1. Introduction to Operational Amplifiers Why do we study them at this point??? 1. OpAmps are very useful electronic components 2. We have already the tools to analyze practical circuits using OpAmps 3. The linear models for OpAmps include dependent sources TYPICAL DEVICE USING OP-AMPS

2. OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD APEX PA03 PIN OUT FOR LM324 DIMENSIONAL DIAGRAM LM 324 LM324 DIP LMC6294 MAX4240

3. CIRCUIT SYMBOL FOR AN OP-AMP SHOWING POWER SUPPLIES LINEAR MODEL OUTPUT RESISTANCE INPUT RESISTANCE GAIN TYPICAL VALUES

4. CIRCUIT WITH OPERATIONAL AMPLIFIER DRIVING CIRCUIT LOAD OP-AMP

5. TRANSFER PLOTS FOR SOME COMERCIAL OP-AMPS SATURATION REGION LINEAR REGION IDENTIFY SATURATION REGIONS OP-AMP IN SATURATION

6. CIRCUIT AND MODEL FOR UNITY GAIN BUFFER WHY UNIT GAIN BUFFER? BUFFER GAIN PERFORMANCE OF REAL OP-AMPS

7. THE IDEAL OP-AMP

8. THE UNITY GAIN BUFFER – IDEAL OP-AMP ASSUMPTION USING LINEAR (NON-IDEAL) OP-AMP MODEL WE OBTAINED PERFORMANCE OF REAL OP-AMPS IDEAL OP-AMP ASSUMPTION YIELDS EXCELLENT APPROXIMATION!

9. WHY USE THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER? THE VOLTAGE FOLLOWER ACTS AS BUFFER AMPLIFIER THE SOURCE SUPPLIES POWER THE SOURCE SUPPLIES NO POWER THE VOLTAGE FOLLOWER ISOLATES ONE CIRCUIT FROM ANOTHER ESPECIALLY USEFUL IF THE SOURCE HAS VERY LITTLE POWER CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER

10. LEARNING EXAMPLE FOR COMPARISON, NEXT WE EXAMINE THE SAME CIRCUIT WITHOUT THE ASSUMPTION OF IDEAL OP-AMP

11. REPLACING OP-AMPS BY THEIR LINEAR MODEL WE USE THIS EXAMPLE TO DEVELOP A PROCEDURE TO DETERMINE OP-AMP CIRCUITS USING THE LINEAR MODELS

12. 1.Identify Op Amp nodes 2. Redraw the circuit cutting out the Op Amp 3. Draw components of linear OpAmp (on circuit of step 2) 4. Redraw as needed

13. INVERTING AMPLIFIER: ANALYSIS OF NON IDEAL CASE NODE ANALYSIS CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES USE LINEAR ALGEBRA

14. KCL @ INVERTING TERMINAL THE IDEAL OP-AMP ASSUMPTION PROVIDES EXCELLENT APPROXIMATION. (UNLESS FORCED OTHERWISE WE WILL ALWAYS USE IT!) GAIN FOR NON-IDEAL CASE SUMMARY COMPARISON: IDEAL OP-AMP AND NON-IDEAL CASE NON-IDEAL CASE REPLACE OP-AMP BY LINEAR MODEL SOLVE THE RESULTING CIRCUIT WITH DEPENDENT SOURCES

15. LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER THE OP-AMP IS DEFINED BY ITS 3 NODES. HENCE IT NEEDS 3 EQUATIONS THINK NODES! KCL AT V_ AND V+ YIELD TWO EQUATIONS (INFINITE INPUT RESISTANCE IMPLIES THAT i-, i+ ARE KNOWN) OUTPUT CURRENT IS NOT KNOWN DON’T USE KCL AT OUTPUT NODE. GET THIRD EQUATION FROM INFINITE GAIN ASSUMPTION (v+ = v-)

16. LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER NODES @ INVERTING TERMINAL NODES @ NON INVERTING TERMINAL IDEAL OP-AMP CONDITIONS

17. LEARNING EXTENSION

18. “inverse voltage divider” INFINITE INPUT RESISTANCE NONINVERTING AMPLIFIER - IDEAL OP-AMP LEARNING EXTENSION SET VOLTAGE INFINITE GAIN ASSUMPTION

19. FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP NOW RE-DRAW CIRCUIT TO ENHANCE CLARITY. THERE ARE ONLY TWO LOOPS DETERMINE EQUIVALENT CIRCUIT USING LINEAR MODEL FOR OP-AMP COMPLETE EQUIVALENT FOR MESH ANALYSIS MESH 1 MESH 2CONTROLLNG VARIABLE IN TERMS OF LOOP CURRENTS

20. INPUT RESISTANCEGAINMESH 1 MESH 2CONTROLLNG VARIABLE IN TERMS OF LOOP CURRENTS MATHEMATICAL MODEL REPLACE AND PUT IN MATRIX FORMTHE FORMAL SOLUTIONTHE SOLUTIONS

