Module 12 Operational Amplifiers – Part II

1. Module 12Operational Amplifiers – Part II

2. Review from Operational Amplifiers I: Negative input Output Positive input Power SupplyVoltages Anatomy of an “Op-Amp” VPOS –VNEG

3. vOUT v+ Av(v+– v–) rin v– These features motivate the Ideal Op-Ampapproximation Equivalent model for the circuit inside an op-amp Dependent Source Model • rin is on the order of several Megohms: • Av is on the order of 105 to 106

4. Vpos Upper Limit Range Lower Limit –Vneg Dependent Source Model • vOUT must lie between Vpos and –Vneg VPOS vOUT –VNEG • Otherwise, the op-amp becomes saturated. • Saturated op-amp  vOUT = Vpos or –Vneg limit

5. VPOS –VNEG vOUT v+ rin =  Av = Very Large v– The Ideal Op-Amp Approximation rin=  –Vneg < VOUT < Vpos This model greatly simplifies op-amp analysis

6. i+ = 0 i= 0 A Consequence of Infinite rin VPOS rin=  VNEG Currents i+ and ito (or from) input terminals are zero

7. Defines the Linear Region of operation A Consequence of Large Av If vOUT lies between Vpos and –Vneg … (v+ v–)  0 VPOS –VNEG

8. vOUT vOUT = = R1 +R2 R1 +R2 vIN vIN R1 R1 (works because i = 0) R1 R1 Via voltage division v– = vOUT vIN= vOUT R1 +R2 R1 +R2 When vOUT in linear region: –vneg< vOUT < vPos  Example: The Non-Inverting Amplifier Revisited vIN vOUT R2 i = 0 R1 Use the Ideal Op-Amp approximation: v  vIN  Done!

9. i1 i = 0 vIN = vIN  v i1 = R1 R1 Via KCL R2 vOUT = vIN R1 vOUT =  vIN R2 R1 Example: The Inverting Amplifier Revisited + – i2 R2 vIN vOUT R1 Use the Ideal Op-Amp approximation: v  0 v+ = 0   Ohm’s Law i1 = i2 (with i = 0) vOUT =  i2 R2 =  i1 R2  Done!

10. iF i1 i2 RF R1 R2 + _ v1 + _ v2 + – vOUT i1 + i2 = iF KCL: v1 v1 v2 v2   RF RF vOUT =  v1 + v2 i2 = i1 = iF = + R2 R1 R1 R2 R1 R2 vOUT =  iFRF Another Example: The Summation Amplifier + – Use the Ideal Op-Amp Approximation… Output is weighted, inverted sum of inputs

11. iF i2 i1 RF R1 R2 v3 vn v1 R3 + + + + . . . _ _ _ _ v2 + – Rn vOUT vOUT =  v1 + v2 + v3 + …+ vn RF RF RF RF R1 R3 Rn R2 Can extend result to arbitrary number of input resistors: Output is weighted, inverted sum of inputs: iF = i1 + i2 + i3+ … + in

12. R2 R1 + _ + – v1 R1 + vOUT _ v2 R2 Another Example: DifferenceAmplifier

13. + – vOUT vOUT =  v1 R2 R1 1st Partial result for vOUT Use Superposition: R2 R1 + _ v1 R1 + _ v2 R2 Set v2 to zero i+ = 0  v+ = 0  v = 0 We have an inverting amplifier

14. i+ + – vOUT R2 v+ = v2 R1 +R2 2nd Partial result for vOUT R2 R1 +R2 R1 +R2 R2 vOUT v+ = v2 = v2 =  R1 +R2 R1 R1 R1 Use Superposition, con’t: R2 R1 + _ v1 R1 + _ v2 R2 Set v1 to zero Via voltage division We have an non-inverting amplifier

15. + – vOUT R2  vOUT = v2 v1 R1 R2 R2 R1 vOUT  = (v2 v1) R1 R2 R1 + _ v1 R1 + _ v2 R2 Add together the 2nd and 1st partial results: Amplifies difference between v2 and v1

16. vOUT =  v1 + v2 + v3 + …+ vn RF RF RF RF R2 R1 R3 Rn R2 vOUT = (v2 v1) R1 Summary • Ideal Op-Amp Approximation simplifies circuit analysis • “Ideal” impliesrin =  andv+ = vin the linear region • Summation Amplifier • Difference Amplifier

17. End of This Module Homework