6.301 Solid-State CircuitsRecitation 15: Op-Amp Non-IdealitiesProf. Joel L. DawsonIt will come as no surprise to you that op-amps are not perfect, and that these imperfections willimpact circuit performance to a certain degree. The challenge for a designer who is using an op-ampis to figure out which performance metrics are most critical for his/her application.An almost complete list of non-idealities for op-amps is given below:Input Offset VoltageV0 SInput Bias CurrentIIBInput Offset CurrentI0 SFinite GainA0Common Mode Rejection RatioCMRRPower Supply Rejection RatioPSRRFinite Gain-Bandwidth Productωµ2πOutput Slew RateSRInput ResistanceRINOutput ResistanceR0Let’s start with offset voltage. An offset refers to the fact that when you ground the input of a DCamplifier you do not get a zero voltage at the output.v0≠ 0A6.301 Solid-State CircuitsRecitation 14: Op-Amp Non-IdealitiesProf. Joel L. DawsonPage 2We model this in ways that you might expect. If, when we ground the input of this DC amp, wemeasure AV output voltage of V0 S, we can adjust our diagram according to:In this case, we speak of V0 S as the “offset referred to the output.” That is to differentiate it from the“offset referred to the input,” as follows:Many times, it will be more convenient analytically to choose one form over the other. For instance,suppose we were interested in the effect of an offset on our familiar inverting op-amp amplifier.V0 SΣ++AV0 SAΣ++Av0+−+−V0 SR2R1vI6.301 Solid-State CircuitsRecitation 14: Op-Amp Non-IdealitiesProf. Joel L. DawsonPage 3Using the approximation that vt v, we havevI− V0 SR1=V0 S− v0R2R2R1vI− V0 S( )− V0 S= −v0v0= V0 S1 +R2R1⎛⎝⎜⎞⎠⎟− vIR2R1So the more gain we ask for, the more offset we can expect at the output.Note that this is a more serious impairment for op-amp integrators.Substituting in the expression we just derived:v0= V0S1 +1R1CS⎛⎝⎜⎞⎠⎟− vI1R1CSv0(t) = V0S+1R1CV0Sdt −1R1CvI(t)dt∫∫v0+−+−V0 S1CSR1vI6.301 Solid-State CircuitsRecitation 14: Op-Amp Non-IdealitiesProf. Joel L. DawsonPage 4This is one reason why you never see op-amp integrators used in a stand-alone, open-loop way:They just integrate their own offset until they hit a supply rail. In the context of a control loop, we’dbe okay:Apply superposition:(1) Y (S) =1R1CS + 1X(S)(2) Y (S) =1R1CS + 1V0SS(3) Y (S) =R1CSR1CS + 1V0SSTreating the offset like a step.Final value theorem y(t)t = x= V0SFinal value theorem y(t)t = ∞= 0Σ+1R1CSΣ++⇒1R1CSΣ1R1CSΣΣY (S)V0 SSX(S)V0 SV0 S+++−321V0 SS6.301 Solid-State CircuitsRecitation 14: Op-Amp Non-IdealitiesProf. Joel L. DawsonPage 5CLASS EXERCISE: Where does this offset come from?Consider the simple op-amp from the last recitation:(1) Remind yourselves of which input is the inverting terminal, and which input is the non-inverting terminal.(2) Notice that there will be an imbalance created by the base currents of Q1 and Q2. Howshould we apply a differential voltage on IN1 and IN 2to correct this?v0− AQ2Q1IN1IN 26.301 Solid-State CircuitsRecitation 14: Op-Amp Non-IdealitiesProf. Joel L. DawsonPage 6So how would you measure the input referred offset of a real op-amp? Pretty easyv0= 1 +999RR⎛⎝⎜⎞⎠⎟V0 S= 1000V0 SOnly caveat: Remember that bipolar op-amps have input bias currents too. They can be modeled:If you’re measuring offset, make sure that IIBR  V0 S.v0++V0 S+−999R−−Rv0++V0