ESE319 Introduction to Microelectronics. BJT Biasing Cont.


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1 BJT Biasing Cont. Biasing for DC Operating Point Stability BJT Bias Using Emitter Negative Feedback Single Supply BJT Bias Scheme Constant Current BJT Bias Scheme Rule of Thumb BJT Bias Design 1
2 Simple Base Biasing Schemes What's wrong with these schemes 2
3 Another Simple Biasing Scheme R C R B I E Is this scheme any better If it is, why 3
4 Biasing for Operating Point Stability A practical biasing scheme must be insensitive to changes in transistor β and operating temperature! Negative feedback is one solution. I B R C I E 4
5 Given the (DC bias) equations: I S e E V T I E = I E = R C E Assume: E =0.7 V Basic relationships: I E =I B = 1 I B I E = 1 = 1 Let: α,, and be specified; Compute: in terms and : = E = 0.7 5
6 Negative Feedback via Negative feedback makes the collector current insensitive to E, I S, and β. If increases due to an increase in I S then E will decrease; thus, limiting the magnitude of the change in. The equations that must be satisfied simultaneously are: =I S e E V T I S insensitivity and E = = 1 if I S => when => E I S insensitivity <= E If = 0 then = E => feedback is absent 6
7 Negative Feedback via =I S e E V T and E = = 1 β insensitivity = E = E 1 β insensitivity E insensitivity = E if >> E => E insensitivity 7
8 Scilab Analysis of Insensitivity to I S Simultaneous equations: =I S e E V T = E Let =4k or: E =I S e E and =4.7V (want >> E ) V T If we plot the exponential function and the straight line function, the solution values of and E for the circuit occur at their intersection. 8
9 Scilab Program //Calculate and plot npn BJT bias characteristic beta=100; alpha=beta/(beta+1); VsubT=0.025; VTinv=1/VsubT; VBB=4.7; Re=4; vbe=0.0:0.01:1; icline=alpha*(vbbvbe)/re;//ma. plot(vbe,icline); ic=0.01:0.01:2; //ma.! IsubS =1E16; //ma. for k= 1:1:8 IsubS=10*IsubS; vbe2=vsubt*log(ic/isubs); plot(vbe2,ic); end //Current in ma. = E E =V T ln I S 9
10 vs I S Results Plot I S =10 11 I S =10 18 = E 0.1mA Insensitive to I S! =I S e E V T 10
11 = 0.7 Example: =100 =4.7V =1 ma = = ESE319 Introduction to Microelectronics Insensitivity to Beta Writing as a function of β: = 0.7 = A Assume: =0.990 ma =0.995 ma CONCLUSION: is insensitive to changes in β! 11
12 Nonzero Base Resistance I B R C R B I E I E = 1 I B = I B =I B R B E 1 I B I B = E R B 1 = I B = V R B 1 R B E E If R B 1 1 E I E = E If R B 1 E = 1 R B = I B R B E Effect of Feedback is minimized! 12
13 Biasing for Operating Point Stability Quick Review What is the purpose What does R B represent What is the condition on the value of What is the condition on the value of What is our rule of thumb for achieving x >> y 13
14 Observations Emitter feedback stabilizes base voltage source bias. To reduce the sensitivity of to E, choose E. R B 0, emitter feedback stabilization works well if R B is not too large, i.e. R B 1 Ideal rule of thumb (if possible): Let R B 1 or 10 1 R B = =100 R B = =10 14
15 EmitterFeedback Bias Design R 1 R C I B R I E Voltage bias circuit Single power supply version 15
16 Thevenin Equivalent I 1 <=> R Th V Th I B I 1 R Th =R B =R 1 R 2 V Th = = R 2 R 1 R 2 16
17 EmitterFeedback Bias Design What is the purpose of What is the purpose of R 1 & R 2 What happened to R B How are & related What is the condition on the values of R 1, R 2 & What is the condition on the values of I 1, I B & I 1 17
18 Using Thevenin Equivalent Lets assume that I 1 >> I B & I 1 <<. Furthermore assume, & R B are independently specified. Solve for resistors R 1 & R 2. R B =R 1 R 2 = R 1 R 2 R 1 R 2 = R 2 R 1 R 2 (2) (1) R B = R 1 R 2 R 1 R 2 R 1 =R B = R 2 R 1 R 2 R 2 =R 1 Can, & R B be independently specified; e.g. R B = =10E 18
19 Can, & R B be independently specified; e.g. =10E R B = I 1 = R T = 10 VC More intuitive! Independently specify, & s.t. R B << (β + 1) & >> E 19
