EE 330 Lecture 25. Amplifier Biasing (precursor) Two-Port Amplifier Model

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1 EE 330 Lecture 25 Amplifier Biasing (precursor) Two-Port Amplifier Model

2 Review from Last Lecture Exam Schedule Exam 2 Friday March 24

3 Review from Last Lecture Graphical Analysis and Interpretation 2 OX Device Model (family of curves) I - 1 DQ GS T DS I D μ C W 2L GS6 DD R 1 M 1 GS5 IN (t) GS4 SS GS3 GS2 GSQ =- SS GS1 DS Saturation region Q-Point Load Line μ C W 2L 2 OX I + DQ SS T Signal swing can be maximized by judicious location of Q-point Often selected to be at middle of load line in saturation region

4 Review from Last Lecture Small Signal Model Simplifications G gs B bs g mb bs g m gs g o i d D ds S Simplification that is often adequate G i d D gs g m gs g o ds S

5 Review from Last Lecture Small Signal Model Simplifications G gs B bs g mb bs g m gs g o i d D ds S Even further simplification that is often adequate G gs S i d g m gs D ds

6 Review from Last Lecture Small Signal BJT Model Summary An equivalent circuit B i b i c C be g π g m be g o ce E g m I CQ t g I CQ β t g o I CQ AF g m g π g o This contains absolutely no more information than the set of small-signal model equations

7 Review from Last Lecture Small Signal BJT Model Simplifications B i b i c C be g π g m be g o ce E Simplification that is often adequate B i b i c C be g π g m be ce E

8 Review from Last Lecture Gains for MOSFET and BJT Circuits BJT MOSFET DD CC R R 1 IN (t) IN (t) M 1 EE SS A B I R C t A M 2I R DQ SS T IN R For both circuits A R g m IN M 1 R Gains vary linearly with small signal parameter g m Power is often a key resource in the design of an integrated circuit In both circuits, power is proportional to I CQ, I DQ

9 Gains for MOSFET and BJT Circuits BJT MOSFET DD CC R R 1 IN (t) IN (t) M 1 - EBQ SS TH EE SS A B I R C t A M 2I R DQ EBQ IN R For both circuits A R g m IN M 1 R For good signal swing at output I CQ R<0.5*( CC - EE ) I DQ R<0.5*( DD - SS )

10 How does g m vary with I DQ? g m 2μC L OX W I DQ aries with the square root of I DQ g m 2I GSQ DQ T 2I DQ EBQ aries linearly with I DQ g m μc L OX W GSQ T Doesn t vary with I DQ

11 How does g m vary with I DQ? All of the above are true but with qualification g m is a function of more than one variable (I DQ ) and how it varies depends upon how the remaining variables are constrained

12 Amplifier Biasing (precursor) CC R 1 out in B C E EE Not convenient to have multiple dc power supplies Q very sensitive to EE

13 Amplifier Biasing (precursor) CC CC =12 R 1 R B =500K R 1 =2K out out in B C E in C 1 =1uF B C E EE Not convenient to have multiple dc power supplies Q very sensitive to EE Compare the small-signal equivalent circuits of these two structures

14 Amplifier Biasing (precursor) CC CC =12 R 1 R B =500K R 1 =2K out out in B C E in C 1 =1uF B C E EE Compare the small-signal equivalent circuits of these two structures R 1 R 1 IN IN R B Since Thevenin equivalent circuit in red circle is IN, both circuits have same voltage gain

15 Amplifier Characterization (an example) Determine Q, A, R IN Determine and (t) if IN =.002sin(400t) CC =12 R B =500K R 1 =2K out in C 1 =1uF B C E In the following slides we will analyze this circuit

16 Amplifier Characterization (an example) Biasing Circuit CC =12 R B =500K R 1 =2K IN (t) C=1uF (biasing components: C, R B, CC in this case, all disappear in small-signal gain circuit) Several different biasing circuits can be used

17 Amplifier Characterization (an example) Biasing Circuit CC =12 R B =500K R 1 =2K IN (t) C=1uF Determine Q and the SS voltage gain, assume β=100

18 Amplifier Characterization (an example) Determine Q CC =12 CC =12 IN (t) R B =500K C=1uF R 1 =2K β=100 R B1 =500K 0.6 I B R 2 =2K βi B Q R B1 =500K CC =12 R 2 =2K dc equivalent circuit I CQ = βi BQ = mA 500K = 12-I R =12-2.3mA 2K 7.4 OUTQ C I B simplified dc equivalent circuit

