Chap 5-Operational Amplifier.pptx. electronics

Electronics course

1. Chapter 5 Operational Amplifiers (OP Amp) 2. Operational Amplifiers 1 Operations: 1- Inverter Amplifier 2- Non inverter Amplifier 3- Adder(Summing) Amplifier 4- Subtractor (Differential) Amplifier 5- Follower 6- Integrator Amplifier 7- Differential(Derivative) Amplifier 8- Comparator Amplifier Applications

2. Figure Circuit symbol for the op amp. 2. Operational Amplifiers 2 Operational amplifier : A differential amplifier with very high voltage gain. Usually realized as integrated circuit. v1= V+: Non inverting input v2 = V - : inverting input vo : Output voltage A OL = A : OpAmp voltage gain Vo = A OL (V+-V-) = A (V+-V-) = A (V1-V2) = , = V+ - V- 5-1 Introduction ε A ε ε

3. 5 . 2 The Ideal Operational Amplifier Ideal operational amplifier: Infinite input impedance Zin ( Zin= ꝏ) Infinite open loop gain A OL for differential signal ( A OL = ꝏ) Zero output impedance Zo ( Zo=0) Infinite bandwidth BW . 3 . : Offset voltage : inverting input : Noniverting input 4 : Negative power supply (-) 5 : Offset voltage 6 : output voltage 7 : Positive power supply (+) 8 : Not connected

4. 5 . 3 The Summing-Point Constraint 2. Operational Amplifiers 4 Operational amplifiers are almost always used with negative feedback, in which part of the op- amp output signal is returned to the input in opposition to the source signal. Ideal op-amp circuits are analyzed by the following steps: Verify that the negative feedback is present. Usually this takes the form of a resistor network connected to the output terminal and to the inverting input terminal. Assume that the differential input voltage and the input current of the op amp are forced to zero. (This is summing - point constraint.) Apply standard circuit analysis principles, such as Kirchhoff’s laws and Ohm’s law, to solve for the quantities of interest. ε

5. 5 . 4 The Inverting Amplifier i 1  v in R 1 i 2  i 1 i 2  v in R 1 v o  R 2 i 2  => vo =- R2 i2 = -R2 i1 Inverting amplifier. 1 R v v in o   R 2 v A  1 1  R i  v in Z in i 2. Operational Amplifiers 5 o v R R v   1 2 v o is independent of the load resistance R L . Thus the output acts as ideal voltage source and output impedance is . Ex: R1 =1k and R2 =10k  Av =-10 Av = 10 absolute value We make use of the summing-point constraint in the analysis of the inverting amplifier. Ideal Op Amp: Zin = infinity I + = I- = 0 A = infinity  vo = A(V+-V-) A= vo /(V+-V-)  V+ = V- Zo = 0 Vin =R1 i1 V+ =0 real GND V-= 0 Virtual GND vin i1 i2 vo

6. 5.5 Summing Amplifier (Adder Amplifier) A circuit known as a summing amplifier is illustrated in Figure below . Ideal Op Amp Use the ideal-op-amp assumption to solve for the output voltage in terms of the input voltages and resistor values. What is the input resistance seen by v A ? By v B ? What is the output resistance seen by R L ?   B   v A  v B  v B  v A A A  R R f R B A A B f i  i  i B B A A R R f v o   i f R f    v  v B  R R i R i Summing amplifier. + - + - + - R A R f R L v o + - v B v A i A i B i f R B v - =0 2. Operational Amplifiers 6 Input impedance seen by v A : R A Input impedance seen by v B : R B o L v doesn’t depend on R , thus the output impedance is 0. If RA = RB = R vo = - Rf /R * (VA +VB) Ex: If Rf =10k and R =1k, v0 = -10(VA+VB)

7. Exercise 2.3. Find an expression for the output voltage of the circuit, shown in Figure . v o 1 . The second Op Amp is connected as summing amplifier  R 2 2 10000 R 4 10000 v o  4 v 1  2 v 2 R o 1 o 1  20000 v   2 v  2 v    R 3   20000 v v o    5 v o 1   5 v 2  Solution: The first Op Amp is connected as an inverting amplifier. Thus 10000 2. Operational Amplifiers 7 R 1 R 20000 v o 1    2 v 1   v 1   2 v 1

