ESE319 Introduction to Microelectronics
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
●
2009 Kenneth R. Laker, update 21Sep12
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ESE319 Introduction to Microelectronics
Simple Base Biasing Schemes
What's wrong with these schemes?
2009 Kenneth R. Laker, update 21Sep12
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ESE319 Introduction to Microelectronics
Another Simple Biasing Scheme
RC
RB
V CC
IE
Is this scheme any better? If it is, why?
2009 Kenneth R. Laker, update 21Sep12
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ESE319 Introduction to Microelectronics
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.
IC
IB
RC
VCC
VB
IE
RE
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ESE319 Introduction to Microelectronics
Given the (DC bias) equations:
I C≈ I S e
V BE
VT
IC
V BE =V B −R E I E =V B − R E

Assume: V BE =0.7 V
Basic relationships:
I E = I B I C =1 I B
1
1
I E=
I C= I C


2009 Kenneth R. Laker, update 21Sep12
Let: VB, VCC and IC be specified;
Compute: RE to obtain IC:
V B −V BE
V B −0.7
R E =
=
IC
IC
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ESE319 Introduction to Microelectronics
Negative Feedback via RE
●
●
●
Negative feedback makes the collector current insensitive to VBE, IS, and β.
If IC increases – due to an increase in IS then VBE will decrease; thus, limiting the magnitude of the change in IC.
The equations that must be satisfied simultaneously are:
I C =I S e
V BE
VT
and
if IS => IC
when IC => VBE
IC <= VBE
if VB >> VBE
IC
1
V BE =V B −R E =V B −
RE I C


VB
=> I C ≈
RE
2009 Kenneth R. Laker, update 21Sep12
IS insensitivity
 V B −V BE 
V B −V BE
I C=
=
1 R E
RE
β insensitivity
VBE insensitivity
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ESE319 Introduction to Microelectronics
Scilab Analysis of IC Insensitivity to IS
Simultaneous equations:
I C= I S e
V BE
VT
V B −V BE
I C =
RE
or:
V B −V BE

=I S e
RE
V BE
VT
Let R E =4k  and V B =4.7V (want VB >> VBE)
If we plot the exponential function and the straight line function,
the solution values of IC and VBE for the circuit occur at their
intersection.
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ESE319 Introduction to Microelectronics
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;
V B −V BE
vBE=0.0:0.01:1;
I C =
RE
iCline=alpha*(VBB-vBE)/Re;//mA.
plot(vBE,iCline);
iC=0.01:0.01:2; //mA.!
IsubS =1E-16; //mA.
IC
for k= 1:1:8
V BE =V T ln
IS
IsubS=10*IsubS;
vBE2=VsubT*log(iC/IsubS);
plot(vBE2,iC); //Current in mA.
end
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ESE319 Introduction to Microelectronics
IC vs IS Results Plot
−11
I S =10
−18
I S =10
V B −V BE
I C =
RE
 I C ≈0.1 mA
Insensitive
to IS!
IC=I S e
2009 Kenneth R. Laker, update 21Sep12
V BE
VT
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ESE319 Introduction to Microelectronics
Insensitivity to Beta
V B −0.7
R E =
IC
Example: =100
V B =4.7 V
I C =1 mA
100
=
≈0.99
101
4.7−0.7
R E =0.99⋅
10−3
R E ≈4000 
2009 Kenneth R. Laker, update 21Sep12
Writing IC as a function of 
β:
V B −0.7

I C =
=
⋅10−3 A
RE
1
Assume: 100 ≤  ≤ 200
100 −3
200 −3
⋅10 =0.990 mA I C 
⋅10 =0.995 mA
101
201
CONCLUSION:
IC is insensitive to changes in β!
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ESE319 Introduction to Microelectronics
Non-zero Base Resistance
IC
IB
RC
RB
VB
VCC
IE
RE
I E =1 I B
V B =I B R B V BE 1 I B R E
IE
V B −V BE
I B=
R B 1 R E

