FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)( ) 8/25/2010

Transcription

1 FE REVEW OPERATONAL AMPLFERS (OP-AMPS)( ) 1

2 The Op-amp 2 An op-amp has two nputs and one output. Note the op-amp below. The termnal labeled l wth the (-) sgn s the nvertng nput and the nput labeled wth the (+) )sgn s the non- nvertng termnal. The standard way to show the devce s wth the nvertng termnal on top but ths s not always the case. BE CAREFUL!!

3 The deal assumptons 3 When we assume deal characterstcs for an op-amp we are able to make several assumptons whch make analyss MUCH easer. The frst of these s that the nput resstance t of each of the two termnals s nfnte. Rn

4 deal assumptons contnued) 4 The next major assumpton s that the output mpedance s equal to zero. R 0 out Ths assumpton allows us to make the further assumpton that V out s ndependent of the load. Ths means that t can not be loaded down.

5 deal assumptons contnued) The next assumpton s that the open-loop gan (A) s nfnte (n real lfe f s n the mllons so ths s a safe assumpton). 5 Open Loop Gan A

6 deal assumptons contnued) The assumpton that A s nfnte allows us to make a further assumpton that the voltage on each termnal s equal to each other. Ths s concept s called a Vrtual Short. Ths assumpton turns out to be the KEY to op-amp crcut analyss. V A(V V ) out AV V V d A A A t follow s that V V out out V so, lm 0 d 6

7 deal assumptons contnued) 7 A Vrtual short s not lke a real short (or t wouldn t be vrtual). A vrtual short only affects voltage, not current.

8 The nvertng Op-amp V- = 0 volts due to the Vrtual 8 R f f Short to the ground + V - - R - = 0A deal + Wth the addton of two resstors and a ground on the non-nvertng termnal we get an nvertng op-amp. Note that snce the non-nvertng termnal s grounded the voltage on the nvertng termnal s also grounded due to the Vrtual short. + V o -

9 The nvertng Op-amp contnued) Note that as long as there sn t a voltage source on the non-nvertng termnal there could be a dozen resstors on t and t stll would not affect the voltage on that termnal. Snce the current s assumed to be zero on the nvertng termnal, KCL proves that current s equal to the feedback current f. 0 f f 9 0 f

10 The nvertng Op-amp contnued) Another aspect of the vrtual short s that the nput resstance of the nvertng op-amp s equal to R. 10 V V V 0 v V R R R V- = 0 volts due to the Vrtual Short to the ground R f f + V - - R - = 0A deal + + V o -

11 The nvertng Op-amp contnued) The closed-loop pgan of the deal nvertng op- R amp s: f f 11 + V - V - R - = 0A deal + A + V o - CL R R f The FE exam uses a (A) for closed-loop gan.

12 The nvertng Op-amp contnued) Fnd V o,, f, and R n for the crcut shown: V o 12 R f 10k 2k - + R deal 2v + - RR f V R 10k 2v 5 2v 10v 2k V V 2v 0v f 1 ma R 2k R R 2kk n f + V o -

13 The Non-nvertng op-amp f you shft the nput voltage down to the non-nvertng termnal and ground R you now have a Non-nvertng op-amp. 13

14 The Non-nvertng op-amp contnued) A 14 The gan of the Non-nvertng op-amp s: R cl 1 f R Rf R R Note that snce there s 0A on the non-nvertng termnal, there wll be 0 volts across the resstor so the voltage on the nvertng termnal wll be V. 5k V - =V V f 0A 1k 10k deal o

15 The Non-nvertng op-amp contnued) The nput resstance stance of an deal non- nvertng op-amp wll be nfnte. 15 V - =V f 10k 5k 0A Rn 1k deal o V

16 The Voltage Follower A Voltage follower s a specal case of the Non-nvertng op-amp wth a voltage gan of A 1 deal CL R n V

17 The deal Op-amp ntegrator c 17 C 1 V R o t Vd t Vo 0 0 V RC deal o t R=10kΩ wth: 3 C=400 μ f 3 v (t) v o (t) () -3 2 v -3 2 v t, s t, s

18 V The deal Op-amp Dfferentator 18 R f dv C Vo t RC f c f deal wth R 50k C.1nf o v (t) v (t) o dt 21 t,, s t, s

19 Example What s the output voltage, V o, of the deal op-amp crcut shown? The crcut s confgured as An nvertng Op-amp. 2V 1k 10k deal o V o R R f V 10k 1k 2V 20V 19

20 Example What s the current, -, (nvertng termnal current) n the crcut shown? 2v f 8 4 a) -0.88A b) -0.25A - c) 0A d) 0.25A 3v 8 deal Snce the nput resstance of the nvertng and non-nvertng nputs s deally nfnte, the current nto those nputs s deally 0A. So the answer s: c) 0A 20

21 Example What s the output voltage of the crcut shown? 2v 3v 4 8 f 8 - deal Rf Rf Vo V1 V2 R1 R2 The crcut s an op-amp 8 8 summng crcut or 2v 3v 4 8 a Lnear Combnaton 2 2v 1 3 Crcut. t Vo 4 3 7v o 21

