BJT Biasing Cont. & Small Signal Model Conservative Bias Design Bias Design Example Small Signal BJT Models Small Signal Analysis 1
Emitter Feedback Bias Design 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 I C /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 = I C to determine R C and R E. 4. Add 0.7 volts to V CC /3 to find V BG. Assume base current through R B is negligible; hence V B V BG 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 RE =V RC =4V. I C =1 ma. Then: R C =R E = 4 10 3 =4 103 =4 k V B =4 0.7=4.7V. R B = 1 R min E = 50 4000 =20 K 10 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 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 this 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 47k, 33k, and 3.3k. 6
BJT Small Signal Models Conceptually, the signal we wish to amplify is connected in series with the bias source and is of quite small amplitude. We will linearize the signal analysis to simplify our mathematics to avoid having to deal with the nonlinear exponential collector 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 C =I S e T =I S e V BE =I S e V BE V e T Identify and substitute the bias current into the expression: i C =I C e Expand the exponential in a Taylor series: I C f x = n=0 f n 0 xn n! 8
Analysis Using Collector Current Model i C =I C e i C I C =e log 10 i C I = v be C log 10e Let i C be doubled, i.e. i : 0.301= v C / I C =2 be V 0.4343 T = 0.301 0.4343 0.025 0.017V. Useful constants: log 10 2 =0.301 log 10 e =0.4343 9
Linearization Using Taylor Series Expand in a Taylor series: For the exponential function: e x = or: e = n=0 V e T 1 1 2 1 v n be n! 2 f x = n=0 n=0 1 v 3 be 6 f n 0 xn n! x n n! 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. = 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 Four term expansion is accurate to about 1.5%, two term expansion is only accurate to about 15%, i.e. 0.17V is not sucha small signal 11
Small Signal Model Using the grouping into bias and signal voltages/currents: i C =I C e And using the first two terms of the Taylor series expansion: i C I C 1 1 =I C I C =I C We define transconductance and incremental (or ac) current as: bias current g m = I C =g m 12
Incremental (small-signal) BJT Model i B =I B = 1 i C= 1 I C i B = 1 i C= 1 I C 1 I C V T Define the incremental base current and base resistance: = 1 = 1 I C V = 1 v T r be bias current = I C = g m 13
Equivalent Models Equations (1): =g m = i e = Equations (2): = = i e 1 i e = 1 = 1 = r e Equations (3): =g m = i e = =g m = I C I C = r e = 1 = 1 I C Choose the model that simplifies the circuit analysis. 14
b Equations (1): r π r 0 e =g m v ce r 0 g m = I C i e = ESE319 Introduction to Microelectronics Equivalent Circuits b Equations (2): Equations (3): g m g m v r 0 be r 0 = V A I C = = I B g m i e = c i e r e e = v ce = i e 1 i e = r e r 0 r e = I E = g m = 1 r 0 c b e =g m v ce r 0 = i e i e = c 15
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. we may include Rs with Rb b c π g m e i e v c Transistor Biased circuit (and all bypass caps are approx short circuits) 16
Biased Circuit Small Signal Analysis v sig = R B i e R E v sig = R B 1 R E i e v c = v sig R B 1 R E Let: =100 and I C =1 ma = =100 0.025 =2.5 k I C 0.001 Note: g m = I C = 0.001 A =0.04 S=40 ms 0.025 V = = v R B 1 R sig E S = siemen 17
Small Signal Analysis - Continued i e v c = = v R B 1 R sig E g m =g m = v R B 1 R sig E R v c = R C = C v R B 1 R sig E g m = I C I C = 100 R E R B v c R C 1 R E v sig A v = v c v sig R C R E 18
Multisim Model of Bias Design A v = v c v sig 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 v sig R C R E = 1 20
Conclusions Conservative voltage bias for best operating point stability and signal swing works, but returns unity voltage gain. How does one obtain operating point stability, and simultaneously achieve a respectable voltage gain? 21