I C = I B =
V −V BE 
R B 1 R E B
If R B ≪1 R E
V B −V BE 
 V B −V BE 
I C = I B ≈
=
RE
RE
1
If R B ≫1 R E

I C = I B ≈ V B −V BE  no feedback!
RB
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ESE319 Introduction to Microelectronics
Biasing for Operating Point Stability –
Quick Review
What is the purpose RE?
What does RB represent?
What is the condition on the
value of VB?
What is the condition on the
value of RE?
What is our rule of thumb for
achieving x >> y?
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ESE319 Introduction to Microelectronics
Observations
●
●
●
Emitter feedback stabilizes base voltage source bias.
To reduce the sensitivity of IC to VBE, choose V B ≫V BE .
RB ≠ 0, emitter feedback stabilization works well if RB is
“not too large,” i.e.
R B ≪1 R E
Ideal rule of thumb (if possible):
1
1
R B ≤ 1 R E or R B = 1 R E
10
10
Let
1=100 R B=
2009 Kenneth R. Laker, update 21Sep12
100
R E =10 R E
10
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ESE319 Introduction to Microelectronics
Emitter-Feedback Bias Design
R1
IC
IB
+
R2
VB
-
Voltage bias circuit
2009 Kenneth R. Laker, update 21Sep12
RC
IE
VCC
RE
Single power supply version
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ESE319 Introduction to Microelectronics
Thevenin Equivalent
I1
RTh
<=>
VTh
I B ≪ I 1≪ I C
R Th=R B =R1∥R 2
R2
V Th =V B =
V CC
R1 R 2
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ESE319 Introduction to Microelectronics
Using Thevenin Equivalent
Use Thevenin's theorem to simplify
base circuit:
R1 R 2
(1)
RTh =R B =R 1∥R 2 =
R 1R 2
I1
R1 R2
R1 R 2=
I B ≈0
RB
R2
V Th=V B =
V CC (2)
R 1R 2
R2 RB
V CC
V B=
V CC ⇒ R 1=R B
R1 R2
VB
R2
V B=
V CC ⇒ R1 V B R 2 V B=R 2 V CC
R 1R 2
VB
R 2=R 1
V CC −V B
2009 Kenneth R. Laker, update 21Sep12
If VCC, VB and RB are independently specified then R1
and R2 are found by solving
Eqs. (1) and (2):
V CC
R1 =R B
VB
VB
R 2=R 1
V CC −V B
Design
Eqs.
Can VCC, VB and RB be independently specified?
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ESE319 Introduction to Microelectronics
Using Thevenin Equivalent
If VCC, VB and RB are independently specified then R1 and R2 are
found by solving Eqs. (1) and (2):
V CC
R1 =R B
VB
VB
R 2=R 1
V CC −V B
Design
Eqs.
RECALL: VB >> VBE and R B ≪1 R E
Can VCC, VB and RB be independently specified?
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ESE319 Introduction to Microelectronics
1st “Rule of Thumb” for Single Supply Biasing
IC
I1
R1
RC
I B ≈0
R2
+
VB
1. Choose RT = R1 + R2 so that I1 << IC, i.e.
I1 is about 1/10 of the desired collector (or
emitter) current (ignoring IB):
V CC
RE
-
V CC
V CC
I 1=
=
 R1  R2  R T
R2
V B =V CC
RT
2009 Kenneth R. Laker, update 21Sep12
V CC I C
I 1=
=
RT 10
V CC
RT =R1 R 2=10
IC
2. Use a voltage divider to give the desired
base voltage VB and solve for R1 and R2:
VB
R 2= RT
V CC
VB
R1 =RT −R 2= RT 1−

