ECE 442 Solid State Devices & Circuits. 11. Operational Amplifiers

Transcription

1 ECE 442 Solid State Devices & Circuits. Operational mplifiers Jose E. Schutt-ine Electrical & Computer Engineering University of Illinois ECE 442 Jose Schutt ine

2 Operational mplifiers Universal importance (e.g. amplification from microphone to loudspeakers) General terminal configuration with bias ECE 442 Jose Schutt ine 2

3 Operational mplifiers Gain= v ( v v ) out 2 Common configuration with bias implied but not shown Signaling. Differential input stage 2. Difference between input is amplified ECE 442 Jose Schutt ine 3

4 Operational mplifiers Ideal Op mp. Infinite input impedance 2. Zero output impedance 3. Infinite open-loop gain inf 4. Infinite CM or zero common-mode gain 5. Infinite bandwidth lso, op amps are dc (or direct coupled) amplifiers since they are expected to amplify signals with frequency as low as DC. ECE 442 Jose Schutt ine 4

5 Differential & Common-Mode Signals - Differential input signal v ID =v 2 -v - Common-mode input signal v ICm =0.5(v +v 2 ) vid v vicm 2 vid v2 vicm 2 ECE 442 Jose Schutt ine 5

6 Operational mplifiers Ideally, v ICM should be zero to achieve high CM. mplifier will amplify the difference between the two input signals ECE 442 Jose Schutt ine 6

7 Practical Considerations The output voltage swing of an op amp is limited by the DC power supply. Since op amp can exhibit high gain, power supply voltage fluctuations must be minimized use decoupling capacitors from power supply ECE 442 Jose Schutt ine 7

8 Inverting Configuration Terminal 2 is tied to ground We introduce (or 2 ) to reduce gain (from inf) When is connected to terminal, we talk about negative feedback. If is tied to terminal 2, we have positive feedback ECE 442 Jose Schutt ine 8

9 Inverting Configuration Need to evaluate v o /v I ssume ideal Op-mp Since gain is infinite: Thus, v v 0 2 v o v2 v ( ) 0 Note: is open-loop gain v is virtual ground i vi v vi ECE 442 Jose Schutt ine 9

10 Inverting Configuration Since input impedance of OP amp is infinite, current through is i i i v I v v 0 v v I o o i I v v v v o I o I ECE 442 Jose Schutt ine 0

11 Inverting Configuration Closed-Loop gain v v o I G Observe that the closed-loop gain is the ratio of external components we can make the closed-loop as accurate as we want. Gain is smaller but more accurate. ECE 442 Jose Schutt ine

12 We assumed that the OP-amp was ideal. If we assume that the gain is finite = Inverting Configuration v v v v v / 2 o o i v ( v / ) v v / in o in o ECE 442 Jose Schutt ine 2

13 v Inverting Configuration Still assume infinite input impedance v v v v / o i o i o o v / G o v / / I G Closed-loop gain for inverting configuration ECE 442 Jose Schutt ine 3

14 ECE 442 Jose Schutt ine 4 / / o o o v v v i v G The reflected impedance of is given by Inverting Configuration G since ( ) G small

15 Inverting Configuration Since the reflected impedance is so small, v is thus very small and the inverting terminal is said to be a virtual ground in this configuration We see that as, G as, 0 Note: To minimize the closed-loop gain (G) on the value of the open-loop gain (), make + / << ECE 442 Jose Schutt ine 5

16 Input and Output Impedances Inverting Configuration i vi vi i v / I If high gain is required, input impedance will be low Output impedance is zero ECE 442 Jose Schutt ine 6

17 Example ind closed-loop gain for =0 3, =0 4 and =0 5 assuming = k and =00 k. ssuming v I =0. V, find v. Using formulas G v mv mv mv Note: Since output of inverting configuration is at terminal of VCVS, output impedance of closed-loop amp is zero. ECE 442 Jose Schutt ine 7

18 Non-Inverting Configuration ssume gain is v v o ID 0 ECE 442 Jose Schutt ine 8

19 v 0 v v virtual short ID Non-Inverting Configuration 2 Infinite input impedance i 0 Therefore v v v 0 o a v a a v o ECE 442 Jose Schutt ine 9

