Example: An op-amp circuit analysis

Transcription

1 /1/11 Example An op amp crcut analyss lecture 1/3 Example: An opamp crcut analyss Let s determne the output voltage (t) of the crcut below: v n (t) R =3K I= ma (t)

2 /1/11 Example An op amp crcut analyss lecture /3 Wthout ths step, your answer (and thus your grade) mean nothng The frst step n EVERY crcut analyss problem s to label all currents and voltages: v n 1 v v 1 R =3K I= ma v v v 3 3

3 /1/11 Example An op amp crcut analyss lecture 3/3 The search for a template Q: I looked and looked at the notes, and I even looked at the book, but I can t seem to fnd the rght equaton for ths confguraton! A: That s because the rght equaton for ths crcut does not exst at least yet. It s up to you to use your knowledge and your sklls to determne the rght equaton for the output voltage!

4 /1/11 Example An op amp crcut analyss lecture 4/3 You have the tools to determne ths yourself no need to fnd a template! Q: OK, let s see; the output voltage s: v =???? out I m stuck. Just how do I determne the output voltage? A: Open up your crcut analyss tool box. Note t conssts of three tools and three tools only: Tool 1: KCL Tool : KVL Tool 3: Devce equatons (e.g., Ohm s Law and the vrtual short). Let s use these tools to determne the rght equaton! Frst, let s apply KCL (I m qute partal to KCL).

5 /1/11 Example An op amp crcut analyss lecture 5/3 The frst KCL Note there are two nodes n ths crcut. The KCL for the frst node s: v n 1 v 1 = I 1 R =3K v I= ma v v v 3 3 Note the potental of ths node (wth respect to ground) s that of the nvertng opamp termnal (.e., v ).

6 /1/11 Example An op amp crcut analyss lecture 6/3 The KCL of the second node s: The second KCL v n 1 v 1 I = 3 R =3K v I= ma v v v 3 3 Note the potental of ths node (wth respect to ground) s that of the nonnvertng opamp termnal (.e., v ).

7 /1/11 Example An op amp crcut analyss lecture 7/3 Now for our second tool KVL. The frst KVL We can conclude: v v = v v = v v n 1 1 n v n 1 v v 1 R =3K I= ma v v v 3 3

8 /1/11 Example An op amp crcut analyss lecture 8/3 The second KVL And also: v v = v v = v v out out v n 1 v v 1 R =3K I= ma v v v 3 3

9 /1/11 Example An op amp crcut analyss lecture 9/3 The thrd KVL And lkewse: v v = v = v 3 3 v n 1 v v 1 R =3K I= ma v v v 3 3

10 /1/11 Example An op amp crcut analyss lecture 1/3 There are seven devce equatons Fnally, we add n the devce equatons. Note n ths crcut there are three resstors, a current source, and an opamp From Ohm s Law we know: v 1 3 = = = 1 3 R R R 1 3 v v And from the current source: I = And from the opamp, three equatons! = = v = v

11 /1/11 Example An op amp crcut analyss lecture 11/3 1 equatons and 1 unknowns! Q: Ykes! Two KCL equatons, three KVL equatons, and seven devce equatons together we have twelve equatons. Do we really need all these? A: Absolutely! These 1 equatons completely descrbe the crcut. There are each ndependent; wthout any one of them, we could not determne! To prove ths, just count up the number of varables n these equatons: We have sx currents: And sx voltages:,,,,, I 1 3 v, v, v, v, v, v 1 3 out Together we have 1 unknowns whch works out well, snce we have 1 equatons! Thus, the only task remanng s to solve ths algebra problem!

12 /1/11 Example An op amp crcut analyss lecture 1/3 Don t ask the calculator to fgure ths out! Q: OK, here s where I take out my trusty programmable calculator, type n the equatons, and let t tell me the answer! A: Nope. I wll not be at all mpressed wth such results (and your grade wll reflect ths!). Instead, put together the equatons n a way that makes complete physcal sense just one step at a tme.

13 /1/11 Example An op amp crcut analyss lecture 13/3 3 = ma Frst, we take the two devce equatons: = and I = And from the second KCL equaton: v n I = = = v v 1 R =3K I= ma v v v 3

14 /1/11 Example An op amp crcut analyss lecture 14/3 So v3 =. V Now that we know the current through R, we can determne the voltage across 3 t (um, usng Ohm s law ). ( ) v = R = 1 = v n 1 v v 1 R =3K I= ma v v

15 /1/11 Example An op amp crcut analyss lecture 15/3 Thus v =. V Thus, we can now determne both v (from a KVL equaton) and v (from a devce equaton): v = v = 3 and v = v = v n 1 v v 1 R =3K v = I= ma v =

16 /1/11 Example An op amp crcut analyss lecture 16/3 And now 1 = Now, nsertng another devce equaton: nto the frst KCL equaton: = = I = = v n v v 1 R =3K v = I= ma v =

17 /1/11 Example An op amp crcut analyss lecture 17/3 From KVL we fnd: So that v = vout and: v = v v v = v v = v 1 n 1 n 1 n v = v v v = v out out v n ( v n ) ( v ) R =3K out v = I= ma v =

18 /1/11 Example An op amp crcut analyss lecture 18/3 Now we can fnd vout So, from Ohm s law (one of those devce equatons!), we fnd: v v = = = v 4 1 n 1 R 1 1 n and: v = = R 3 Equatng these last two results: vout v 4 = v = 14 3v 3 n out n

19 /1/11 Example An op amp crcut analyss lecture 19/3 The rght equaton! Thus, we have at last arrved at the result: v out = 14 3v n v n vn vn 4 R =3K ( v n ) ( 3 1) v n v = I= ma v = v out = 14 3v n

20 /1/11 Example An op amp crcut analyss lecture /3 An alternatve: superposton Note an alternatve method for determnng ths result s the applcaton of superposton. Frst we turn off the current source (e.g., I = ) note that ths s an open crcut!!!!!!!! v n R =3K I=

21 /1/11 Example An op amp crcut analyss lecture 1/3 It s just an nvertng amp Note ths s the same confguraton as that of an nvertng amplfer! Thus, we can quckly determne (snce we already know!) that: R 3 v = v = v = 3v out n n n R 1 1 R =3K v n v out = 3v n

22 /1/11 Example An op amp crcut analyss lecture /3 See f you can prove ths result! Lkewse, f we nstead set the nput source to zero ( v = ground potental!), n we wll fnd that the output voltage s 14 volts (wth respect to ground): R =3K I= ma v = out 14

23 /1/11 Example An op amp crcut analyss lecture 3/3 Look; the answer s the same! From superposton, we conclude that the output voltage s the sum of these two results: v = 14 3v out n The same result as before! v n vn vn 4 R =3K ( v n ) ( 3 1) v n v = I= ma v = v out = 14 3v n