V CC
3. Calculate RB: R B =R 1∥R 2
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ESE319 Introduction to Microelectronics
An Unavoidable Design Tradeoff
R1
IC
RC
B
+
+
VR
+
V CE
-
IE
R2 V R R
E
E
-
V R =V B −V BE
E
Three design goals so far
R B ≪1 R E
V B ≫V BE
I 1 ≪ I C also I 1 ≫ I B
C
V CC
Constraint: V CC =V R V CE V R
●TRADEOFF
and/or VR => Reduce VCE
● Increase V
R
C
C
●
E
E
(VCC fixed)
VR too large => possible saturation & reduced opC
erating range (i.e. VCE -> 0.2 V or smaller).
VR too small => possible cutoff & reduced operating
C
range (i.e. IC -> small or zero).
● NEED A COMPROMISE!
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ESE319 Introduction to Microelectronics
R T = R1
I1
R1
2nd Useful “Rule of Thumb”
“1/3, 1/3, 1/3 Rule”
+R
2
+
VR
IC
RC
+
V CC
V CC
V R = I C RC =
VR =
3
3
C
E
C
+
V CE
-I
V CC
V CC
V CE =
3
where V CC =V R V R V CE
C
E
E
R2 V R R
E
Design Equations
E
-
V R V CC
RC =
=
IC 3 IC
V R V CC  V CC
RE =
=
=
I E 3 I E 3 IC
R2
V CC
V R =V CC
−V BE =
RT
3
C
V B =V BE V R ≈ I 1 R 2
V BE =0.7V
V R = I E R E ≈V B
E
E
VB >> VBE
2009 Kenneth R. Laker, update 21Sep12
E
E
I C V CC
I 1= ≈
10 RT
VB 1
R 2=R T
≈ RT
V CC 3
VB
2
R1 =RT 1−
≈ R T
V CC 3
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ESE319 Introduction to Microelectronics
Biasing for Operating Point Stability –
Quick Review
I1
R1 V R
VB
What is the condition on the value of VB?
IC
RC
C
What is the condition on the value of RE?
V CE
IB
V CC
IE
R2 V R R
E
What is the purpose for R1 & R2?
What happened to RB?
E
What is the constraint on the value of I1?
V CC =V R V CE V R
C
E
V CC ≈V R V CE V B
What is the “1/3, 1/3, 1/3 Rule”?
C
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ESE319 Introduction to Microelectronics
Constant Emitter Current Bias
The current mirror is used to create a current source:
I REF
IIO
Rref
V CC
I C1
I B2
I B1
V CE1 =V BE1
V BE2=V BE1
1. A BJT collector is the current source:
I C= I S e
V BE
VT
2009 Kenneth R. Laker, update 21Sep12
2. A diode-connected transistor sets the
current.
V CC −V BE1
I REF =
R ref
I REF =I C1I B1I B2≈ I C1≈ I E1
I O =I C2
3. Choose Rref for the desired
current:
V CC −0.7 V CC −0.7
R ref =
≈
I REF
I C1
4. If Q1 = Q2 then IC2 = IC1 => I O≈ I ref
matched
and VCE2 = VCE1
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ESE319 Introduction to Microelectronics
Constant Emitter Current
Now: VBE1 = VBE2
If Q1 and Q2 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 Rref – are
equal, as long as Q2 is not saturated.
V CC −0.7
I O = I C2 = I ref ≈
Rref
Since VCE1 ≠ VCE2, the Early Effect needs to be included in simulations.
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ESE319 Introduction to Microelectronics
Constant Emitter Current – Early Voltage
Rref
V CC
Assume Q1 = Q2
I REF
I C2 =I O
I C1
I B2
V BE2 =V BE1=V CE1=V BE
I B1
Early Effect
For Q2:
V CE2
I O = I C2= I S e
1

(1)
VA
Since IB is not effected by VA, i.e.
I S V / V I C2
V CE2
I B2 = e
=
 =1


F where F
VA
For Q1: IB1 = IB2
I REF = I C1  I B1 I B2 = I C1 2 I B
V BE / V T
BE
T
2009 Kenneth R. Laker, update 21Sep12
V BE
I REF = I S e
V BE /V T
V CE1
IS V
1
2 e
VA

T
V BE
= ISe
V BE /V T
V BE
IS V
1
2 e
VA

T
V BE /V T
Solving for I S e
I REF
V /V
ISe
=
V BE 2
1

VA 
Sub (2) into (1)
BE
T
1
I O = I C2= I REF
(2)
V CE2
VA
V BE 2
1

VA 
Ideally
=1
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ESE319 Introduction to Microelectronics
Constant Emitter Current – Early Voltage Cont.
V CE2
1
VA
I O = I C2= I REF
V BE 2
1

VA 
=>
IO
I REF
1
=
1
V BE 2

VA 
IO
Let V A =∞ => Early effect is negligible I
REF
If also =∞ => I O = I REF
V CE2
VA
=
1
2
1

