BJT Biasing Cont. & Small Signal Model Conservative Bias Design (1/3, 1/3, 1/3 Rule) Bias Design Example Small-Signal BJT Models Small-Signal Analysis 1
Emitter Feedback Bias Design R B R C V CC R 1 R C V CC V B R E R 2 R E Voltage bias circuit Single power supply version 2
Thevinen Equivalent Use Thevinen's theorem: h =V B = R 2 R 1 R 2 V CC Let: = R 1 R 2 V B = 1 1 V CC R B = R 1 R 2 R 1 R 2 = R 1 1 Since we specify V B and R B, the inverse is needed: 1 1 = V B V CC = V CC V B 1 R 1 = 1 R B R 2 = R 1 3
5.Or choose R T =R 1 R 2 = V CC I 1 = V CC /10 Conservative Bias Procedure: 1. Bias so that V CC is split equally across R C, V CE (or V CB ), and R E. 2. Select desired collector current. 3. Assume I E = to determine R C and R E. 4. Add 0.7 V to V CC /3 to find V B. Assume base current through R B is negligible. 5. Choose R B approximately equal to R E /10. (Use lowest value of 6. Finally, compute R 1 and R 2. 4
Example 50 100 V CC =12V V R E =V RC =4 V =1mA Then: R C =R E = 4 10 3 =4 103 =4 k V B =4 0.7=4.7V R B = min 1 R E 10 = 50 4000 10 For a single power supply: = V CC 1= 12 V B 4.7 1=1.55 R 1 = 1 R B =2.55 20=51k R 2 = R 1 = 51 1.55 =32.9 k =20 k 5
Completed Bias Design Electronics Workbench simulation results Our design differs from the simulation because we neglected the base current. 4.7 V 1 ma There is no point in including the base current, since we will build the circuit using resistors that come only in standard sizes and with 5% tolerances attached to their values. The closest available resistors in the RCA Lab are 47 kω, 33 kω, and 3.3 kω. 6
BJT Small-Signal Models R B i C + - v BE R C V CC Conceptually, the signal we wish to amplify is connected in series with the bias source and is of quite small amplitude. V B R E i C =I S e v BE We will linearize the signal analysis to simplify our mathematics to avoid having to deal with the nonlinear exponential collector current i C vs. v BE characteristic. 7
Linearization Process Split the total base-emitter voltage and total collector current into bias and signal components: v BE V i C = =I S e T =I S e V BE V BE V =I S e T V e T Identify and substitute the bias current into the expression: i C = e Expand the exponential in a Taylor series: f x = n=0 f n 0 xn n! 8
Analysis Using Collector Current Model i C = e i C =e Useful constants: log 10 2 =0.301 log 10 e =0.4343 log 10 i C = log 10 e Let i C be doubled, i.e. i : 0.301= v C / =2 be V 0.4343 T = 0.301 0.4343 0.025 0.017V 2 0 i C time 9
Linearization Using Taylor Series Expand in a Taylor series: For the exponential function: e x = or: e e = n=0 1 1 2 1 n! f x = n=0 n n=0 2 1 6 x n n! f n 0 xn n! 3 where x= non-linear terms 10
Linearization Continued Recall: of about 17 mv causes a 2x change in collector current. Let's expand in Taylor series for this value of. linear approximation = 0.017 0.025 =0.68 V e T =1 0.68 1 2 0.68 2 1 6 0.68 3 V e T 1 0.68 0.2312 0.0524=1.9444 Compare with: e 0.68 =1.973 exact! Four term expansion is accurate to about 1.5%, two term (linear) expansion is accurate to about 15%, i.e. 0.17V may not be sucha small signal 11
Small-Signal Model Using the grouping into bias and signal voltages/currents: i C = e And using the first two terms of the Taylor series expansion: i C 1 1 = = We define transconductance and small-signal (or ac) current as: bias current g m = =g m 12
Small-Signal BJT Model i B =I B = 1 i C = 1 i B = 1 i C= 1 1 Define the small-signal base current and base resistance: = 1 = 1 = 1 bias current = = g m 13
Small-Signal BJT Equivalent Models Equations (1): =g m = = Equations (2): = = 1 = 1 = 1 = r e Equations (3): =g m = = =g m = = r e = 1 = 1 Choose the model that simplifies the circuit analysis. 14
ESE319 Introduction to Microelectronics Small-Signal BJT Equivalent Circuits Equations (1): =g m v ce r 0 = = = I B g m = Equations (2): r e e = v ce = 1 = r e r 0 Equations (3): b c b c b c r π g r 0 i m r 0 c g m v r 0 be e g m = r 0 = V A r e = I E = g m = 1 e =g m v ce r 0 = = 15
b π ESE319 Introduction to Microelectronics Small-Signal Circuit Analysis The small signal model will replace the large signal model and be used for (approximate) signal analysis once the transistor is biased. c g m we may include R S with R B R B R C g m e R E v c Transistor Biased circuit amplifier circuit 16
Biased Circuit Small Signal Analysis R B =20k R C =4k = R B R E =2.5k gm = R B 1 R E R E =4k v c = R B 1 R E Let: =100 and =1 ma = =100 0.025 0.001 =2.5 k Note: = = g m = = 0.001 A =0.04 S=40 ms 0.025 V R B 1 R E S = siemen 17
Small Signal Analysis - Continued v c = = v R B 1 R s E g m =g m = v R B 1 R s E v c = R C = R C R B 1 R E g m = = 100 R E R B v c R C 1 R E A v = v c R C R E 18
Multisim Model of Bias Design R C =4k R B =20k v c V B =4.7V R E =4k V CC =12V A v = v c R C R E = 1 1. R E needs to be large to achieve good op. pt. Stability! 2. Consequence: A v is low. 19
Multisim Input-output Plot A v = v c R C R E = 1 20
Conclusions Conservative voltage bias for best operating point stability and signal swing works, but returns unity or low voltage gain. How does one obtain operating point stability, and simultaneously achieve a more respectable voltage gain? 21