19 Amplifier Characterization (an example) Determine the SS voltage gain CC =12 i B R B =500K R 1 =2K IN R B g m BE BE g π R 1 IN (t) C=1uF β=100 ss equivalent circuit g R OUT m BE 1 IN R B ss equivalent circuit R 1 A IN BE R g 1 m ICQR1 A - t 2.3mA 2K A m This basic amplifier structure is widely used and repeated analysis serves no useful purpose Have seen this circuit before but will repeat for review purposes

20 Amplifier Characterization (an example) Determine R IN CC =12 R B =500K R 1 =2K i IN i B IN (t) C=1uF β=100 IN R B g m BE BE g π R 1 R IN IN Rin i IN IN R B ss equivalent circuit R 1 Rin RB // r Usually R B >>r π Rin RB // r r R in r I CQ β t

21 Examples Determine and (t) if IN =.002sin(400t) R B =500K CC =12 R 1 =2K out A OUT IN 177 OUT.002sin(400 t ) 0.354sin(400 t ) t +A OUT OUTQ IN in C 1 =1uF B C E sin(400 t) OUT

22 Two-Port Representation of Amplifiers CC =12 R B =500K R 1 =2K IN (t) C=1uF R L R 1 IN R B R L i B IN R B g m BE BE g π R 1 R L Two-Port Network Two-port model representation of amplifiers useful for insight into operation and analysis Internal components to the two-port can be quite complicated but equivalent two-port model is quite simple

23 Two-port representation of amplifiers Amplifiers can be modeled as a two-port for small-signal operation 1 y 12 2 y 11 y 22 2 y 21 1 Amplifier often unilateral (signal propagates in only one direction: wlog y 12 =0) One terminal is often common 1 y 11 y 22 2 y 21 1

24 Two-port representation of amplifiers 1 y 11 y 22 2 y 21 1 Thevenin equivalent output port often more standard R IN, A, and R OUT often used to characterize the two-port of amplifiers 1 R OUT A 1 R IN 2

25 Amplifier input impedance, output impedance and gain are usually of interest Example 1: Assume amplifier is unilateral R S Why? IN Amplifier R L R S R OUT IN 1 A 1 R IN 2 R L R L R IN OUT RL RIN OUT = A A IN AMP = = A R L +ROUT R S +RIN IN R L +ROUT R S +RIN Can get gain without recondsidering details about components internal to the Amplifier!!! Analysis more involved when not unilateral

26 Amplifier input impedance, output impedance and gain are usually of interest Why? Example 2: Assume amplifiers are unilateral R S IN Amplifier 1 Amplifier 2 Amplifier 3 R L R S R OUT1 R OUT2 R OUT3 IN 11 A 1 11 R IN A 2 12 R IN A 3 31 R IN3 23 R L R L RIN3 RIN2 RIN1 = A A A R L +ROUT3 R OUT2 +RIN3 R OUT1 +RIN2 R S +RIN1 OUT IN OUT RL RIN3 RIN2 RIN1 A AMP = = A3 A2 A1 IN R L +ROUT3 R OUT2 +RIN3 R OUT1 +RIN2 R S +RIN1 Can get gain without recondsidering details about components internal to the Amplifier!!! Analysis more involved when not unilateral

27 Two-port representation of amplifiers Amplifier usually unilateral (signal propagates in only one direction: wlog y 12 =0) One terminal is often common Amplifier parameters often used I 1 I 2 R OUT 1 y 11 y 22 2 y 21 1 y parameters R IN 1 A 1 2 Two Port (Thevenin) Amplifier parameters Amplifier parameters can also be used if not unilateral One terminal is often common I 1 I 2 R IN R OUT 1 y 12 2 y 11 y 22 2 y 21 1 A R 2 1 A 1 2 Two Port (Thevenin) y parameters Amplifier parameters

28 Determination of small-signal model parameters: y 12 2 y 11 y 22 2 y 21 1 I 1 I 2 R IN R OUT A R 2 Two Port (Thevenin) A 1 In the past, we have determined small-signal model parameters from the nonlinear port characteristics 1,2, I f f, I 1 2 f y ij i 1 j 2 Q Will now determine small-signal model parameters for two-port comprised of linear networks Results are identical but latter approach is often much easier

29 Two-Port Equivalents of Interconnected Two-ports Example: i 1 i 1A i 2A i 2 1A y 11A y 12A 2A y 21A 1A y 22A 2A 1 i 1B i 2B 2 1B y 11B y 12B 2B y 21B 1B y 22B 2B I 1 I 2 R IN R OUT A R 2 1 A 1 2 Two Port (Thevenin) could obtain two-port in any form often obtain equivalent circuit w/o identifying independent variables Unilateral iff A R =0 Thevenin-Norton transformations can be made on either or both ports

