Vi Gain 3. Homework In the circuit below: a) Determine the gain of this amplifier. b) Is it inverting or non-inverting?

Transcription

1 Hmewrk 5. In the crcut belw: a) Determne the gan f ths amplfer. b) Is t nvertng r nnnvertng? 3 9 Ans: Treatng the () termnal f the pamp as a nde, we see that ts vltage s zer, snce the () termnal s at grund. (emember: =) Therefre, we can wrte the ndal equatn: 3 9 emember: n current flws n r ut f the () r () termnals n the pamp. educng and rearrangng varables: 9 Gan 3 3 Ths gves us the gan and the mnus sgn dentfes ths as an nvertng amplfer.. In the crcut belw: a) Determne the gan f the amplfer. b) Is t nvertng r nnnvertng?

2 Ans: We use the () termnal f the pamp as a nde agan. The vltage at the () termnal s the same as the vltage n the () termnal,.e.. The ndal equatn s: earrangng and cmbnng: Gan.5 The pstve gan dentfes ths as a nnnvertng amplfer. 3. In the crcut belw: a) Determne the gan f the amplfer. b) Is t nvertng r nnnvertng? Ans: As n Prblem, treatng the () termnal f the pamp as a nde, we see that ts vltage s zer, snce the () termnal s at grund. (emember: = ). Therefre we can wrte the ndal equatn: emember, n current flws n r ut f the () r () termnals n the pamp. educng and rearrangng varables: Gan.

3 Ths gves us the gan and the mnus sgn dentfes ths as an nvertng amplfer.. What wuld have t be t make the gan f ths amplfer? Ans: As n the nvertng pamp prblems abve, we treat the () termnal f the pamp as a nde and see that ts vltage s zer, snce the () termnal s at grund. (emember: = ). Therefre we can wrte the ndal equatn: emember, n current flws n r ut f the () r () termnals n the pamp. educng and rearrangng varables t slve fr : Gan 5. Desgn a nnnvertng amplfer wth a gan f 5. Ans: We wuld use the nnnvertng amplfer tplgy shwn belw:

4 We use the () termnal f the pamp as a nde agan. The vltage at the () termnal s the same as the vltage n the () termnal,.e.. The ndal equatn s: earrangng and cmbnng: Gan 5 Snce, we nly have ne equatn and tw unknwns, we can chse the value f r and then calculate the value f the ther resstr based n the relatn abve. S we chse t be and that makes =. 6. What s n terms f a, b and c? a b Ans: Snce the vltage at the () termnal f the pamp s, the vltage at the () termnal s als. We can use superpstn t buld up the utput relatn by applyng each nput vltage and shrtng ut the ther nputs n turn. T start, we apply a and shrt b and c nputs t grund. Snce the () termnal s als at grund, n current flws thrugh the resstrs t the nputs shrted t grund. They can be gnred n the ndal analyss. (EMEMBE: Yu CANNOT always d ths. ONLY when the vltage s acrss the resstrs.) Usng ndal analyss as n the nvertng amplfers abve, the relatn fr a s then smply:

5 a a Smlarly, we can develp the relatns fr b and c. b b Superpstn gves us the ttal relatn as the sum f the respnses: a b c 7. What s n terms f a and b? a 5 b Ans: We can break the crcut nt tw parts, analyzng the frst amplfer as an nvertng amplfer wth a gan f. Ths places a at the utput f the frst pamp and the nput t the secnd part f the crcut. Usng superpstn, we can carry ths nput t the fnal utput,, recgnzng the secnd part f the crcut as an nvertng amplfer wth gan: ( a ) a 5 Cntnung the superpstn wth b : b b S the ttal respnse s: a b 8. Desgn an pamp crcut t carry ut ths equatn: ( ) a b c Ans: There are a number f crcuts that culd carry ths ut. But the crcut n Prblem 7 gves us the easest slutn:

6 b 5 a The frst pamp s cnfgured as an nvertng amplfer wth a gan f fr bth b and c. The utput f the frst pamp s carred thrugh the secnd pamp wth a gan f ( dvded by 5). S the fnal respnse t b and c s *=. The secnd pamp s als an nvertng amplfer fr a wth a gan f (/)=. S the ttal respnse s: = a ( b c ). Nte: Dfferent valued resstrs culd be used, hwever, the rats must be the same. 9. What s n terms f a, b and c? a b 3 3 Ans: As n the prblem abve, the () termnal f the pamp frms a nde t analyze the crcut. Hwever, the vltage at ths nde s NOT. We lk at the () termnal and see that the 3 hm and hm resstrs frm a vltage dvder wth c. The vltage at the () termnal s then:.5 c ( 3) Ths s als the vltage at the () termnal. We can nw wrte the ndal equatn: a.5 c a.5 c.5 c 3 Slvng fr n terms f the nputs gves:.5.5 a b c

7 . Try prblem 9 wth superpstn. That s, turn n each f the nputs n turn wth the thers shrted t grund, analyze the respnse and sum the resultng utputs. Ans: Startng wth a, we shrt the ther nputs t grund. The key thng t remember here s that the 3 hm and hm resstr have n current thrugh them t the () termnal snce n current can flw n r ut f the nput termnals n an pamp. Ths places the () termnal at grund and cnsequently the () termnal s als at vlts. We can then analyze the crcut as a smple nvertng amplfer frm a. Nte: snce the b nput s grunded and the () termnal s als at vlts, n current flws n the hm resstr t b and t des nt enter nt the calculatn. The gan f the nvertng amplfer s as we saw abve: 3/ r.5. S the respnse t a s.5 a. In the same way, we can shrt a and c and then apply b. The analyss s the same as fr a and the utput due t b s.5 b. Nw, thngs can get dffcult f yu are nt careful. Shrtng a and b t grund and applyng c, we see that the 3 hm and hm resstrs frm a vltage dvder and the vltage at the () termnal s.5 c. Ths vltage s als at the () termnal. The key s t remember that ths vltage causes current t flw n BOTH hm resstrs t the shrted nputs fr a and b. The ttal current s:.5.5 I.5 Ndal analyss says that ths current flws frm the 3 hm resstr cnnected t. Ths gves us the drp acrss the 3 hm resstr and that plus the vltage at the () termnal gves us : 3I.5 c 3(.5 ) c c c c c Tgether wth the results fr the nputs abve, the ttal utput s:.5.5 a b c The same result as befre.