I REF
≈1
Let V A = finite and VA = 50 V, VBE = 0.7 V, =100
IO
10.02V CE2
= f V CE2 =
=0.9710.02V CE2 
I REF
1.034
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ESE319 Introduction to Microelectronics
BJT Emitter Current Source Bias
RB
We thus can use a current mirror
to provide stable control of transistor collector current. Rref sets
the emitter and collector currents
and the collector-ground voltage
for Qamp.
vin
2009 Kenneth R. Laker, update 21Sep12
V CC
iE
RRrefC
I ref
Q amp
RB
vin is the ac input voltage source.
RB can be any “reasonable” value
– this is not voltage biasing!
RC
Rref
Q1
v in
VB
iE
V CC
Q2
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ESE319 Introduction to Microelectronics
Emitter Current Bias Quick Review
What is the circuit in the green box?
What is the relation between Rref and IREF?
IREF
What is the approximate relation between
IREF and IE?
Does VBE1 = VBE2?
VBE1
VBE2
Does the Early Voltage effect the relation
between IREF and IE? How?
What is the output resistance of Q2?
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ESE319 Introduction to Microelectronics
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.
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ESE319 Introduction to Microelectronics
Emitter-Feedback Bias Design
1. Use single supply for base bias and collector sources.
2. Use the IC /10 rule for the current I1 through the base bias
network (R1 and R2).
3-1. Try less negative feedback using a smaller emitter
resistor RE “saving” more of the VCC supply voltage for the
RC voltage drop VRC.
or
3-2. Use the “1/3, 1/3, 1/3 Rule”.
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ESE319 Introduction to Microelectronics
Let's Try “Emitter-Feedback 3-1” Bias Design
I 1=
IC
10
R1
RC
V CC
R2
VB
RE
VC
Complete the bias design given
the following design values:
=100
V CC =12 V I C =1 mA V C =6 V
It follows:
V Rc V CC −V C
RC =
=
=6 k 
IC
IC
V Rc =6 V
IC
V CC
12
I 1= =0.1 mA ⇒ R1 R2 =
= −4 =120 k 
10
I 1 10
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” Bias Design - Continued
I1 = 0.1 mA
R2
R1
VB
Let's choose a small feedback voltage, say
VRe = 1 V. Ignoring the base current:
V Re 1V
RE=
≈
=1 k 
IE
IC
V CC
RC
RE
Previous I C =1 mA R = 6 kΩ
C
slide
R1 R2 =120 k 
Then the voltage across R2 is 1.7 V
1.7V
R 2= −4 =17 k 
V B =1.7V
10 A
R1 =120k−17k=103 k 
Recall: R B ≤
1
1 R E
10
1
R B =R 1∥R 2≈14.5 k   1 R E ≈10 k 
10
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ESE319 Introduction to Microelectronics
3-2 (“1/3, 1/3, 1/3 Rule”) Bias Design
V CC
Let's choose V Rc =V B =
=4 V
3
V Rc
4V
RC=
=
=4 k 
I C 10−3 A
4
4V
R E=
=4 k 
IC
The voltage across R2 is VB = VRE + VBE = 4.7 V
4.7 V
R 2= −4 =47 k 
10 A
R1=120k−47k=73 k 
1
Recall: R B ≤ 1 R E
10
1
R B =R1∥R 2≈28.6 k  1 R E ≈33 k 
10
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ESE319 Introduction to Microelectronics
RECALL: Bias Stability Condition Argument
V B = I B R B V BE 1 R E I B
IC
IB
RC
RB
VB
V CC
IE
RE
I E =1 I B
V B −V BE
I B=
R B 1 R E

I C = I B =
V B −V BE 
R B 1 R E
if R B ≪1 R E
V B −V BE 
 V B −V BE 
I C = I B≈
=
RE
RE
1
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” β Sensitivity
IC
RC
RB
VB
IB
RE
RE
VCC = 12V
VB = 1.7 V
VBE = 0.7 V
IC = 1 mA
β = 100
RC = 6kΩ

I C = I B =
V B −V BE 
R B 1 R E
RE = 1 kΩ
RB = 14.5 kΩ
β = 100;
IC = 0.873 mA
β = 200;
IC = 0.932 mA
β = ∞;
IC = 1 mA
β = 100 & RB = 0 Ω;
IC = 0.99 mA
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” Bias Scilab Simulation
//3-1
/RuleBias
of thumb
BJTwith
bias
sensitivity
Scheme
Re=1
K
Beta=100;
Sensitivity to IS
VsubT=0.025;
VB=1.7;

Rb=14.5;
I
=
V B −V BE 
C
Re=1;
BetaPlusRe=101;
R B 1 R E
vBE=0.0:0.01:1;
iCline=Beta*(VB-vBE)/(Rb+BetaPlusRe);//mA.
plot(vBE,iCline);
iC=0.01:0.01:2; //mA.!
IsubS =1E-16; //mA.
for k= 1:1:8
IsubS=10*IsubS;
vBE2=VsubT*log(iC/IsubS);
plot(vBE2,iC); //Current in mA.
end
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ESE319 Introduction to Microelectronics
Scilab Plot (Zoomed)
−18
I S =10
−11
I S =10
 I C ≈0.4 mA
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ESE319 Introduction to Microelectronics
Compare to IC vs IS Results Plot with
Ideal Emitter Feedback
−11
I S =10
−18
I S =10
 I C ≈0.1 mA
R B ≈0
or
R B ≪1 R E
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ESE319 Introduction to Microelectronics
Conclusion
“1/3, 1/3, 1/3 Rule” Provides a Good Compromise Base
Voltage Bias Scheme
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