At point G V = = = = = = RB B B. IN RB f

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1 Common Emitter

2

3

4 At point G CE RC I C RC 116. R 1k C 116. ma I IC 116. ma β 100 F 116µ A I R ( 116µ A)( 20kΩ) 2. 3 R IN R f

5 Gain in Constant Current Region I I I C F β ( IN F ) R I R OUT CC C C R OUT CC C β ( ) R F IN F d d OUT IN R R C β F

6 FET Inverter (Common Source - CS)

7

8

9 At Point C DS 55. RTH i D RTH 2. 0 R 500 TH 4 ma i k( ) D GS TR 2 GS id + k TR 4 ma + 2 1mA /

10 Output Gain in Constant Current Region I k( ) D IN TR 2 R I OUT TH TH D R k( ) 2 TH TH IN TR d d OUT IN 2R k( ) TH IN TR

11 Emitter Follower Input and output loops share load element x IN OUT x OUT Feedback

12

13 Emitter Follower I X R F X IN OUT I R ( I + I ) R ( 1+ β ) I R OUT E E C E F E Assuming Constant Current Operation I R IN OUT F I ( 1+ β ) I R R IN F E F [ R + ( 1+ β ) I R ] I F E IN F

14 I R IN F + ( 1+ β ) R F E R OUT ( )( 1+ β ) R R + ( 1+ β ) R << ( 1+ β ) R F E IN F F E F E IN CE > F > SAT OUT IN F Current Gain of Emitter Follower I I + I LOAD C I + β I F ( 1+ β ) I F I I IN

15 I I LOAD IN ( 1+ β ) F "Current Gain" R IN IN I R IN IN F + ( 1+ β ) R F E R IN IN R R IN F [ + ( 1+ β ) ] F E

16 FET Follower OUT I S R S I D R S GS IN - OUT IN - I D R S GS changes with I D (feedback) can use iterative graphical approach to solve

17 R DD IN TR S 15 8 k 05. ma / 2 1kΩ 2

18 Common ase (C) (Tracking Configuration)

19 I I I + I IN E C ( 1+ β ) I F I IIN ( 1+ β ) F β F IC β F I I ( 1+ β ) F IN I R OUT CC C L β F CC RL I ( 1 + β ) F IN

20 Common ase Transfer Function

21 Cascode Configuration

22 iasing Allows non-linear elements to be treated as linear elements (under certain conditions) Different techniques for Discrete and Integrated designs Avoid non-linearities of -I characteristics by choosing a portion of the curve over which the device will operate

23

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33 For JT of Fig R 1kΩ R 100 kω β CC C F F with A 0 ias levels I β I β C F F ( F ) ( ) R 100 kω 5mA I R 10 ( 5mA)(1k Ω) 5 CE CC C C Nonlinear Regions egin at 10 I 0 CE CC C (Cutoff) CE SAT CE SAT 0. 2 IC ma (Saturation) R C

34 Minimum A A MIN A MIN A MIN Maximum I A MAX IC ( SAT ) 9. 8 ma β ma + A MAX F R I MAX I R + A MAX MAX R F A MAX A MAX ( ma)(100 k)

35 Fixed oltages and Current iasing A simple way of eliminating a separate biasing source is to replace it with one of the power supply buses ias values are adjusted by selecting proper resistors in the input loop

36

37 10 R 1kΩ β 100 CC C F Find R that will result in bias value C 5 I R C CC C C I I C 5mA IC β F 0. 05mA I CC R F R CC I F kω

38 GS R A R + R DD 0. 5MΩ MΩ

39 Assume transistor in constant current region ( ) I k( ) 1mA / 4 2 D GS TR 4 ma I R DS DD D D 16 ( 4 ma)(2 kω) 8 i.e. > ( ) DS GS TR Constant Current Note: No current through gate of transistor

40 Parameter Independent iasing Previous biasing techniques are sensitive to device parameters such as β F, k and TR which are in turn sensitive to temperature and fabrication variances. One configuration called feedback biasing is virtually independent of device parameters

41

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43 I R ( I + I ) R E I I + I I R ( I + I ) R 2 CC 2 C 2 1 E Input loop and output loop share the voltage drop (I 2 + I 1 )R E. This feedback mechanism is responsible for stabilizing bias levels against device parameters. Difficult to use graphical analysis to analyze circuit Example: Given 2, k 0.5 ma /, R 1MΩ, R 2 MΩ, R 1kΩ, R TR A D E 5kΩ 2 (A) Find I, () Find I if k is changed to 1 ma / 2 D D

44

45 Solution: R A R + R DD 2 MΩ MΩ + 2 MΩ ( ) Assume FET operates in the constant current region (must confirm later). I k( ) 2 D GS TR I k( I R ) 2 D D E TR

46 I D RE I D + 2RE ( TR ) TR 0 k + ( ) I 2 2 D ( 5kΩ) 25I 2 D 1 I D + 2( 5kΩ)( 8 2 ) mA / 62 I D + ( 8 2) 0 2 Applying quadratic formula I D 0. 93A I 155. ma D

47 First value gives 3.36, second value gives GS GS 0.24 (not valid cutoff) I ( R + R ) DS CC D D E Note: mA ( 1kΩ + 5kΩ) > DS GS TR Therefore constant current region 2 If k 10. ma / then 3, I 10. ma GS 100% change in k results in only a 7.5% change in I D D

48 JT Feedback ias

49 Problem: Find the vlaue of I C if βf varies from 50 to 200. F 0.7 CC R2 R + R kω kω + 20 kω 4 R R R kω

50 β F 50: I C 50( ) 6.67 kω + 51( 1kΩ) ma β F 200: I C 200( ) 6.67 kω + 201( 1kΩ) 318. ma β Changes by a factor of 4 I Changes by 11 % C

51 Same problem without R E 4 R kω I R F I R F kω ma

52 β 50: I β I 50( ma) 24. 5mA F C β 200: I β I 200( ma) 98 ma F C β Changes by a factor of 4 I Changes by a factor of 4 C Note: Current gain without R is greater (i.e. Sacrifice some gain for stability) E

53 iasing with ipolar Supplies The use of bipolar supplies can improve bias designs Facilitates DC - coupled input signals Allows outputs to be set to a bias level of zero In some cases can reduce the number of resistors in the bias circuit A bipolar supply bus is formed by positive and negative DC voltage sources each connected to a common ground

54

55 For a periodic AC waveform with no DC component the transistor input looks like a DC ground Applying KL F + I E RE + EE 0 For large βf I E ( EE F IC ) R E I R R C CC C C CC + ( EE + F ) R C E E F

Biasing the CE Amplifier

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