30 Two-Port Equivalents of Interconnected Two-ports Example: i 1 i 1A i 2A i 2 1A y 11A y 12A 2A y 21A 1A y 22A 2A R XX I 1B I 2B 1 R 1 1B 2B R B g21b2b g12b1b g11b g22b 2 Two Port (Norton) i 1C v 1C H-parameters (Hybrid Parameters) h i h v 1C 11C 1C 12C 2C h 22 h i i1 v 2 21C C 2 C C C i 2C v 2C Linear Two Port I 1 I 2 R IN R OUT A R 2 1 A 1 2 Two Port (Thevenin)

31 RB i1c v1c R1 H-parameters (Hybrid Parameters) 1 C h11 C i1 C h12 Cv2C i 2C h21 C i1 C h22 Cv 2C Linear Two Port i2c v2c I1B 1B g11b g21b2b g12b1b Two Port (Norton) RXX g22b I2B 2B Determination of two-port model parameters (One method will be discussed here) i1a y12a2a 1A y11a y21a1a y22a i2a 2A A method of obtaining R in test i test i 1 1 R in i 2 A vr2 R o A v01 2 Terminate the output in a short-circuit 1 AR i Rin Rin A0 1 i2 1 2 R0 R0 i i test test R in i test test

32 RB i1c v1c R1 H-parameters (Hybrid Parameters) 1 C h11 C i1 C h12 Cv2C i 2C h21 C i1 C h22 Cv 2C Linear Two Port i2c v2c I1B 1B g11b g21b2b g12b1b Two Port (Norton) RXX g22b I2B 2B Determination of two-port model parameters i1a 1A y11a y12a2a y21a1a y22a i2a 2A A method of obtaining A 0 test i 1 1 R in i 2 A vr2 R o A v01 2 out-test Terminate the output in an open-circuit 1 AR i1 1 2 Rin Rin A0 1 i2 1 2 R0 R0 i test out-test A out-test 0 test

33 RB i1c v1c R1 H-parameters (Hybrid Parameters) 1 C h11 C i1 C h12 Cv2C i 2C h21 C i1 C h22 Cv 2C Linear Two Port i2c v2c I1B 1B g11b g21b2b g12b1b Two Port (Norton) RXX g22b I2B 2B i1a i2a 1A y11a y22a 2A Determination of two-port model parameters y12a2a y21a1a A method of obtaining R 0 i 1 1 R in i 2 A vr2 R o A v01 2 i test test Terminate the input in a short-circuit 1 AR i1 1 2 Rin Rin A0 1 i2 1 2 R0 R0 1 0 R 0 i test test

34 RB i1c v1c R1 H-parameters (Hybrid Parameters) 1 C h11 C i1 C h12 Cv2C i 2C h21 C i1 C h22 Cv 2C Linear Two Port i2c v2c I1B 1B g11b g21b2b g12b1b Two Port (Norton) RXX g22b I2B 2B Determination of two-port model parameters i1a i2a A method of obtaining A R 1A y11a y12a2a y21a1a y22a 2A out-test i 1 1 R in i 2 A vr2 R o A v01 2 test Terminate the input in an open-circuit 1 AR i1 1 2 Rin Rin A0 1 i2 1 2 R0 R0 i1 0 A out-test R test

35 Determination of Amplifier Two-Port Parameters Input and output parameters are obtained in exactly the same way, only distinction is in the notation used for the ports. Methods given for obtaining amplifier parameters R in, R OUT and A for unilateral networks are a special case of the non-unilateral analysis by observing that A R =0. In some cases, other methods for obtaining the amplifier paramaters are easier than what was just discussed

36 Examples Biasing Circuit CC =12 R B1 =50K R 2 =2K IN (t) C=1uF R B2 =10K R 1 =0.5K Determine Q and the SS voltage gain (A ), assume β=100 (A is one of the small-signal model parameters for this circuit)

37 Examples CC =12 CC =12 R B1 =50K R 2 =2K R B1 =50K C=1uF R 2 =2K IN (t) C=1uF R B2 =10K R 1 =0.5K IN (t) R B2 =10K R 1 =0.5K IN R B 10K//50K R 1 0.5K R 2 2K Determine Q and the SS voltage gain (A ), assume β=100 (A is one of the small-signal model parameters for this circuit)

38 Examples Determine Q CC =12 CC =12 R B1 =50K I BB R 2 =2K Q IN (t) R B1 =50K C=1uF B R B2 =10K I BB I B R 2 =2K R 1 =0.5K β=100 B 0.6 R B2 =10K R 1 =0.5K I B βi B dc equivalent circuit CC =12 This circuit is most practical when I B <<I BB With this assumption, R B1 =50K B R B2 =10K R 2 =2K I Q BB I B simplified R 1 =0.5K dc equivalent circuit RB2 B = R B1+R B B I CQ = I EQ = = = 2.8mA R 1.5K OUTQ C =12-I R = 6.4 Note: This Q-point is nearly independent of the characteristics of the nonlinear BJT!

