ES 330 Electronics II Homework # 5 (Fall 2016 Due Wednesday, October 4, 2017)

Transcription

1 Pag1 Na olutions E 33 Elctonics II Howok # 5 (Fall 216 Du Wdnsday, Octob 4, 217) Pobl 1 (25 pots) A coon-itt aplifi uss a BJT with cunt ga = 1 whn biasd at I =.5 A. It has a collcto sistanc of = 1 k. (a) alculat, o and AO. W bg with calculatg g g A and I.5 A 2 A/ (o units of ).25 TH 1 5kΩ g 2 A/ 5kΩ 1 kω g 2 A/ 1 kω = -2 o (b) Th aplifi is d with a nal souc has sistanc = 1 k and has a load sistanc L = 1 k connctd to th put tal. alculat both A and. W now clud th load sisto, 1 kω kω 1 kω L A A L 5kΩ A ( 1 ) kΩ 1 kω L (c) If th pak-to-pak oltag of a susoidal nal acoss th bas-itt junction is liitd to 5, what nal souc oltag is allowd?

2 Pag2 Fo 5, oltag souc can b found usg 5kΩ 5kΩ 1 kω 5kΩ 5 kω 1 kω Hnc, ; 5 35 = 15 5kΩ 1 kω 5 kω (d) ontug on with pat (c) abo, what pak-to-pak put oltag appas acoss th load sisto L? ospondgly, th put oltag will b =.5 Pobl 2 (15 pots) A MOFET connctd a coon-souc configuation has tansconductanc g = 5 A/ (). Whn a souc sisto is connctd to th souc tal, th ffcti tansconductanc g is ducd to 2 A/ (). Estiat th alu of this sisto. [Not: ouc sisto is not bypassd with a capacito.] With th us of a souc dgnation sisto, w ha a duction aplifi ga, o quialntly th ffcti tansconductanc, g. 1 This duction facto is always of th fo, 1 g, wh g dic tansconductanc with th sisto. In sybols, g ffcti 1 g g 1g 1g 5 W a told that g 2 ; olg fo 1 (5 ).3 kω 3 ffcti is th Pobl 3 (2 pots) A bipola coon-bas (B) aplifi is opatd with a load sistanc of L = 1 k, a collcto sistanc of = 1 k and a nal souc sistanc of = 5.

3 Pag3 (a) What collcto cunt I will gi an put sistanc of = 5? Fo both and qual to 5, w qui = 5,wh and with 1(assus y high ), thn I (Not: b that 1/ g ) I TH 25.5 A 5 (b) What is th sultg oall oltag ga assug 1 (i.., unity)? To calculat th ga w nd to know g, I g 2 A/; W know that g TH L 5 2 A/1 k 1 k A/5 k 5 2 (c) Fo th coon-bas aplifi shown schatically blow, Fd th put oltag o if >> and 1. [b that is th itt sistanc ( pag 44 of da & ith).] TH

4 Pag4 W a told that is uch gat than ( ) That ans that that alost all of cunt i th tansisto. In sybols, i Th collcto cunt is thn i i i i flows to th itt of (th itt cunt). c Thus th put oltag i i c Pobl 4 (2 pots) An itt follow (i.., coon-collcto) stag is opatg at a collcto cunt bias of.5 A and is usd to connct a 1 k souc sistanc to a 1k load sistanc. Assu th noal cunt ga of th tansisto to b 1. (a) What put sistanc is obtad at th itt follow s put and what is its oall oltag ga? Lookg to th itt, w ha fo TH TH 25 5 I I.5 A E 1, L L 1 1 (b) If is ducd to 5 (fo 1), fd th cospondg and alus.

5 Pag5 W know fo pat (a) that E 1, L L 1 TH TH 25 5 ; But now = 5 (not 1 pat (a)) I I.5 A and 1 Pobl 5 (2 pots) Two idntical coon-itt ga stags a cascadd togth. Th collcto tal sistanc is idntical fo both stags, naly, 1 = 2 = 1 k. Th fist stag is d fo oltag souc hag souc sistanc = 1 k. A load sistanc of L = 1 k is connctd to th collcto of th scond stag. Both BJT tansistos (i.., idntical stags) a biasd at I =.25 A and both ha cunt ga = 1. But kp d th loadg sistos a diffnt btwn stags. (a) ktch th sall-nal quilalnt cicuit fo th oall aplifi. Labl th ipotant oltags. Th quialnt cicuit (usg hybid- odl) is shown blow: 1 k 1 2 _ = 1 k 1-1 g g L o - (b) alculat th oall oltag ga, = /. W bg by calculatg th tansconductanc g and E put sistanc. W know that I =.25 A and = 1. W also not that fo tansistos 1 & 2 that g and ( g = g & )

6 Pag6.25 A g 1 g2 g 1 A/ o.1 A/ kω 1, g.1 A/ Nxt, w st up th quation fo th oall oltag ga. Usg th quialnt cicuit fo pat (a) abo, k.5 1 k 1 k 1 k1 k g g g 1 k 1 k 2 1 1( 1 2) k 1 k1 k 2 g2 2( 2 L) g2 2 g 2(5 k ) 1 k 1 k obg ths quations by substitution to gt / yilds g g 5 k 5 k.5 W fd that g 5 k (.1 A/ 5, ) 5, thfo, , 25