1 Module 12 Operational Amplifiers – Part II

2 Review from Operational Amplifiers I: Negative inputPositive inputOutput V POS –V NEG Power Supply Voltages Anatomy of an “Op-Amp”

3 Dependent Source Model r in is on the order of several Megohms: A v is on the order of 10 5 to 10 6 These features motivate the Ideal Op-Amp approximation v+v+ v–v– A v (v + – v – ) Equivalent model for the circuit inside an op-amp r in v OUT

4 Dependent Source Model v OUT must lie between V pos and –V neg V POS –V NEG v OUT V pos –V neg Upper Limit Lower Limit Range Otherwise, the op-amp becomes saturated. Saturated op-amp  v OUT = V pos or –V neg limit

5 The Ideal Op-Amp Approximation V POS –V NEG r in =  v+v+ v–v– A v = Very Large v OUT This model greatly simplifies op-amp analysis –V neg < V OUT < V pos

6 A Consequence of I nfinite r in V POS V NE G r in =  i + = 0 i  = 0 Currents i + and i  to (or from) input terminals are zero

7 A Consequence of Large A v V POS –V NEG If v OUT lies between V pos and –V neg … (v +  v – )  0 Defines the Linear Region of operation

8 Example: The Non-Inverting Amplifier Revisited Use the Ideal Op-Amp approximation: v – = v OUT R 1 + R 2 R1R1 Via voltage division (works because i  = 0) v IN R2R2 R1R1 v OUT R 1 + R 2 R1R1 v OUT v IN = i  = 0 v   v IN When v OUT in linear region: –v neg < v OUT < v Pos v IN = v OUT R 1 + R 2 R1R1   R1R1 v OUT v IN = Done !

9 + – v IN v OUT R2R2 R1R1 v IN = Example: The Inverting Amplifier Revisited R2R2 R1R1 i  = 0 Use the Ideal Op-Amp approximation: v   0 v + = 0   i1i1 i 1 = v IN  v  R1R1 v IN R1R1 = i2i2 i 1 = i 2 Via KCL (with i  = 0) Ohm’s Law v OUT =  i 2 R 2 =  i 1 R 2 v OUT = R2R2 R1R1  v IN  Done!

10 + – The Summation Amplifier v OUT +–+– + _ R2R2 R1R1 + _ v1v1 v2v2 RFRF Another Example: Use the Ideal Op-Amp Approximation… i2i2 i1i1 iFiF KCL: i 1 + i 2 = i F i 1 = v1v1 R1R1 i 2 = v2v2 R2R2  v OUT =  i F R F i F = + v1v1 R1R1 v2v2 R2R2 v OUT = RFRF R1R1 RFRF R2R2 v2v2 v1v1 +   Output is weighted, inverted sum of inputs

11 Can extend result to arbitrary number of input resistors: Output is weighted, inverted sum of inputs: v OUT +–+– R2R2 R1R1 + _ v2v2 RFRF i2i2 i1i1 iFiF... + _ v1v1 + _ v3v3 + _ vnvn R3R3 RnRn v OUT = RFRF R1R1 RFRF R2R2 v2v2 v1v1 +  RFRF R3R3 RFRF RnRn v3v3 +vnvn +…+ i F = i 1 + i 2 + i 3 + … + i n

12 Difference Amplifier Another Example: v OUT +–+– + _ R1R1 R2R2 + _ v2v2 v1v1 R2R2 R1R1

13 v OUT +–+– + _ R1R1 R2R2 + _ v2v2 v1v1 R2R2 R1R1 Use Superposition: Set v 2 to zero i + = 0  v + = 0  v  = 0 We have an inverting amplifier v OUT = R2R2 R1R1  v1v1 1 st Partial result for v OUT

14 v OUT +–+– + _ R1R1 R2R2 + _ v1v1 R2R2 R1R1 Use Superposition, con’t: Set v 1 to zero We have an non-inverting amplifier 2 nd Partial result for v OUT v + = v 2 R 1 + R 2 R2R2 Via voltage division R 1 + R 2 R1R1 v OUT v+v+ == v 2 R 1 + R 2 R1R1 R2R2  = v 2 R2R2 R1R1 v2v2 i+i+

15 v OUT +–+– + _ R1R1 R2R2 + _ v2v2 v1v1 R1R1 Add together the 2 nd and 1 st partial results: R2R2 = v 2 R2R2 R1R1 v OUT  R2R2 R1R1 v1v1  = (v 2  v 1 ) R2R2 R1R1 v OUT Amplifies difference between v 2 and v 1

16 Summary Ideal Op-Amp Approximation simplifies circuit analysis “Ideal” implies r in =  and v + = v  in the linear region Summation Amplifier v OUT = RFRF R1R1 RFRF R2R2 v2v2 v1v1 +  RFRF R3R3 RFRF RnRn v3v3 +vnvn +…+ = (v 2  v 1 ) R2R2 R1R1 v OUT Difference Amplifier

17 End of This Module Homework