6.301 Solid State CircuitsRecitation 10: Op-Amp ApplicationsProf. Joel L. DawsonWe’re going to treat op-amps in some detail in this class. They are very useful as all-around circuitblocks, as we shall see in the next few lectures. From the standpoint of learning circuit design,though, they are extremely good for sensitizing us to the issues surrounding DC amplification.Let’s pause and take stock of some of the things that AC coupling has given us.(1) We don’t worry about what our reference level is:(2) DC offsets? What DC offsets?(3) “Bias Picture” vs. “Midband Picture” tSymmetric aboutground.RealityBias PictureMidband Picture6.301 Solid State CircuitsRecitation 10: Op-Amp ApplicationsProf. Joel L. DawsonPage 2But suppose our input has a DC component that we care about?CLASS EXERCISEConsider the following circuit, for which VBE is 0.6V.Suppose we have chosen our reference output level to be +2V. That is, when VIN is 0, V0 is 2V.(1) Pick RL to achieve this.(2) What range is RL restricted to if we demand accuracy to within ±1mV?(Workspace)−5.6V5kΩV0−+RL+10VVIN6.301 Solid State CircuitsRecitation 10: Op-Amp ApplicationsProf. Joel L. DawsonPage 3When we design for DC performance, then, we go to a whole new bag of circuit tricks. Differentialstructures and op-amps are big parts of that bag…So what are op-amps useful for? First, recall how to analyze op-amp circuits quickly:(1) Non-inverting amplifierStart by making no assumptions about the gain, a, of op-amp.vN=R1R1+ R2v0 ; v0= a vp− vN( )v0= a vi−R1R1+ R2v0( )Solve for V0:1 + aR1R1+ R2⎛⎝⎜⎞⎠⎟v0= aviv0vi=a1 + aR1R1+ R2And of course, if we are lucky enough to have large gain a,v0vi= 1 +R2R1R1R2v0−+vNvpvi6.301 Solid State CircuitsRecitation 10: Op-Amp ApplicationsProf. Joel L. DawsonPage 4Before we leave this, let’s go back and look atv0= a vp− vn( )Notice that infinite gain a, together with finite values v0, imply vp− vn→ 0. Put (yet) another way:vp− vn=v0a=11 + aR1R1+ R2As a goes to infinity, it must be true that vp≈ vN.(2) Inverting amplifier and Summer+−v0R2R1vi−+RfR1R2v1v2vi− 0R1=0 − v0R2⇒v0vi= −R2R1vi− 0R1+v2− 0R2+vN− 0RN=0 − v0RfAllows you to performweighted sum!6.301 Solid State CircuitsRecitation 10: Op-Amp ApplicationsProf. Joel L. DawsonPage 5v0= −RfviRii =1N∑(3) Differential AmplifierThose of you who continue on to do advanced circuit design, particularly for ICs, will probably workwith differential signal paths. An example is a fully differential amplifier: Sometimes it becomes necessary to convert to a single-ended signal, that isIf we’re concerned only with AC signals, we can use a transformer (sometimes called a balun):But if we care about the DC component, we must find another solution. Once again, the op-amprides to the rescue:vi+vi−−−++vIN= vi+ −vi −( )vOUT= v0+ −v0 −( )DifferentialtoSingle-Endedv0= vi+ −vi −vi −vi +v0vi −vi +6.301 Solid State CircuitsRecitation 10: Op-Amp ApplicationsProf. Joel L. DawsonPage 6We can lower our heads and grind on this one, or take advantage of superposition and our previousresults:V0=z4z3+ z4⎛⎝⎜⎞⎠⎟z1+ z2z1⎛⎝⎜⎞⎠⎟Vb−z2z1VaNow suppose we choose z4z3=z2z1V0=z4z31 +z4z3⎛⎝⎜⎞⎠⎟1 +z2z1⎛⎝⎜⎞⎠⎟Vb−z2z1VaV0=z4z3Vb−z2z1Va=z2z1Vb− Va( )The op-amp has many, many useful applications. Soon, we will learn how they work on the