8. Positive Feedback Schmitt trigger circuit. The high gain increases the input voltage v i , this increases further the output voltage and so on. Very soon the output voltage reaches the supply voltage, the amplifier enters in switching mode of operation and doesn’t function any more as amplifier. The current equation at the noninverting input is  v i  v in  v i  v o R R o in OL i 2. Operational Amplifiers 8 i in 2 2 v  1  v  v   1  v  A v 

9. 5 . 6 The Noniverting Amplifier V +  v in = V- = v1 5.7 The Follower Amplifier A v = 1 when R 2 = 0 and/or R 1 =  . The circuit is called voltage follower . Noninverting amplifier. Voltage follower. Vin = V- =V+ = vo  Av = vo /vin = 1 unity gain o v R 1  R 2 R 1 1 v  R 1 2. Operational Amplifiers 9 v in A v  v o  1  R 2 Ex: If R1 =1k and R2 = 10K  Av = 11 . Ideal OP Amp: Zin = infinity  I + = I- =0 A = infinity V+ =V- Zo =o

10. + - + - + - R 1 R 1 R L v o + - v 2 v 1 i 1 R 2 5.8 Subtractor amplifier. Find an expression for the output voltage in in terms of the resistance and input voltages for the differential amplifier shown in Figure below . i 1 v - v + R 2 Differential amplifier. Since Op Amp input voltage is 0, v - = v + and 1 v 1  v o R 1  R 2 v 1  v o 1 R R  R v   v  i R  v  1 1 1 1 1 2 From voltage divider principle i  1 2 R 2 2 R  R v   v 2 1 2 1 1 o 2 R  R 1 2 1 R  v R  R v  v R v  2 1 2. Operational Amplifiers 10  v  v  1 2 R R o v  Cicruit Ex: If R1 =R2 = R, vo = v2-v1 If R1=1k and R2 =4 k, vo = 4(v2-v1)

11. 5.9 Op-amp Imperfections in the Linear Range of Operation Input Impedance and Output Impedance Input impedance BJT input stage: > 100k  , typically few M  ; FET input stage: ~10 12  Output impedance: ~100  or less. If the gain the Op Amp is high, the influence of the input and output impedance is small. 2. Operational Amplifiers 11 The nonideal characteristics of real op amps fall into three categories: Nonideal properties in the linear range of operation. Nonlinear characteristics. DC offsets.

12. Gain and Bandwidth Limitations When the Op Amp is included in a feedback loop in order to realise a finite gain amplifier, the bandwidth of the finite gain amplifier is extended proportionally to the feedback. Bode plot of open-loop gain for a typical op amp. A OL ( f )  2. Operational Amplifiers 12 1  j( f / f BOL ) A OL Bode plots.

13. 5.10 Large Signal Operation Output Voltage Swing Output Current Limits The maximum that an Op Amp can supply to a load is restricted. For  A741 this limitation is  25mA. If a small-value load resistance drew a current outside this limits, the output waveforme would become clipped. For a real op amp, clipping occurs if the output voltage reaches certain limits. For  A741: If the supply voltages are +15V and –15V the amplitude of the output voltage without clipping is 14V typically (guaranteed is 12V). 2. Operational Amplifiers 13

14. 5 .11 Integrators and Differentiators R 2. Operational Amplifiers 14 i in ( t )  v in ( t ) t c v ( t )  i ( x )dx C  in 1 v o ( t )   v c ( t ) t v ( x )dx o RC  in v ( t )   1 Integrator. i in =(vin-V-)/R = Cdvc / dt = Cd(V— vo )/ dt Since V- = V+ = 0 , V- : Virtual GND v in/R= - Cdvo / dt  vo = -1/RC integral ( dvin / dt ) + cste i/o o/p Square Triangulare Sinwave sinwave

15. 5.12 Differentiators Circuit (Derivative) dt v o ( t )   RC dv in Figure 2.18 Differentiator. 2. Operational Amplifiers 15