20 Non-Inverting Configuration Virtual short v I a v v v a 2 I v G o v I v a o G a ECE 442 Jose Schutt ine 20

21 If The Buffer Stage 0, G a lthough voltage gain is low, current gain can be quite high. Buffer stage can be used to interface between processors and switches. ECE 442 Jose Schutt ine 2

22 The Voltage ollower Unity gain amplifier 00% negative feedback in out 0 ECE 442 Jose Schutt ine 22

23 requency esponse Non-Inverting No feedback with feedback description (f) ni (f) Gain MBOa MBni Midband gain f 2oa f 2ni 3 db freq pt GBW oa GBW ni Gain-BW prod MBni MBoa MBoa a a ECE 442 Jose Schutt ine 23

24 requency esponse Non-Inverting If MBoa, MBni ni a a ( f) a ( f) MBoa jf / f 2oa ECE 442 Jose Schutt ine 24

25 requency esponse Non-Inverting ni ( f) MBoa MBoa j f f MBoa 2 oa MBni MBoa MBoa f f 2ni 2oa MBoa ECE 442 Jose Schutt ine 25

26 requency esponse Non-Inverting ni ( f) MBni jf / f 2ni MBoa GBW ni MBni f2ni f2oa MBoa MBoa GBW f f GBW ni MBni 2ni MBoa 2oa oa Gain-Bandwidth product is constant ECE 442 Jose Schutt ine 26

27 requency esponse Non-Inverting Midband voltage gain is reduced from MBoa to MBni The upper 3-dB frequency will be greater than that of the op amp by the same factor of gain reduction. If the low-frequency gain of the op amp is MBoa = 200,000 and with resistors MBni = 40, the gain is reduced by a factor of 5,000. If the basic 3-db frequency is 5 Hz, the noninverting 3dB frequency will be 25 khz. ECE 442 Jose Schutt ine 27

28 requency esponse Inverting OP mp i ( ) Using MBoa jf / f 2oa i MBoa ( ) j f / f MBoa 2oa ECE 442 Jose Schutt ine 28

29 requency esponse Inverting OP mp Using ( ) MBoa MBoa and neglecting i f j f /[ ] 2oa MBoa ECE 442 Jose Schutt ine 29

30 requency esponse Inverting OP mp MBi f f /[ ] 2i 2oa MBoa f f MBi 2i 2oa MBoa ECE 442 Jose Schutt ine 30

31 requency esponse Inverting OP mp f f MBi 2i 2oa MBoa f f MBi 2i 2oa MBoa if >> ( f) i MBi jf / f 2i ECE 442 Jose Schutt ine 3

32 requency esponse Inverting OP mp if >> gain-bandwidth is constant if ~, GBW f f i MBi 2i MBoa 2oa GBW GBW i oa ECE 442 Jose Schutt ine 32

33 requency esponse Inverting OP mp GBW f oa 2i GBW oa MBi ECE 442 Jose Schutt ine 33

34 Example Design an amplifier to couple a microphone to a resistive load. The microphone generates a peak output of 50 mv for a typical voice input level and has a 0-k output impedance. The output voltage across the 2-k load is to have a peak value of 0 V. the bandwidth of the voltage gain should be at least 40 khz. If the GBW of the op amp used is 30 6 Hz, calculate the bandwidth of the final design The midband voltage gain is: f 2ni MBni 0 MBo or a single noninverting stage with this gain, the upper corner frequency is: 6 GBWni GBWoa 30 5 khz 200 MBni ECE 442 Jose Schutt ine 34

35 Example (cont ) This value of BW will not work need 2 stages Pick first stage gain 0; second stage gain 20. We must then have: and 0 2 ECE 442 Jose Schutt ine 35

36 Example (cont ) Choose 2 and 22 arbitrarily and use above equations to extract and 2 ; we get: = 8 k, 2 = 38 k Next, find 3-dB bandwidth of each stage by dividing respective gains into GBW oa or GBW ni f 2ni khz f 2ni khz ECE 442 Jose Schutt ine 36

37 The overall gain is: o Example (cont ) 0 20 ( ) jf /3 0 jf / t 3dB point, magnitude squared of denominator must be 2 rom which 2 2 f 2o f 2o f2o 26 khz ECE 442 Jose Schutt ine 37