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 28Sep09
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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
VB
RC
VCC
IE
RE
2009 Kenneth R. Laker, update 28Sep09
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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 =0.7V.
Basic relationships:
I E =I B  I C =1 I B
1
1
I E=
I C= I C


2009 Kenneth R. Laker, update 28Sep09
and VB and VCC are given. Then
compute RE to obtain the desired
collector current IC:
V B −0.7
R E =
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 28Sep09
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.
2009 Kenneth R. Laker, update 28Sep09
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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
2009 Kenneth R. Laker, update 28Sep09
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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!
I C=I S e
2009 Kenneth R. Laker, update 28Sep09
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 28Sep09
Writing IC as a function of 
β:
V B −0.7

I C =
=
⋅10−3 A
RE
1
Assume: 50200
50 −3
200 −3
⋅10 =0.980 mAI C 
⋅10 =0.995 mA
51
201
So IC is insensitive to changes in β!
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ESE319 Introduction to Microelectronics
Voltage Source With Internal Resistance
V B =I B R B 1 R E I B V BE
IC
IB
RC
RB
VB
VCC
IE
RE
I E =1 I B
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
2009 Kenneth R. Laker, update 28Sep09
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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 is not possible, but can be approximated quite
well if the voltage source resistance is “not too large,”
i.e.
R B ≪1 R E
Ideal rule of thumb (if possible):
R B≤
1
100
1 R E ≈
R E =10 R E
10
10
2009 Kenneth R. Laker, update 28Sep09
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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 28Sep09
RC
IE
VCC
RE
Single power supply version
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ESE319 Introduction to Microelectronics
Thevenin Equivalent
RTh
<=>
VTh
RTh =R B =R 1∥R 2
R2
V Th =V B=
V CC
R 1R 2
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
Thevenin Equivalent
Use Thevenin's theorem to simplify
base circuit:
R2
V Th =V B=
V CC
R 1R 2
I B ≈0
R1
Let: =
R2
V B=
1
V CC
1
R1 R2
R1
RTh =R B =R 1∥R 2 =
=
R1R 2 1
2009 Kenneth R. Laker, update 28Sep09
Since we will specify VCC, VB
and RB, the inverse is needed
for design:
VB
1
=
1 V CC
V CC
=
−1
VB
R1 =1 R B
Design
Eqs.
R1
R2=

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ESE319 Introduction to Microelectronics
A “Rule of Thumb” for Single Supply Biasing
IC
I1
R1
RC
I B ≈0
R2
+
VB
1. Choose RT so that the current 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 R 2  RT
R2
V B =V CC
RT
2009 Kenneth R. Laker, update 28Sep09
V CC I C
I 1=
=
RT 10
V CC
RT = R1 R2 =10
IC
2. Use a voltage divider to give the desired
base voltage:
VB
R 2= RT
V CC
Solve for R1
VB
R1 =RT − R2 =RT 1−

V CC
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ESE319 Introduction to Microelectronics
An Unavoidable Design Tradeoff
R1
R2
IC
RC
V CB+ +
IE
VB
-
Two design goals so far
+
V Rc
R B ≪1 R E
V B ≫V BE
V CC
Constraint: V CC =V RcV CB V B
RE
TRADEOFF
● Increase V => Reduce V
+ VCB
B
Rc
● V
large => large signal swing & large voltage gain
Rc
(before cutoff).
● V
(or VCE) large => large voltage swing
CB
(before saturation).
● NEED A COMPROMISE!
●
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
RT = R1
R1
R2
IC
Another Useful “Rule of Thumb”
“1/3, 1/3, 1/3 Rule”
+R
2
RC
+
V Rc
V CB+ +
IE
VB
-
V CC
V B=
3
V CC
RE
V B =V BE I E R E
V BE =0.7 V
V Re =I E R E
VB >> VBE
2009 Kenneth R. Laker, update 28Sep09
V CC
V Rc =I C RC =
3
V CC
V CB =V CC −V Rc −V B =
3
V CC
or V CE =V Rc−V Re ≈
3
where V CC =V Rc V CB V B
Design Equations
V Rc V CC
RC =
=
I C 3 IC V
CC

