ch18_solutions - 18-705 CHAPTER 18 18.1a( 29T=Ab= | Av=1b=5...

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Unformatted text preview: 18-705 CHAPTER 18 18.1a( 29T=Ab= | Av=1b=5 | FGE=b( 29A =108620=20000 | T=20000 0.2(29=4000Av=A1+Ab=200001+4000=5.00 | FGE=100%1+Ab=100%4001=0.025%c( 29T=10 0.2(29=2 | Av=A1+Ab=101+2=3.33 | FGE=100%1+2=33.3%18.2a( 29b=R1R1+R2=1k101k=1101b( 29T=Ab=1080201101=99.0 | Av=A1+Ab=104100=10018.3a( 29bs( 29=R1R1+R2=1k101k=1101| T s( 29=Ab=1080201101 =99.0Av= -R2R1Ab1+Ab= -100k1k99100 = -9918.4bs( 29=Z1s( 29Z1s( 29+Z2s( 29=RR+1sC=ss+1RC=ss+104| A=108020=104T s( 29=Ab=104ss+104| Av= -Z2Z1Ab1+Ab= -1sRC104ss+1041+104ss+104= -1RC1s+1Instead of a pole at the origin,the integrator has a low- pass response with a pole a w=1 rad/s.18-706 18.5SAAv=AAvAvAAv=A1+AbAvA=1+Ab(291-Ab1+Ab(292=11+Ab(292SAAv=AA1+Ab11+Ab(292=11+Ab1AbSAAv=11+1050.01(29=11001AvAv=SAAvAA=1100110%=9.99x10-3%18.6AV=A1+Ab=A1+A| From Chapter 12, GE =11+Ab=11+A11+A10-4A9999 | A80 dB18.7-2.0V-1.0V0V1.0V2.0VVS2.0V0V-2.0VOutput VoltageInput VoltageFeedbackNo feedback*Problem 18.7 Figure 18.74 - Class-B Amplifiers VCC 3 0 DC 10 VEE 4 0 DC -10 VI 1 0 DC 0 Q1 3 1 2 NBJT Q2 4 1 2 PBJT RL1 2 0 2K RID 1 7 100K E1 5 0 1 7 5000 RO 5 6 100 Q3 3 6 7 NBJT Q4 4 6 7 PBJT RL2 7 0 2K .MODEL NBJT NPN .MODEL PBJT PNP .OP .DC VS -10 10 .01 .PROBE V(1) V(2) V(7) .END 18-707 18.8Av=A1+Ab| From Chapter 12, GE=11+Ab22451Ab1b=200 | GE2245200A0.002A2000.002=105| A100 dB18.9(a) Series-series feedback (b) Shunt-series feedback (c) Shunt-shunt feedback (d) Series-shunt feedback 18.10(a) Series-shunt feedback (b) Shunt-series feedback (c) Series-series feedback (d) Shunt-shunt feedback 18.11(a) Series-shunt and series-series feedback (b) Shunt-series and shunt-shunt and feedback 18.12(a) Shunt-series and series-series feedback (b) Shunt-shunt and series-shunt feedback 18.13A=108620=20000a( 29Rin=Rid1+Ab(29| For b=1, Rin=40k1+20000(29=800 Mb( 29Rin=Rid1+Ab(29| For b=1, Rin=40k1+20000(29=2.00 c( 29Rout=Ro1+Ab(29| For b=1, Rout=1k1+20000(29=20 Md( 29Rout=Ro1+Ab(29| For b=1, Rout=1k1+20000(29=50.0 m18.14a( 29Av=108620=20000 | Ai=ioii| io=ii40k(29200001kAi=8.00 x105With resistive feedback, the closed - loop gain cannot exceed the open- loop gain.Therefore, Ai8.00 x105.b( 29Atr=iovi=ioii40k(29=Ai40k(29| Atr8x1054x104=20 S18-708 18.15(a) vo+R1vIR2+A-RLRIFeedback Networkb( 29h11A=v1i1v2==15k| h11F=4.3k39k=3.87k| h11T=18.9kh22A=i2v2i1==1k(29-1= 1k(29-1| h22F=39k+4.3k(29-1= 43.3k(29-1| h22T= +1.02mSh21A=i2i1v2== -15k5000(291k= -75,000 | h21F=i2i1v2== -4.3k39k+4.3k= -0.0993h12A=v1v2i1==0 | h12F=v1v2i2==4.3k39k+4.3k=0.0993(c) A=-h21ARI+h11T(29h22T+GL(29=- -75000(291k+15k+3.87k(2915.6k+11k+143.3k=3140b=0.0993(d) Av=31401+3140 0.0993(29=10.0e( 29h21Ah21F| h12Fh12A| Note : Rin=6.22 M, Rout=2.66 (2918.16The circuit topology is identical to Fig. 18.8. The circuit topology is identical to Fig....
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This note was uploaded on 01/17/2011 for the course ECE 3040 taught by Professor Hamblen during the Fall '07 term at Georgia Institute of Technology.

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ch18_solutions - 18-705 CHAPTER 18 18.1a( 29T=Ab= | Av=1b=5...

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