Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10EE201 Lecture 28/29 P. 1Thevenin Equivalent Circuits with Op Ampsand First Order Op-amp Circuits (RC)Thevenin and Norton Equivalent Circuits for Op AmpsConsider the following op amp circuit. Assuming ideal op amp behavior, find its Norton and Thevenin equivalents.+_200 10V2k+_ vOC+_1k3kAi200i2kEE201 Lecture 28/29 P. 2Step 1: Find the node voltage VA.v+ = v_ = 0 i200 = (10 V / 200 ) = 0.05 Ai2k = [(VA-0) / 2000] = -i200 VA = -100 VStep 2: Find vOC.Using voltage division,vOC = (3k / 4k) (-100 V) = -75 VStep 3: Find iSC.From Ohm’s law, iSC = VA / 1 k = -0.1 AStep 4: Calculate RTH from vOC and iSC.RTH = (vOC / iSC) = 750 EE201 Lecture 28/29 P. 3750 The Norton equivalent network is,0.1 A750 The Thevenin equivalent network is,75 V+_EE201 Lecture 28/29 P. 4First Order Op-amp Circuits (RC)Capacitors can be added to op-amp circuits for increased control of circuit outputs. However, there are subtleties concerning this approach:1. All resistors do not determine time constant2. Op-amp circuits may deviate from ideal response3. Stray capacitances (a capacitance between a conductor and ground) may influence results.EE201 Lecture 28/29 P. 5Example: Find vo(t) for the op amp circuit shown assuming ideal op amp behavior.+_+--+vo(t)icRCiRvsInput characteristics for op-amp v- = v+ = 0Initial condition for capacitor voltagevc(0- ) = vc(0+ ) = vo(0) vo(t) = vc(t ) = vc(0+ ) + ic()d (1)1Ct0ic(t) = - iR(t) (Property of ideal op amp i-=0 )EE201 Lecture 28/29 P. 6xample: Find vout(t) assuming vc(0-) = 0Step 1: Find voltage across capacitor vout(t) = v+Solution for voltage across capacitor vc(t) = vc() +[vc(0+) - vc()] exp{- t/(RTHC)}iC(t) = - vs(t)/ R put this into (1) vo(t) = vc(0+ ) - vs()d0 1 RCt+_+_+vout(t)v+(t)1k1k0.1F5u(t)V_1k+_EE201 Lecture 28/29 P. 7Since i+= 0, resistors and capacitor are in series RTH = 2 kRTHC = 0.2 x 10 -3secvC(0+) = vC(0-) = 0vC() = 5VvC(t) = 5 - 5e-5000t VStep 2: Find current through 1k resistor between capacitor and ground.Differentiate voltage across capacitoriC = C = (0.1F)(-5000)(-5e-5000t)iC(t) = 2.5 x 10-3 e-5000t Ad vC(t) dtEE201 Lecture 28/29 P. 8Step 3: Calculate vout(t) from iC(t) vout(t) = (1k)(iC(t))vout(t) = (103)(2.5x10-3) e-5000tvout(t) = 2.5 e-5000tBecause properties of op amp (v+ = v-) , vout(t) does not depend on load resistorEE201 Lecture 28/29 P. 9Example: Find vout(t) for the op amp circuit shown assuming ideal op amp behavior.+_+--+vout(t)if0.01Fvin(t)i2Step 1: Realize v+= v-Therefore i1= i2= -if i1= i2= vin/2RAlso, v+ =v-=vin/2100k100k100ki1EE201 Lecture 28/29 P. 10Step 2 : Find feedback current from capacitor equation if = C [vout - vin/2 ] = [vout – vin/2 ]Integrating,vout = vin 2ddt-vin2RCddtStep 3: Solve for vout(t) d vout d vin vin dt 2dt 2RC =_ 12RC0tvin() d_ vin 20t_vin() dvout
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