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Unformatted text preview: ECE 2100 Homework 5 Due April 14, 2011 Professor Alyosha Molnar Subjects: Complex power, complex conjugate power matching, transformers, opamp circuits, nonideal opamps. 1) Prelab: Consider the following differentiating Op Amp circuit. a. Derive a relationship between the input voltage and the output voltage. Assume the Op Amp is ideal. Iin=CdV in /dt. Vout=R F Iin = R F CdV in /dt = 1msdV in /dt b. Assume the input voltage is a triangular wave (shown below). What is the output waveform? Explain! Output is a square wave : a triangle wave alternates slope between positive and negative 2Vpp/T, where Vpp is the peaktopeak amplitude and T is the period. Thus, the output will have an alternating sign every half cycle, with equal plus and minus amplitudes. c. Find a relationship between the frequency, f and peaktopeak amplitude, Vpp , of the triangular wave and the peaktopeak voltage of the output waveform. Hint: based on frequency and peaktopeak voltage, what is the slope of the triangle wave Slope = +/2Vpp/T, and frequency f=1/T, so the peaktopeak amplitude of Vout will be 4msVppf (4 instead of 2 because we want the output peakto peak) d. Predict the output voltage peaktopeak value if the input is driven by a 1 volt peaktopeak, 200 Hz sine wave. V in =0.5cos(400 t), Vout = 1msdV in /dt=0. 2 sin(400 t): peaktopeak: 1.256V 10k Vin  + Vout 0.1 F Deleted:  2) Prelab: consider the inverting amplifier shown. a. Assume the opamp is ideal (Zin = , Zout = 0) but with finite gain Av. Write equations for the amplifier gain, input and output impedance as a function of A. As A gets very large, what do these converge to? Because Zout =0, Vout = Av(V1) b V1 = Vout/Av Start with KCL on V1: 0 = (VinV1)/R1+(VoutV1)/R2 b R2Vin = Vout(R2/Av + R1 + R1/Av) Vout = Vin(AvR2/(R1+R2+AvR1)) b gain = Av100/(101+Av) Iin = (VinV1)/R1 = Vin/R1  Vin/(AvR1)(AvR2/(R1+R2+AvR1)) Iin = Vin/R1(1 R2/(R1+R2+AvR1)) = Vin(1+Av)/(R1+R2+AvR1) b Rin = Vin/In = (R1+R2+AvR1)/(1+Av) = (101+Av)/(1+Av) 1k Rout = 0 (looking into an ideal voltage source on output) As Av b gain b100 Rin b 1k b. Now include finite input and output impedances (assume Rin = Rout = 1k ) and rewrite equations for gain, input and output impedance as a function of A. As A gets very large, what do these converge to? Now, KCL again, for two nodes (V1, Vout) To find gain: V1: 0 = (VinV1)/R1+(VoutV1)/R2V1/Rin Vout : 0 = (VoutV1)/R2 + (Vout +AvV1)/Rout First solve for V1 in terms of Vout : V1(RoutR2Av)=Vout(Rout+R2) V1=Vout(Rout+R2)/(RoutR2Av) = Vout101/(1100Av) Now sub in : 0 = (VinV1)/1k +(VoutV1) /100k V1/1k 0 =100Vin201V1+Vout b 100Vin=Vout(201101/(1100Av)1) 100(1100Av)Vin = Vout(203011+100Av) Gain = Vout/Vin = (1100Av)/(Av+203) Input impedance: Iin = (VinV1)/R1 = Vin/1k Vout/1k 101/(1100Av) Vout = Vin*Gain, so In1k = Vin(1(1100Av)/(Av+203) 101/(1100Av))=Vin(1101/(Av+203))...
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This note was uploaded on 06/10/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Spring '05 term at Cornell University (Engineering School).
 Spring '05
 KELLEY/SEYLER
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