21. A SEMI-IDEAL OP-AMP MODEL This is an intermediate model, more accurate than the ideal op-amp model but simpler than the linear model used so far Non-inverting amplifier and semi-ideal model

22. Sample Problem Find the expression for Vo. Indicate where and how you are using the Ideal OpAmp assumptions Set voltages? Use infinite gain assumption Use infinite input resistance assumption and apply KCL to inverting input

23. Sample Problem 1. Locate nodes 2. Erase Op-Amp 3. Place linear model 4. Redraw if necessary TWO LOOPS. ONE CURRENT SOURCE. USE MESHES DRAW THE LINEAR EQUIVALENT CIRCUIT AND WRITE THE LOOP EQUATIONS MESH 1 MESH 2CONTROLLING VARIABLE

24. “INVERSE VOLTAGE DIVIDER” LEARNING EXTENSION

25. LEARNING EXAMPLE UNDER IDEAL CONDITIONS BOTH CIRCUITS SATISFY DETERMINE IF BOTH IMPLEMENTATIONS PRODUCE THE FULL RANGE FOR THE OUTPU EXCEEDS SUPPLY VALUE. THIS OP-AMP SATURATES! POOR IMPLEMENTATION

26. COMPARATOR CIRCUITS Some REAL OpAmps require a “pull up resistor.” ZERO-CROSSING DETECTOR

27. LEARNING BY APPLICATION OP-AMP BASED AMMETER NON-INVERTING AMPLIFIER

28. LEARNING EXAMPLE DC MOTOR CONTROL - REVISITED CHOOSE NON-INVERTING AMPLIFIER (WITH POWER OP-AMP PA03) Constraints: Power dissipation in amplifier Simplifying assumptions: Significant power losses Occur only in Ra, Rb Worst case occurs when Vm=20 One solution: Standard values at 5%!

29. DESIGN EXAMPLE: INSTRUMENTATION AMPLIFIER DESIGN SPECIFICATIONS “HIGH INPUT RESISTENCE” “LOW POWER DISSIPATION” OPERATE FROM 2 AA BATTERIES ANALISIS OF PROPOSED CONFIGURATION SIMPLIFY DESIGN BY MAKING DESIGN EQUATION: USE LARGE RESISTORS FOR LOW POWER Infinite gain MAX4240

30. DESIGN EXAMPLE DESIGN SPECIFICATION Power loss in resistors should not exceed when Design equationS: Max Vo is 20V Solve design equations (by trial and error if necessary)

31. DESIGN EXAMPLE IMPLEMENT THE OPERATION DESIGN CONSTRAINTS AS FEW COMPONENTS AS POSSIBLE MINIMIZE POWER DISSIPATED USE RESISTORS NO LARGER THAN 10K Given the function (weighted sum with sign change) a basic weighted adder may work ANALYSIS OF POSSIBLE SOLUTION DESIGN EQUATIONS SOLVE DESIGN EQUATIONS USING TRIAL AND ERROR IF NECESSARY ANALYZE EACH SOLUTION FOR OTHER CONSTRAINTS AND FACTORS; e.g. DO WE USE ONLY STANDARD COMPONENTS?

32. DESIGN EXAMPLE DESIGN 4-20mA TO 0 – 5V CONVERTER 1. CONVERT CURRENT TO VOLTAGE USING A RESISTOR CANNOT GIVE DESIRED RANGE! 2. CHOOSE RESISTOR TO PROVIDE THE 5V CHANGE … AND SHIFT LEVELS DOWN! MUST SHIFT DOWN BY 1.25V (SUBSTRACT 1.25 V)

33. DOES NOT LOAD PHONOGRAPH LEARNING BY DESIGN

34. LEARNING EXAMPLE UNITY GAIN BUFFER COMPARATOR CIRCUITS ONLY ONE LED IS ON AT ANY GIVEN TIME

35. MATLABSIMULATION OF TEMPERATURE SENSOR WE SHOW THE SEQUENCE OF MATLAB INSTRUCTIONS USED TO OBTAIN THE PLOT OF THE VOLTAGE AS FUNCTION OF THE TEMPERATURE »T=[60:0.1:90]'; %define a column array of temperature values » RT=57.45*exp(-0.0227*T); %model of thermistor » RX=9.32; %computed resistance needed for voltage divider » VT=3*RX./(RX+RT); %voltage divider equation. Notice “./” to create output array » plot(T,VT, ‘mo’); %basic plotting instruction » title('OUTPUT OF TEMPERATURE SENSOR'); %proper graph labeling tools » xlabel('TEMPERATURE(DEG. FARENHEIT)') » ylabel('VOLTS') » legend('VOLTAGE V_T' )