39 Examples Determine SS voltage gain CC =12 R B1 =50K R 2 =2K IN (t) i B R 1 =0.5K R 2 =2K IN (t) C=1uF B R B2 =10K I BB I B R 1 =0.5K β=100 IN (t) i B BE g π R B =10K//50K g m BE 8.3K R 1 =0.5K R 2 =2K This voltage gain is nearly independent of the characteristics of the nonlinear BJT! This is a fundamentally different amplifier structure It can be shown that this is slightly non-unilateral A g R OUT m BE 2 R g +g IN BE 1 BE m R2gm BE R2gm R g +g R g +g 1 BE 1 BE m 1 m A R g R g R R 2 m 2 1 m 1

40 Examples Biasing Circuit CC =12 R B1 =50K R 2 =2K C 1 =1uF IN (t) R B2 =10K R 1 =0.5K C 2 =100uF Determine Q, R IN, R OUT,and the SS voltage gain, and A R assume β=100

41 Examples CC =12 R B1 =50K R 2 =2K CC =12 C 1 =1uF IN (t) R B1 =50K C 1 =1uF R B2 =10K R 2 =2K R 1 =0.5K C 2 =100uF IN R B2 =10K R 1 C 2 0.5K 100uF R IN R OUT IN A R 2 R 1 A 1 0.5K Two Port (Thevenin) Determine Q, R IN, R OUT,and the SS voltage gain, and A R ; assume β=100 (A, R IN, R OUT, and A R are the small-signal model parameters for this circuit)

42 Examples Determine Q CC =12 R B1 =50K R 2 =2K This is the same as the previous circuit! C 1 =1uF OUTQ = 6.4 IN (t) R B2 =10K CC =12 R 1 =0.5K C 2 =100uF 5.6 I CQ = 2.8mA 2K R B1 =50K R 2 =2K I BB Q B R B2 =10K I B R 1 =0.5K Note: This Q-point is nearly independent of the characteristics of the nonlinear BJT! The dc equivalent circuit

43 Examples Determine the SS voltage gain CC =12 This is the same as another previous-previous circuit! IN (t) R B1 =50K C 1 =1uF R B2 =10K R 2 =2K R 1 =0.5K C 2 =100uF A -g R m 2 I A - R CQ 2 t IN R B1 //R B2 R A m The SS equivalent circuit Note: This Gain is nearly independent of the characteristics of the nonlinear BJT!

44 IN (t) Examples R B1 =50K C 1 =1uF R B2 =10K CC =12 R 2 =2K R 1 =0.5K Determination of R IN β=100 C 2 =100uF The SS equivalent circuit R B1 //R B2 R 2 IN i IN i b i c IN be g π g m be g o ce R 2 R B1 //R B2 R R //R //r r IN B1 B2 π π -1 1 I CQ 2.8mA r π = 928 βt m RIN R B1//R B2 //rπ rπ 930

45 Examples Determination of R OUT R B1 =50K CC =12 R 2 =2K The SS equivalent circuit R B1 //R B2 IN (t) C 1 =1uF R B2 =10K R 1 =0.5K β=100 C 2 =100uF IN R 2 i b i c i TEST be g π g m be g o ce R 2 TEST R B1 //R B2 R TEST OUT =R 2//ro i TEST 1 1 ICQ 2.8mA 1 ro 1.4E-5 71K AF 200 R R //r R OUT 2 o 2 2K

46 Examples Determine A R CC =12 The SS equivalent circuit R B1 =50K R 2 =2K R B1 //R B2 IN (t) C 1 =1uF R B2 =10K R 1 =0.5K β=100 C 2 =100uF IN R 2 i b i c TEST be g π g m be g o ce R 2 TEST R B1 //R B2 TEST =0 A R 0

47 Determination of small-signal two-port representation CC =12 R B1 =50K R 2 =2K IN (t) C 1 =1uF R B2 =10K R 1 =0.5K C 2 =100uF 1 R OUT A 1 R IN 2 A 215 RIN rπ 930 ROUT R2 2 This is the same basic amplifier that was considered many times K

48 End of Lecture 25 End of Slide 30

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