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Homework 5_soln

Homework 5_soln - ECE 2100 Homework 5 Due Professor Alyosha...

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ECE 2100 Homework 5 Due April 14, 2011 Professor Alyosha Molnar Subjects: Complex power, complex conjugate power matching, transformers, op-amp circuits, non-ideal op-amps. 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=C·dV in /dt. Vout=-R F Iin = -R F C·dV in /dt = -1ms·dV 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 peak-to-peak 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 peak-to-peak amplitude, Vpp , of the triangular wave and the peak-to-peak voltage of the output waveform. Hint: based on frequency and peak-to-peak voltage, what is the slope of the triangle wave Slope = +/-2Vpp/T, and frequency f=1/T, so the peak-to-peak amplitude of Vout will be 4ms·Vppf (4 instead of 2 because we want the output peak-to- peak) d. Predict the output voltage peak-to-peak value if the input is driven by a 1 volt peak-to-peak, 200 Hz sine wave. V in =0.5cos(400 π t), Vout = -1ms·dV in /dt=0. 2 π sin(400 π t): peak-to-peak: 1.256V 10k Vin - + Vout 0.1 μ F Deleted: -

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2) Prelab: consider the inverting amplifier shown. a. Assume the op-amp 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) barb2right V1 = -Vout/Av Start with KCL on V1: 0 = (Vin-V1)/R1+(Vout-V1)/R2 barb2right R2·Vin = -Vout(R2/Av + R1 + R1/Av) Vout = -Vin(Av·R2/(R1+R2+Av·R1)) barb2right gain = -Av·100/(101+Av) Iin = (Vin-V1)/R1 = Vin/R1 - Vin/(AvR1)·(Av·R2/(R1+R2+Av·R1)) Iin = Vin/R1(1- R2/(R1+R2+Av·R1)) = Vin(1+Av)/(R1+R2+Av·R1) barb2right Rin = Vin/In = (R1+R2+Av·R1)/(1+Av) = (101+Av)/(1+Av) ·1k Rout = 0 (looking into an ideal voltage source on output) As Av barb2right gain barb2right -100 Rin barb2right 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 = (Vin-V1)/R1+(Vout-V1)/R2-V1/Rin Vout : 0 = (Vout-V1)/R2 + (Vout +AvV1)/Rout First solve for V1 in terms of Vout : V1(Rout-R2Av)=Vout(Rout+R2)
V1=Vout(Rout+R2)/(Rout-R2Av) = Vout·101/(1-100Av) Now sub in : 0 = (Vin-V1)/1k +(Vout-V1) /100k -V1/1k 0 =100Vin-201V1+Vout barb2right 100Vin=Vout(201·101/(1-100Av)-1) 100(1-100Av)Vin = Vout(20301-1+100Av) Gain = Vout/Vin = (1-100Av)/(Av+203) Input impedance: Iin = (Vin-V1)/R1 = Vin/1k – Vout/1k ·101/(1-100Av) Vout = Vin*Gain, so In·1k = Vin(1-(1-100Av)/(Av+203) ·101/(1-100Av))=Vin(1-101/(Av+203)) Vin/In = 1k /(1-101/(Av+203))= 1k (Av+203)/(Av+102)=Zin Now Rout: test by injecting a current into Vout, holding Vin = 0 So rewrite KCL: V1: 0 = (-V1)/R1+(Vout-V1)/R2-V1/Rin Vout : 0 = (Vout-V1)/R2 + (Vout +AvV1)/Rout-Iout Now use first equation to find V1 in terms of Vout :

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