−0.7V 
V B −V BE 
3
R E=
=
IE
IE
R2
V CC
V CC
=V B =
R 1R 2
3
V CC I C
I 1=
=
RT 10
VB
R 2 =RT
V CC
VB
R1 =RT 1−

V CC
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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
2. A diode-connected transistor sets the
current.
V CC −V BE1
I REF =
Rref
I REF =I C1 I B1I B2≈ I C1≈ I E1
3. Choose Rref for the desired
current:
V CC −0.7 V CC −0.7
1. A BJT collector is the curR ref ≈
≈
rent source:
I E1
I C1
V CE1=V BE1
V BE2=V BE1
I C =I S e
V BE
VT
2009 Kenneth R. Laker, update 28Sep09
4. If Q1 = Q2 then IC2 = IC1 => I O≈ I ref
matched
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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 ≈
R ref
VCE1 ≠ VCE2, the Early Effect needs to be included in simulations.
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
Constant Emitter Current – Early Voltage
Rref
V CC
Assume Q1 = Q2
I ref
I C1
I C2 =I O
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 CE
I B2= e
=
 =1


 F where F
VA
For Q1:
I ref = I C1 I B1 I B2 = I C1 2 I B
V BE /V T
BE
T
2009 Kenneth R. Laker, update 28Sep09
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

VT 
Sub (2) into (1)
V CE2
1
VA
I O = I C2 = I ref
V BE 2
1

VA 
BE
T
(2)
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ESE319 Introduction to Microelectronics
Constant Emitter Current – Early Voltage Cont.
I O = I C2 = I ref
V CE2
1
VA
V BE 2
1

VA 
=>
IO
=
I ref
1
1
V CE2
VA
V BE 2

VA 
IO
1
=
≈1
Let V A=∞ => Early effect is negligible I
2
ref
1

If also =∞ => I O = I ref
Let V A = finite and VA = 50 V, VBE = 0.7 V, =100
IO
10.02V CE2
= f V CE2 =
=0.9710.02 V CE2 
I ref
1.034
2009 Kenneth R. Laker, update 28Sep09
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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 28Sep09
V CC
iE
RC
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
Summary
Two practical methods for achieving stable
bias for a BJT are:
●
●
●
Use a voltage source in the base with a feedback resistance in the emitter circuit.
Place a current source directly in the emitter circuit.
2009 Kenneth R. Laker, update 28Sep09
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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 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.
or
3-2. Use the 1/3, 1/3, 1/3 rule.
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” Bias Design
I 1=
IC
10
R1
RC
V CC
R2
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=6V
IC
V CC
12
I 1 = =0.1 mA ⇒ R1 R2 =
= −4 =120 k 
10
I 1 10
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” Bias Design - Continued
I1 = 0.1 mA
R1
RC =6 k 
V CC
R2
RE
I C =1 mA =100
R1 R2=120 k 
Let's choose a small feedback voltage,
say 1 V. Ignoring the base current:
1V
R E=
=1 k 
IC
=12V
Then the voltage across R2 is 1.7 V
1.7 V
R 2= −4 =17 k 
V B =1.7V
10 A
R1 =120k−17k=103 k 
Recall: R B ≤
1
1 R E
10
103
1
R B =R1∥R 2=
⋅17≈14.5 k  1 R E ≈10 k 
120
10
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
RECALL: Voltage Source With Internal Resistance
V B = I B R B 1 R E I B V BE
IC
IB
RC
RB
VB
V CC
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
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” β Sensitivity
IC
RC
RB
VB
IB
V CC
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 = 50;
IC = 0.775 mA
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
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
“Emitter-Feedback 3-1” Bias Scilab Simulation
//Rule of thumb BJT bias sensitivity
Beta=100;
VsubT=0.025;
VB=1.7;

I C=
V B −V BE 
Rb=14.5;
R B 1 R E
BetaPlusRe=101;
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
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
Scilab Plot (Zoomed)
−18
I S =10
−11
I S =10
 I C ≈0.4 mA
2009 Kenneth R. Laker, update 28Sep09
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ESE319 Introduction to Microelectronics
Compare with Ideal IC vs IS Results Plot
−11
I S =10
−18
I S =10
 I C ≈0.1 mA
R B =0
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