20 1st Rule of Thumb for Single Supply Biasing I 1 R 2 R 1 I B +  R C R 2 R T Iff I 1 >> I 1 V I B CC R 1 R 2 = R T 1. Choose R T = R 1 + R 2 so that I 1 >> I B & I 1 <<, i.e. I 1 is about 1/10 of the desired & I B is ignored when compared to I 1 ): R T =R 1 R 2 =10 I 1 = = R T Use a voltage divider to give the desired base voltage and solve for R 1 and R 2 : R 2 =R T R 1 =R T R 2 =R T 1 3. Calculate R B : R B =R 1 R 2 20
21 I 1 R 1 R 2 + V R C V CE  I E + ESE319 Introduction to Microelectronics An Unavoidable Design Tradeoff V RC V RE = E With, & specified, 3 design constraints, i.e. where: TRADEOFF R B 1 E I 1 also I 1 I B =V RC V CE V RE Increase V RC and/or V RE => Reduce V CE V RC too large => possible saturation & reduced operating range (i.e. V CE > 0.2 V or smaller). V RC too small => possible cutoff & reduced operating range (i.e. > small or zero). NEED A COMPROMISE! ( fixed) 21
22 I 1 R T = R 1 + R 2 R 1 ESE319 Introduction to Microelectronics 2nd Useful Rule of Thumb 1/3, 1/3, 1/3 Rule V RE = 3 V RC = R C = 3 V CE = where =V RC V RE V V CE = R 2 & V V CE  CC R CC T + I E If >> E => V RC V CE R 2 V  R C + V RC =E V RE I 1 R 2 E =0.7V Design Equations R C = V R C = 3 = V I E = 3 I E = 3 I 1 = 10 R 2 =R T R T 1 3 R T If >> E V RE =I E R 2 V RE = V R BE = T 3 R 1 =R T R T 22
23 Constant Emitter Current Bias R C R B Q amp I REF I E I E Specified: & I E E1 E2 23
24 ESE319 Introduction to Microelectronics Constant Emitter Current Bias The current mirror is used to create a current source: 1 R ref I REF I I O 2. A diodeconnected transistor sets the current. I REF = E1 R ref V CE1 =E1 I B1 1. A BJT collector is the current source: =I S e E V T I B2 E2 =E1 I REF =1 I B1 I B2 1 I O =2 3. Choose R ref for the desired current: Rref = I REF 1 4. If Q1 = Q2 matched and V CE2 = V CE1 then 2 = 1 => I O I ref 24
25 Constant Emitter Current Now: E1 = E2 If Q 1 and Q 2 have the same saturation current: I S1=I S2 And the transistors are at the same temperature: T 1 =T 2 The two collector currents set primarily by R ref are equal, as long as Q2 is not saturated. I O =2 =I ref 0.7 R ref Since V CE1 V CE2, the Early Effect needs to be included in simulations. 25
26 Constant Emitter Current Early Voltage R ref 1 I REF I B2 Assume Q1 = Q2 2 =I O I REF =I S e E /V T 1 V CE1 V A 2 I S e E V T For Q2: I B1 E2 =E1 =V CE1 =E Early Effect = I S e E /V T 1 E V A 2 I S e Solving for I S e E /V T I O =2 =I S e E /V T 1 V CE2 I V S e E /V T (1) = A Since I B is not effected by V A, i.e. 1 E 2 (2) V A I B2 = I S ee /V T = 2 F = 1 V CE2 Sub (2) into (1) where F V V CE2 A 1 For Q1: I B1 = I V A B2 I O =2 =I REF I REF =1 I B1 I B2 =1 2 I B 1 E 2 V A I REF E V T Ideally = 1 26
27 Constant Emitter Current Early Voltage Cont. I O =2 =I REF 1 V CE2 V A 1 E V A 2 => I O I REF = 1 V CE2 V A 1 E V A 2 I REF Let V A = => Early effect is negligible I O I REF = If also = => I O =I REF Let V A = finite and V A = 50 V, E = 0.7 V, =100 I O = f V E2 = V CE2 = V REF CE2 27
28 Emitter Current Bias Quick Review What is the circuit in the green box What is the relation between R ref and I REF I REF What is the approximate relation between I REF and I E Does E1 = E2 E1 E2 Does the Early Voltage effect the relation between I REF and I E How What is the output resistance of Q2 28
29 Emitter Current Bias Quick Review What is the circuit in the green box Current mirror to realize a fixed, stable current source. I REF What is the relation between R ref and I REF E1 E2 What is the approximate relation between I REF and I E Does E1 = E2 Does the Early Voltage effect the relation between I REF and I E How What is the output resistance of Q2 29
30 Emitter Current Bias Quick Review What is the circuit in the green box Current mirror to realize a fixed current source. I REF What is the relation between R ref and I REF I REF = E1 R ref E1 E2 What is the approximate relation between I REF and I E Does E1 = E2 Does the Early Voltage effect the relation between I REF and I E How What is the output resistance of Q2 30
