Operational Amplifiers

Transcription

1 Operational Amplifiers Here we see two matched differential amps cascaded to form a basic OPAMP. The differential pair cancel temperature drifts and common mode noise at the input. First built to perform math operations, analogue computers. 1

2 Differential Amplifiers Vcc R C R C V OUT =A V (V 1 V 2 ) Vcc and V EE bias the inputs and output to 0 volts. V 1 V C1 Q1 V C2 Q2!V V 2 Common mode noise refers to same sign pickup on V 1 and V 2. The differential pair cancels commonmode signals in transistors Q 1 and Q 2 eg, temperature drifts, noise,etc. R E V EE Tdrift noise V 1 V

3 LM741 3

4 Ideal Op Amp Z in Z out Very high or Infinite open loop gain. A V ~100K Very high or Infinite input impedance ~100MΩ. Very low or zero output impedance. ~1Ω Zero DC output voltage when inputs grounded. 4

5 β Z in Z out Negative Feedback A v = Vout Vin Vout Vin βvin = High gain amplifiers can become unstable in open loop configuration. A portion of the output β can be feed back to the input to correct this situation. For Av>>1 the closed loop gain A v β = 1/β is only a function of the feedback loop and β easily fixed. ( Vout Vin ) ( ) = A v 1 β Vout Vin β 1 βa v β A v β = A v A β v! 1 1 βa v β when A v >> 1 5

6 Voltage Follower Zin hi V A ΔV Z in Z out V out Zout low V in 1. ΔV =0 no current flows no voltage drop! 2. V in =V A =V out 3. Voltage Follower is used as an impedance matching device from a hi to low impedance devise. It also acts as a current booster or buffer amp. 6

7 Inverting Amp R 2 I V in R 1 Virtual ground I A 0V Z in V out =(R 2 /R 1 )V in 1. V =V = 0 ( since no current flows thru Zin ) 2. V in I R 1 = I R 2 =V out 4. A V β = Vout / Vin = R2/R1 ( combining 2 and 3 ) 5. Z in = R 1 (pointa at virtual ground). R 1 should be fixed and not too small! 7

8 NonInverting Amp R 2 I R 1 V A A V out = V in (1R 2 /R 1 ) V in!v Z in 1. V A =V in ( No current flows thru Z in so!v = 0 ) 2. V out I R 2 = V A =V in 3. V in I R 1 = 0 4. V out = V in (1R 2 /R 1 ) combining 1,2 5. Input impedance = Z in R1 ~ Z in 8

9 Difference Amp R F V 1 R 1 A V out =(V 2 V 1 ) R F /R 1 V 2 R 1 Z in R F 9

10 Integrator/Differentiator C V 1 R integrator V out = 1/RC V in dt V 2 R differentiator V 1 C V out = RC d dt V in V 2 10

11 Voltage Summer R 1 R F I V 1 V 2 R 2 V out = (V 1 /R 1 V 2 /R 2 V 3 /R 3 )R F R 3 V 3 Inputs V 1,V 2,V 3 are summed with gain factor R i /R F I=1,2,3. 11

12 Frequency Response A V Band width ~ 1MHz f hi ~ (106 /gain) Hz 6db f lo f hi We build an amplifier with gain of 1000, what will f hi be? Ans. Based on the graph: f hi ~ (106 /gain) > f hi ~1 khz 12

13 Frequency Response(2) With a voltage gain of 190, an amplifier will have that gain up to the frequency response of about 5.3 KHz below. Not very good for HiFi reproduction! But if we were to reduce the voltage gain to just 20 then the frequency response would extend to 50 KHz. If we needed an amplifier to have a gain of 190 then we would be far better off to have two cascaded amplifier; each with a voltage gain of just 14. The total gain would then be a little over 190 and the frequency response would be flat to over 70KHz. f max ~ (106 /gain)

14 Instrumentation Amp buffer low input impedance Lowtohigh impedance match high output impedance low output impedance high input impedance Lowtohigh impedance match gain low input impedance high output impedance buffer An unwanted common mode noise signal V CM appears on inputs 1 and 2 with a weak transducer signal V S. The V CM is cancelled by the Difference Amp resulting in a noisefree V S signal. Vout = A v (Vs Vs ) A CMRR (Vcm Vcm ) 14

15 INSRUMENATION AMPLIFIER Instrumentation amplifiers are actually made up of 2 parts: a buffered amplifier XOP1, XOP2 and a basic differential amplifier XOP3. The differential amplifier part is often essential when measuring sensors. Why? A sensor produces a signal between its terminals. However, for some applications, neither terminal may be connected to the same ground potential as your measuring circuit. The terminals may be biased at a high potential or riding on a noise voltage. The differential amplifier rescues the signal by directly measuring the difference between the sensor s terminals. The buffered amplifier XOP1 and XOP2 not only provides gain, but provides impedance matching from a low Impedance transducer to the high impedance Differentiral amplifier stage XOP3. SIGNAL GAIN The instrumentation amp offers two useful functions: amplify the difference between inputs and reject the signal that s common to the inputs. The latter is called Common Mode Rejection (CMR). The signal gain is accomplished by XOP1 and XOP2 while XOP3 typically forms a differential gain of 1. You can calculate the overall gain by where R1=R3 and R5/R4 = R7/R

16 CMRR The common mode rejection ratio (cmrr) reflects the ability of the opamp to reject noise common to both inputs. cmrr = A v /A cmrr = Ndb Vout = A v (V V ) A CMRR (V V ) x Av Consider an amplifier with cmrr=80db and Av=180 gain. If a differential signal of 2mV,2mV is applied along with an unwanted common mode signal of 10mV, what is the amplitude of each at the output Vdiff = 180 (4mV) = 720mV A CMRR = Common mode gain =Av/80db = 180/10 4 = Vcmrr = (20 mv ) = 0.36 mv Vout = Vdiff Vcmrr = 720mV 0.36mV 16

17 d 2 y dt 2 Analogue Computation dy 2 dt 2 k 2 y = 0 R C 1 RC!y R C dy dt 1 RC 2 y k 2 y d 2 y dt 2 R 1 R F R F y k 2 y 0 RC = 1 R F /R 1 = k 2 17

18 Analogue Computation!!y a!y by c = 0 V CC!!y = (a!y by c) c by a!y R 1 R 2 R 3!!y!### "### $ C!y R ' C ' y loop backs!#" # $ R R y!# " $# V OUT Sum min g Integrator a = 1 / R 3 C b = 1 / R 2 C c = 1 / R 1 C Integrator 1 = 1 / R ' C Inverter 1 = R / R 18

19 Amp Distortion 19

20 gain 360 o 180 o 1)High gain for feedback cancellation 2) 180 o 60 o 60 o 60 o =360 o 60 o 60 o 60 o Music Tremolo Flasher 20

21 Summary 21