || 11280K = 0.956 KΩAnalysis of Q2 [CC] :Check for approximation, hoeRL’ 0.10.0239 0.1 , which is less than 0.1Approximate analysis :AI2 = = 1 + hFe = 51Ri2 = hie + (1 + hFe)RL’ = 1.1 + 51 × 0.956 = 49.856 KΩAV2 = = 0.9779Analysis of Q1 [CC] :RL1 = Ri = 49.856 KΩChecking approximation: hoeRL1 0.1 which is greater than 0.1Exact analysis:AI1 = Ri1 = hie + AI1RI1 = 1.1 + 22.70299 × 49.856 = 1132.98 KΩ Ans. AV1 = = AV = AV1 × AV2 = 0.976 Ans.R = 19583.33 KΩFrom Fig. (a):Ri = R || Ri = 19583.33 K || 1132.98 K = 1071.017 KΩ
AVS = = 0.976 × = 0.8927 Ans.AI = AI
AI = AI = 1047.30 RS1 = 100 K || 19583.33 K = 99.491 KΩRO1 = , VO1 = hOe + 0.532 mʊRO1 = 1.8796 KΩ RS2 = RO1 = 1.8796 KΩRO2 = KΩRO’ = RO2 || RL’ = 0.0584 || 0.956 = 55 Ω Ans. Q9) For the two-stage cascade shown, find AI , AV , Ri and RO’.
A9) Given: RB = 100 KΩ , Re = 0.05 KΩ , RL = 2 KΩ Amplifier configuration: Q2 → CE with Re , Q1 → CE with ReAC analysis: For a.c. analysis of amplifier we consider the d.c. supply as ground and apply Miller’s theorem across (RB = 100 KΩ) then modified circuit is given as :
Analysis of Q2 [CE with Re] :Check for approximation, hoe(Re+RL) 0.1 hoe(Re + RL) = ×[0.5 + 2] = 0.05125 < 0.1Approximate analysis :AI2 = = - hFe = -50 Ri2 = hie + (1 + hFe)Re = 1.1 + 51 × 0.05 = 3.65 KΩAV2 = = -27.397Analysis of Q1 [CE with Re] : RL’ = || 10 || Ri2Assume |AV1| = 1 for CE, << 1 , 1RL’ = 100 || 10 || 3.65 = 2.604 KΩCheck for approximation hoe(RL’ + Re) 0.1hoe(RL’ + Re) = = 0.078 < 0.1 Approximate analysis :AI1 = = -50Ri1 = hie + (1 + hFe)Re = 1.1 + 51 × 0.5 = 26.6 KΩ AV1 = = AV = AV1 × AV2 = (-4.895) × (-27.397) = 134.108 (Ans)Ri = || Ri1 = 16.964 || 26.6 = 10.358 KΩ (Ans)
AI = AI1 = 50AI2 = AI = (-1) × (50) × From fig. (b) ,
From fig. (d) : AI = 50 × 0.71 × 50 × 0.389 = 690.475 Ans.RO2 = ∞ , RO’ = 2 || ∞ = 2 KΩ (Ans) Q10) Three cascaded stages have an overall upper 3dB frequency of 16kHz and a lower 3dB frequency of 25 HZ. What are the value of FL and FH of each stage? Assume that all the stages are identical Also calculate the bandwidth of each stage A10) Given: FL(n) = 25 HZ , FH(n) = 16 KHZ n = 3 Lower 3dB frequency, FL for each stage: Therefore FL(n) FL = FL (n) × = 25 × = 12.75 HZUpper 3dB frequency FH for each stage: Therefore FH(n) = FH FH= FH= = 31.38 KHZBandwidth of each stage: BW = FH – FL =31.38 KHZ – 12.75HZ= 31.37 KHZ
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