31 Emitter Current Bias Quick Review What is the circuit in the green box Current mirror to realize a fixed current source. What is the relation between R ref and I REF I REF I REF = E1 R ref E1 E2 What is the approximate relation between I REF and I E I E I REF Does E1 = E2 Does the Early Voltage effect the relation between I REF and I E How What is the output resistance of Q2 31
32 Emitter Current Bias Quick Review What is the circuit in the green box Current mirror to realize a fixed current source. What is the relation between R ref and I REF I REF I REF = E1 R ref What is the approximate relation between I REF and I E I E I REF Does E1 = E2 Yes E1 E2 Does the Early Voltage effect the relation between I REF and I E How Yes, V CE2 V CE1 What is the output resistance of Q2 32
33 Emitter Current Bias Quick Review What is the circuit in the green box Current mirror to realize a fixed current source. What is the relation between R ref and I REF I REF I REF = E1 R ref What is the approximate relation between I REF and I E I E I REF Does E1 = E2 Yes E1 E2 Does the Early Voltage effect the relation between I REF and I E How Yes, V V CE2 CE1 What is the output resistance of Q2 r o = V A ' ' = I S e v BE /V T 33
34 BJT Emitter Current Source Bias R C R B We thus can use a current mirror to provide stable control of transistor collector current. R ref sets the emitter and collector currents and the collectorground voltage for Q amp. I ref v in R ref C Q amp i E v in is the ac input voltage source. R ref v in R B i E R B can be any reasonable value this is not voltage biasing! Q 1 Q 2 34
35 Summary Two practical methods for achieving stable bias for a BJT are: Use a dc voltage source in the base with a feedback resistance in the emitter circuit. Place a dc current source directly in the emitter circuit. 35
36 EmitterFeedback Bias Design 1. Use single supply for base bias and collector sources. 2. Use the /10 rule for the current I 1 through the base bias network (R 1 and R 2 ) Try less negative feedback using a smaller emitter resistor saving more of the supply voltage for V CE & the R C voltage drop V R C. or 32. Use the 1/3, 1/3, 1/3 Rule. 36
37 I 1 = 10 ESE319 Introduction to Microelectronics Let's Try EmitterFeedback 31 Bias Design R 2 R 1 R C V C Complete the bias design given the following design values: =12V =100 =0.99 It follows: =1 ma V C =6 V & E = 0.7 V R C = V Rc = V C =6k V I Rc =6V C I 1 = 10 =0.1 ma R R = 1 2 = 12 I =120 k 37
38 EmitterFeedback 31 Bias Design  Continued I 1 = 0.1 ma Previous slide R 1 R C Let's choose a small feedback voltage, say V Re = 1 V (i.e. V CE = 5 V). Ignoring the base current: = V Re I E = 1V / =990 R 2 Then the voltage across R 2 is V R 2 = 1.7V 10 4 A =17 k =1mA R C = 6 kω R 1 R 2 =120 k R 1 =120k 17k=103 k Recall: R B R B =R 1 R k R 9.9 k E B = 1.7 V =1.7V 38
39 EmitterFeedback 31 β Sensitivity R B I B R C = 12V = 1.7 V E = 0.7 V = 1 ma β = 100 R C = 6kΩ = 1 kω R B = 14.5 kω = I B = R B 1 E β = 100; β = 200; β = ; β = 100 & R B = 0 Ω; = ma = ma = 1 ma = 0.99 ma 39
40 Scilab Plot (Zoomed) I S =10 11 I S = ma 40
41 32 ( 1/3, 1/3, 1/3 Rule ) Bias Design Recall: 4 R B Let's choose V Rc = = 3 =4V R C = V Rc = 4V 10 3 A =4 k = 4V / =3.96 k The voltage across R 2 is = V + V = 4.7 V BE R 2 = 4.7 V 10 4 A =47 k R 1 =120k 47k=73 k R B =R 1 R k k 41
42 RECALL: Bias Stability Condition Argument =I B R B E 1 I B I B R B I E R C I B = = I B = E R B 1 R B 1 E I E = 1 I B if = I B R B 1 1 E = E 42
43 EmitterFeedback 32 β Sensitivity R B I B R C = 12V = 1.7 V E = 0.7 V = 1 ma β = 100 R C = 4 kω = 3.96 kω R B = 28.6 kω = I B = R B 1 E EmitterFeedback 31 β = 100; β = 200; β = ; β = 100 & R B = 0 Ω; = ma = ma = 1 ma = 1 ma 43
44 Compare to vs I S Results Plot with 32 Emitter Feedback I S =10 11 I S = ma Recall for 31 Emitter Fdbk 0.4 ma 44
45 Conclusion 1/3, 1/3, 1/3 Rule Provides a Good Compromise Base Voltage Bias Scheme for circuits implemented with discrete BJTs. Emitter current bias scheme using a current mirror is an effective bias scheme for BJT circuits implemented as ICs. 45
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