Back-of-EnvelopeInspectionMethodAssumethen ValidateMethodIdeal Operational Amplifier (Op Amp)10i =20i =3()ivv= F3,0isatiE fvv=>_+vii1+_v1v2i2v3i31234invertingterminalnon-invertingterminaloutputterminalThe nonlinear relationship Fdepends only on 21ivvv−F (vi) is defined by:3,0sat iifv E v=− <3,0sat sat iE vE v if−<< =satEsatE−3v21ivvv−Esatis called the saturation voltage. Typically 15 .satE V≤0()3sgnsat iv E v=+_ValidatingInequalityop ampEquationop ampCircuitModel- Saturation Region+ Saturation RegionLinear RegionOperatingRegion_+vi =0ia+_ibvoabcd_+_+vi > 0iaibvoabcd_+_+vi < 0iaibvoabcd_+0ai =0bi =0iv =sat o satEvE−<<0iv > 0iv<0ai=0bi=osatvE=0ai=0bi=osatvE=−satEsatE0∞OP AMP in Linear Regionebib= 0+_+_i0v0bia= 0ib= 0vi eb–ea = 0≡eaia= 0a_+vi= 0ebib= 0+_+_i0v0beaia= 0a_+vi= 00∞Remarki0and v0can assume any valuedetermined by the external circuitOP AMP Element Laws(Linear Region)Step 1. Assume Op Amp RegimeSimple Op Amp Circuit Analysis :Step 2. Solve the resulting linear resistive circuitStep 3. Derive the dynamic range of the Op Amp’s assumed operating regime3 Common StepsReplace Op Amp by circuit model for the assumed regime.3 typical problems:• find operating point• derive and sketch DP plot• derive and sketch TC plot Derive the op-amp regime validation variable (vi,for linear regime,and vo, for ± saturation regimes) and apply the regime validationinequality.5 Analysis Steps within each Regime1. Express the current of each resistor in terms of its node-to-datum voltages, via Ohms law and KVL.2. Apply KCL to all nodes except the ground (datum) node, andthe op amp output node.3. Solve for the desired variable.4. Solve for the op amp regime validation variable (vi, for linear regime, vi, for ± saturation regimes) as a function of the driving-point voltage (or current).5. Apply the appropriate regime validation inequality and calculate thedynamic range in terms of the driving-point voltage v (i.e., va≤ v ≤ vb), or in terms of the driving-point current i (i.e., ia≤ i ≤ ib).Find Output Voltage vofor Op amp operating in the Linear Region+-R1= 5 Kvo_+R2= 10 K10+-R1= 5 KvoR2= 10 K10Replace Op amp by Model in Linear Regime0∞+-R1= 5 K+vovi_R2= 10 K+_vivo021oiRvvR=−12satRER−12satREREsat-EsatValidationFind a dynamic range of the driving-pointvoltage viwhere op amp is operatingin the assumed linear regime.Validation Inequality:- Esat< v0< Esatsince21oiRvvR=−21sat i satREvER−<− <Right inequality :12isatRvER>−Left inequality :12isatRvER<CCCSVCVSCCVSVCCSOp amp circuit realization of linear controlled sources+_A=∞v1+_R+_v2i1i2v1+_+_v2i1i2v1+_R1i1v1+_+_v2i1i2R+_v2i2R2R2+_A=∞+_A=∞+_A=∞R11gR10i =21igv=:dynamic range212sat satvE vvE−<<+rR−10v =21vri=:dynamic range1sat satEEiRR−<<211RRµ+10i =21vvµ=:dynamic range11112 12sat satRREvERR RR −<< ++ 121RRα+10v =21iiα=:dynamic range22111sat satvE vEiRR−+<<−+viR+−0satEE−ERivDP Plot of an op-amp VCCS CircuitLinear Region0i−=0i+=−0iv =+1: Derive an equation relating and viStep 1across terminals Rii=RRvERi==0EEivRR==+i(1)(2)(3)This equation describes the Norton equivalent circuit across terminals0,eq scEGiR∴==(4)+ov2431−satEE+Linear RegionE−+viR+−1+ov2431−ERi+−RRvRi=1 3Substitute (2) into (1) :1 3(),eq sciGvi=+(DP plot in the linear region) :: : sat o satValidation E v E−<<Step 2(5)(6)(7)Apply KVL around closed node sequence1 3 42:0oivv Ev+−−=ovEv⇒=−impliessat o satEvE−<<sat satEEvEE−− << +0i−=0i+=−0iv =+−+viR+−1+ov2431−ERi+−RRvRi=To test the above condition for the op amp to be operating in the linear region, we must derive an equation relating voand v :1Hence,−+viR+−+ Saturation Region0i−=0i+=−0iv >+1: Derive an equation relating and viStep 1satvRiE=−(8)+ov2431−E−+viR1+ov2431−ERi+−RRvRi=0satEE−ERivsatEE+Linear RegionsatE−R1satEApply KVL around closed-node sequence1 2 43:1: : 0iValidation v >Step 2(9)(10)Apply KVL around closed node sequence1 2 43:0isatvv EE−− + − =isatvvEE⇒=−+−0iv >satvEE<−To test the above condition for the op amp to be operating in the + saturationregion, we must derive an equation relating viand v :1Hence, implies(DP plot in the + saturation region) :−+viR+−- Saturation Region0i−=0i+=−0iv >+1: Derive an equation relating and viStep 1satvRiE=+(11)+ov2431−E−+viR1+ov2431−ERi+−RRvRi=0satEE−ERivsatEE+Linear RegionsatE−R1satEApply KVL around closed-node sequence1 2 43:1: : 0iValidation v <Step 2(12)(13)Apply KVL around closed node sequence1 2 43:0isatvv EE−− + + =isatvvEE⇒=−++0iv <satvEE>+To test the above condition for the op amp to be operating in the - saturationregion, we must derive an equation relating viand v :1Hence, impliesR1(DP plot in the - saturation region) :(16)Upon defining122110iiGv==2sat satEEvEE−− < < +The above op-amp circuit functions as a VCCS defined byIt is valid over the voltage
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