ECE3050AnalogElectronicsQuiz5February 11, 2009Professor Leach Last Name:First Name:Instructions. Print your name in the spaces above. Place a box around any answer. Honor Code Statement:I have neither given nor received help on this quiz. InitialsFor the transistors in this problem, assume the parameters β = 100, gm=1/25 S, rx=0, rπ=2.5kΩ, re=24.75 Ω,and r0= ∞ (an open circuit).rπ=VTIBre=VTIEgm=ICVTr0π= rx+ rπ+(1+β) Rter0=VA+ VCEICr0e=Rtb+ rx1+β+ reric=r0+ r0ekRte1 − αRter0e+ Rte(a) The figure on the left shows the signal circuit of a CE amplifier. It is given that RS=1kΩ, R1=10kΩ,R2=3kΩ,andR3=75Ω. Solve for the small-signal Thévenin equivalent circuit seen looking into its outputterminal. The circuit should be in the form of a voltage source A1vsin series with a resistor rout, where you mustgive numerical values for A1and rout.(b) A load resistor RL= 300 Ω is connected from the output to ground. Use the Thévenin equivalen t circuit foundin part (a) to solv e for the new value of vo.(c) The figure on the right show s the signal circuit of the CE amplifier with a CC stage added between the CEstage and the 300 Ω load resistor. If R4=2kΩ, solve for the new value of vo.(d) Why does the voltage gain increase with the addition of a CC stage which has a voltage gain b y itself that isless than unity ?Over for solutions.1Quiz 5RS1000 R110000 R23000 R375 R42000 RL300β100 gm125rπ2500 re24.75 vs1Part (a)First Solution:r'πrπ1β()R3.r'π1.008 104=vtb1vsR1RSR1.vtb10.909=Rtb1Rp2RSR1,Rtb19.091 102=ib1vtb1Rtb1r'πib18.276 105=i'c1βib1.i'c18.276 103=vo1i'c1R2.vo124.829=routR2rout3 103=Second Solution:r'e1Rtb11βrer'e133.751=i'e1vtb1r'e1R3i'e18.359 103=i'c1β1βi'e1.i'c18.277 103=vo1i'c1R2.vo124.83=Part (b)vo2vo1RLroutRL.vo22.257=Part (c)r'e2rout1βrer'e254.453=vo3vo1Rp2R4RL,r'e2Rp2R4RL,.vo320.542=Part (d)Because the CC stage has a much lower loutput resistance than the CE
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