22 Example For the Dfference Amplfer crcut shown, determne the output voltage. a) -28.3v 30v b) -6.07V 25v c) 15.45v d) v We haven t talked about a dfference amp but t can be easly solved wth KCL, the Vrtual short, and Ohm s law V + 3 V 25v 5v 5 3 V - f 20-3 deal Frst, the 5v n the equaton s the voltage on the nonnvertng leg voltage dvded between the 5 and the 3 ohm resstors to gve us V +. Contnued on the next page o 22

23 Next, due to the vrtual short, the nvertng termnal also has 5 volts on t. Now we can determne, whch h s the dfference between the voltages on each sde dvded by 15 ohms. (Example contnued) 30v 25v 23 V V 15 30v 5v 15 V - =5v f A v - 3 deal o

24 (Example contnued) 24 Now we note that snce the current nto the nvertng termnal s zero, f = 1.667A by KCL. 30v 25v A 0 A f 1.667A f f

25 (Example contnued) 25 30v 25v Fnally we wrte an equaton for Vo n terms of f. V Vo f 20 5v Vo 1.667A 20 Vo 5v 1.667A 20 5v v 28.33v

26 Example For the deal operatonal amplfer below, what should the value of R f be n order to obtan a gan of 5? 26 Example contnued on the next page.

27 (Example contnued) 27 1k 2k deal R F o 3k F Frst, note that the op-amp s upsde down from the way we normally look at t. Ths s the way that the exam seems to show t a Lot, but t s not ndustry standard. So, n the next slde we wll flp t over. Example contnued on the next page

28 (Example contnued) 28 1k 2k deal R F o 3k F Now that t s flpped over, note that the voltage on the non-nvertng termnal s a voltage dvded V. V 2k 1k 2k 2 V 3 Example contnued on the next page

29 (Example contnued) The voltage dvder has been replaced by the result of the dvder. Next, note that due to the vrtual short the voltage on the nvertng termnal, V - s also 2/3 V V V 3 due to the vrtual short 2 V 3 Example contnued on the next page

30 (Example contnued) We can now fnd an equaton for. The voltage on the butt of the current arrow s 2/3 V and the voltage at the 2 tp of the current arrow s 0 2 V V volts. 3 due to the Dvde that dfference vrtual short voltage by 3k ohms and you get V V 0 V 3 Example contnued 3k 3k 3k V 3 Example contnued on the next page

31 (Example contnued) 31 B y KCL, 0 - f 0-0 f f - 3k f 0A 2 V 3 R F deal o Now, remember that the NFNTE mpedance of the nvertng and non-nvertng termnals means that the current out of those termnals s assumed to be zero. U Usng ths and KCL (above) we fnd that t the feed back current, f s equal to. Problem contnued on the next page.

32 (Example contnued) 32 f V o V R f V 3k f 0A 2 V 3 R F deal o Now we can defne ths current n terms of V o. The butt of the arrow s V o whle voltage at the tp of the arrow s (2/3)V. Problem contnued on the next page.

33 (Example contnued) The problem statement states that the gan s 5. So, that t means that t V o =+5V Vo V f Rf 2 5V V 3 f So, the answer to Rf the queston s that 2 2 an R V 5V 3 V f value of 19.5k 3 ohms wll result n 3k R a gan of 5. f Rf k 19. 5k 3 k R 3 R f 33

34 Example Evaluate the followng amplfer to determne the value of R 4 requred to obtan a voltage gan V o of k V 34 R M deal o Contnued on the next page

35 (Example contnued) 35 Note that the nvertng termnal voltage s 0 volts due to vrtual short whch exsts between t and the non-nvertng termnal. Contnued on the next page

36 (Example contnued) We can now fnd an equaton for the current. Note that the voltage at the butt of the arrow s +V and the voltage at the tp of the arrow s 0v. We use ths voltage dfference and Ohms law to fnd an equaton for. 36 V 0v V Contnued on 1M 1M the next page

37 (Example contnued) 37 Now, usng the fact that the current nto both the nvertng and non-nvertng termnals s assumed to be 0A due to the assumed nfnte nput resstance, we can use KCL to show that 2 equals A 2 Contnued on the next page

38 (Example contnued) 38 We can use the same technque as before to defne 2 n terms of V x. and then equate agan to. 2 0v Vx 500k V x so Vx 500 k 500k Contnued on the next page

39 (Example contnued) 39 Next defne the current 4 n terms of V o and V x. 4 V x R 4 V o Contnued on the next page

40 (Example contnued) And now use KCL to defne the currents n terms of voltage. 40 By KCL V V V V x o x x R 500k Contnued on the next page

41 (Example contnued) 41 From the problem statement: V V o 120 V 120V o and 2 = 1 so we can expand that to V Vx 500k Vx V 1M 500k 1M Vx Vo Vx Vx R4 500k k 500kk 500kk V 1M 120V V 1M V 1M R4 500k 100 Contnued on the next page

42 (Example contnued) 500 k 500 k 500 k 120 1M 1M 1M R 500k k 500 k 500 k 120 1M 1M 1M 1M R4 500k k120M 500k 500k R 500k 100 R M M 1 5k 1 5k k 24k So, the value of R 4 requred to have a voltage gan of -120 s 24k ohms.