Lecture 10 OUTLINE. Reading: Chapter EE105 Spring 2008 Lecture 10, Slide 1 Prof. Wu, UC Berkeley
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1 Lecture 0 OUTLIN BJT Aplifiers (3) itter follower (Coon-collector aplifier) Analysis of eitter follower core Ipact of source resistance Ipact of arly effect itter follower with biasin eadin: Chapter Sprin 2008 Lecture 0, Slide Prof. Wu, UC Berkeley
2 itter Follower (Coon Collector Aplifier) 05 Sprin 2008 Lecture 0, Slide 2 Prof. Wu, UC Berkeley
3 itter Follower Core When the input is increased by ΔV, output is also increased by an aount that is less than ΔV due to the increase in collector current and hence the increase in potential drop across. Howeer the absolute alues of input and output differ by a V B. 05 Sprin 2008 Lecture 0, Slide 3 Prof. Wu, UC Berkeley
4 Sall Sinal Model of itter Follower V A = out in = rπ β As shown aboe, the oltae ain is less than unity and positie. 05 Sprin 2008 Lecture 0, Slide 4 Prof. Wu, UC Berkeley
5 Unity Gain itter Follower V A A = = The oltae ain is unity because a constant collector current (= I ) results in a constant V B, and hence Vout follows Vin exactly. 05 Sprin 2008 Lecture 0, Slide 5 Prof. Wu, UC Berkeley
6 Analysis of itter Follower as a Voltae Diider V A = 05 Sprin 2008 Lecture 0, Slide 6 Prof. Wu, UC Berkeley
7 itter Follower with Source esistance V A = out in = S β 05 Sprin 2008 Lecture 0, Slide 7 Prof. Wu, UC Berkeley
8 Input Ipedance of itter Follower V i A X X = = r ( β ) π The input ipedance of eitter follower is exactly the sae as that of C stae with eitter deeneration. This is not surprisin because the input ipedance of C with eitter deeneration does not depend on the collector resistance. 05 Sprin 2008 Lecture 0, Slide 8 Prof. Wu, UC Berkeley
9 itter Follower as Buffer Since the eitter follower increases the load resistance to a uch hiher alue, it is suited as a buffer between a C stae and a heay load resistance to alleiate the proble of ain deradation. 05 Sprin 2008 Lecture 0, Slide 9 Prof. Wu, UC Berkeley
10 Output Ipedance of itter Follower s out = β itter follower lowers the source ipedance by a factor of β iproed driin capability. 05 Sprin 2008 Lecture 0, Slide 0 Prof. Wu, UC Berkeley
11 05 Sprin 2008 Lecture 0, Slide Prof. Wu, UC Berkeley itter Follower with arly ffect Since r O is in parallel with, its effect can be easily incorporated into oltae ain and input and output ipedance equations. ( )( ) O s out O S in S O O r r r r r A = = = β β β π
12 Current Gain There is a current ain of (β) fro base to eitter. ffectiely speakin, the load resistance is ultiplied by (β) as seen fro the base. 05 Sprin 2008 Lecture 0, Slide 2 Prof. Wu, UC Berkeley
13 itter Follower with Biasin A biasin technique siilar to that of C stae can be used for the eitter follower. Also, V b can be close to V cc because the collector is also at V cc. 05 Sprin 2008 Lecture 0, Slide 3 Prof. Wu, UC Berkeley
14 Supply Independent Biasin By puttin a constant current source at the eitter, the bias current, V B, and I B B are fixed reardless of the supply alue. 05 Sprin 2008 Lecture 0, Slide 4 Prof. Wu, UC Berkeley
15 Suary of Aplifier Topoloies The three aplifier topoloies studied so far hae different properties and are used on different occasions. C and CB hae oltae ain with anitude reater than one, while follower s oltae ain is at ost one. 05 Sprin 2008 Lecture 0, Slide 5 Prof. Wu, UC Berkeley
16 Aplifier xaple I out in = S β 2 The keys in solin this proble are reconizin the AC round between and 2, and Theenin transforation of the input network. 05 Sprin 2008 Lecture 0, Slide 6 Prof. Wu, UC Berkeley C S
17 Aplifier xaple II out C = S in β 2 Aain, AC round/short and Theenin transforation are needed to transfor the coplex circuit into a siple stae with eitter deeneration. 05 Sprin 2008 Lecture 0, Slide 7 Prof. Wu, UC Berkeley S
18 05 Sprin 2008 Lecture 0, Slide 8 Prof. Wu, UC Berkeley Aplifier xaple III The key for solin this proble is first identifyin eq, which is the ipedance seen at the eitter of Q 2 in parallel with the infinite output ipedance of an ideal current source. Second, use the equations for deenerated C stae with replaced by eq. 2 2 C in A r r = = β π π
19 Aplifier xaple IV A = S The key for solin this proble is reconizin that CB at frequency of interest shorts out 2 and proide a round for. appears in parallel with C and the circuit siplifies to a siple CB stae. 05 Sprin 2008 Lecture 0, Slide 9 Prof. Wu, UC Berkeley C
20 Aplifier xaple V in B = β β 2 The key for solin this proble is reconizin the equialent base resistance of Q is the parallel connection of and the ipedance seen at the eitter of Q Sprin 2008 Lecture 0, Slide 20 Prof. Wu, UC Berkeley
21 Aplifier xaple VI out 2 ro = S in 2 ro β S = r S out 2 O β The key in solin this proble is reconizin a DC supply is actually an AC round and usin Theenin transforation to siplify the circuit into an eitter follower. 05 Sprin 2008 Lecture 0, Slide 2 Prof. Wu, UC Berkeley
22 05 Sprin 2008 Lecture 0, Slide 22 Prof. Wu, UC Berkeley Aplifier xaple VII Ipedances seen at the eitter of Q and Q 2 can be luped with C and, respectiely, to for the equialent eitter and collector ipedances. ( ) B B C B C out B in A r = = = β β β β β π
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