Ideal Op Amps

Op Amp CircuitsObjective of LectureApply the almost ideal op amp model in the following circuits:Inverting AmplifierNoninverting AmplifierVoltage FollowerSumming AmplifierDifference AmplifierCascaded Amplifiers

Almost Ideal Op Amp Modeli2 = 0i1 = 0Linear Region: When V+< vo< V- , vo is determined from the closed loop gain Av times v2 as v1 = v2 (vd = 0 V).Saturation:When Av v2 V+, vo = V+.When Av v2 V-, vo = V-.Ri = W and Ro = 0 WExample #1: Inverting Amplifier

i2 = 0i1 = 0isifV+ = 15VV- = -10ViExample #1 (cont)

i2 = 0i1 = 0isifV+ = 15VV- = -10ViExample #1Closed loop gains are dependent on the values of R1 and Rf.Therefore, you have to calculate the closed loop gain for each new problem.Example #1 (cont)

i2 = 0i1 = 0isifivo

Example #1 (cont)Since AV = -10If Vs = 0V, V0 = -10(0V) = 0VIf Vs = 0.5V, Vo = -10(0.5V) = -5VIf Vs = 1V, Vo = -10(1V) = -10VIf Vs = 1.1V, Vo = -10(1.1V) < V-, Vo = -10VIf Vs = -1.2V, V0 = -10(-1.2V) = +12VIf Vs = -1.51V, Vo = -10(-1.51V) > V+, Vo = +15VExample #1 (cont)Voltage transfer characteristic

Slope of the voltage transfer characteristic in the linear region is equal to AV.Example #2: Noninverting Amplifier

Example #2 (cont)

Example #2 (cont)

Example #2 (cont)

Example #2: Noninverting Amplifier

Example #2 (cont)

AV = +11If Vs = 0V, V0 = 11(0V) = 0V

If Vs = 0.5V, Vo = 11(0.5V) = +5.5VIf Vs = 1.6V, Vo = 11(1.6V) > V+, Vo = +15V

If Vs = -0.9V, V0 = 11(-0.9V) = -9.9VIf Vs = -1.01V, Vo = 11(-1.01V) > V+ Vo = +15V

Example #2 (cont)Voltage transfer characteristic

Slope of the voltage transfer characteristic in the linear region is equal to AV.Example #3: Voltage FollowerA voltage follower is a noninverting amplifier where Rf = 0W and R1 = W.

Vo /Vs = 1 +Rf/R1 = 1 + 0 = 1

Example #4: Summing AmplifierA summing amplifier is an inverting amplifier with multiple inputs.V+ = 30VV-=-30V

Example #4 (cont)i2 = 0v1v2i1 = 0iA iB iC if We apply superposition to obtain a relationship between Vo and the input voltages.19Example #4 (cont)

A virtual ground

Example #4 (cont)

Example #4 (cont)Note that the voltages at both nodes of RC are 0V.

Example #4 (cont)

Example #4 (cont)

Example #4 (cont)

Example #4 (cont)

Example #5: Difference Amplifier

Example #5 (cont)i2 = 0v1v2i1 = 0iA iB iC if

Example #5 (cont)i2 = 0v1v2i1 = 0iA iB iC if

Example #5 (cont)i2 = 0v1v2i1 = 0iA iB iC if

Example #5 (cont)

iAifExample #5 (cont)

Example #5 (cont)If RA/Rf = RB/RC

And if RA = Rf

Example #6: Cascading Op Amps

Example #6 (cont)Treat as two separate amplifier circuitsExample #6 (cont)First CircuitSecond Circuit

Example #6 (cont)It is a noninverting amplifier.

Example #6 (cont)It is a inverting amplifier.

Example #6 (cont)The gain of the cascaded amplifiers is the multiplication of the two individial amplifiers

SummaryThe almost ideal op amp model:Ri = W.i1 = i2 = 0A; v1 = v2 Ro = 0W.No power/voltage loss between the dependent voltage source and vo.The output voltage is limited by the voltages applied to the positive and negative rails.V+ vo V- This model can be used to determine the closed loop voltage gain for any op amp circuit.Superposition can be used to solve for the output of a summing amplifier.Cascaded op amp circuits can be separated into individual amplifiers and the overall gain is the multiplication of the gain of each amplifier.