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ESE319 Introduction to Microelectronics
12009 Kenneth R. Laker, update 23Sep11
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
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ESE319 Introduction to Microelectronics
22009 Kenneth R. Laker, update 23Sep11
Previous Simple Base BiasingSchemes
What's wrong with these schemes?
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ESE319 Introduction to Microelectronics
32009 Kenneth R. Laker, update 23Sep11
Biasing for Operating Point StabilityA practical biasing scheme must be insensitive to
changes in transistor and operating temperature!Negative feedback is one solution.
IC
IB
IE RE
RC
VCC
VB
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ESE319 Introduction to Microelectronics
42009 Kenneth R. Laker, update 23Sep11
ICISe
VBE
VT
VBE=VBRE IE=VBREIC
Basic relationships:
IE=IBIC=1IB
IE=1IC= 1
IC
Given the (DC bias) equations:
Assume: VBE=0.7Vbe given andI
Cbe specified. Then
computeRE
to obtain the specified
collector currentIC:
RE=VB0.7
IC
, Let: VB,VCC
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ESE319 Introduction to Microelectronics
52009 Kenneth R. Laker, update 23Sep11
Negative feedback makes the collector current insensitive to VBE,I
S, and.
IfIC
increases due to an increase inIS
then VBE
will decrease; thus, limit-
ing the magnitude of the change inIC.
The equations that must be satisfied simultaneously are:
Negative Feedback via RE
IC=VBVBE
1RE=
VBVBERE
VBE=VBREIC
=VB
1
RE ICIC=ISe
VBE
VT and
ifIS
=>IC whenIC => VBE
IC> V
BE VBE
insensitivityICVB
RE=>
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ESE319 Introduction to Microelectronics
62009 Kenneth R. Laker, update 23Sep11
Scilab Analysis of IC Insensitivity to IS
Simultaneous equations:
IC=ISe
VBE
VT
IC=VBVBE
RE
or: VBVBERE
=ISe
VBE
VT
If we plot the exponential function and the straight line function,the solution values ofI
Cand V
BEfor the circuit occur at their
intersection.
Let RE=4k VB=4.7Vand (want VB >> VBE)
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ESE319 Introduction to Microelectronics
72009 Kenneth R. Laker, update 23Sep11
Scilab Program
//Calculate and plot npn BJT bias characteristicbeta=100;alpha=beta/(beta+1);VsubT=0.025;VTinv=1/VsubT;
VBB=4.7;Re=4;vBE=0.0:0.01:1;iCline=alpha*(VBB-vBE)/Re;//mA.plot(vBE,iCline);iC=0.01:0.01:2; //mA.!IsubS =1E-16; //mA.
for k= 1:1:8IsubS=10*IsubS;
vBE2=VsubT*log(iC/IsubS);plot(vBE2,iC); //Current in mA.
end
IC=VBVBE
RE
VBE=VTlnIC
IS
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ESE319 Introduction to Microelectronics
82009 Kenneth R. Laker, update 23Sep11
IC
vs IS
Results Plot
IS=1011
IS=1018
IC0.1
mA InsensitivetoIS!
IC=VBVBE
RE
IC=ISe
VBE
VT
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ESE319 Introduction to Microelectronics
92009 Kenneth R. Laker, update 23Sep11
RE=VB0.7
IC
Example: =100
VB=4.7V
IC=1mA
RE=0.994.70.7
103
=100
1010.99
IC=VB0.7
RE
=
1103A
WritingIC
as a function of:
Assume: 100200
100
101103=0.990mAIC
200
201103=0.995mA
SoIC
is insensitive to changes in !
RE4000
Insensitivity to Beta
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ESE319 Introduction to Microelectronics
102009 Kenneth R. Laker, update 23Sep11
Voltage Source With Internal Resistance
IB
IC
VB=IB RBVBE1IBRE
IB=VBVBE
RB1RE
IE=1IB
IC=I
B=
RB1REV
BV
BE
IC=IB
1
VBVBE
RE=
VBVBE
RE
If RB1RE
IE
RC
RE
RB
VB
VCC
If RB1RE
IC=IB
RBVBVBE no feedback!
IE
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ESE319 Introduction to Microelectronics
112009 Kenneth R. Laker, update 23Sep11
Biasing for Operating Point Stability
Quick ReviewWhat is the purposeR
E?
What doesRB
represent?
What is the condition on thevalue ofV
B?
What is the condition on the
value ofRE?
What is our rule of thumb forachievingx >> y?
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ESE319 Introduction to Microelectronics
132009 Kenneth R. Laker, update 23Sep11
Emitter-Feedback Bias Design
Voltage bias circuit Single power supply version
R1
R2
VCC
RC
RE
IC
IE
IB
VB
+
-
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ESE319 Introduction to Microelectronics
142009 Kenneth R. Laker, update 23Sep11
RTh=RB=R1R2
RTh
VTh
VTh=VB=R
2
R1R
2
VCC
Thevenin Equivalent
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ESE319 Introduction to Microelectronics
152009 Kenneth R. Laker, update 23Sep11
Thevenin EquivalentUse Thevenin's theorem to simplifybase circuit:
RTh=RB=R1R2=R1R2
R1R
2
= R1
1
Since we will specify VCC
, VB
andRB
, the inverse is needed
for design:
1
1=
VB
VCC=
R1
R2
=VCC
VB1
R1=1RB
VB=1
1VCC
Let:
R2=R
1
DesignEqs.
IB0
VTh=VB=R
2
R1R
2
VCC
determine
determineR1
R1R
2
R1R
2
=R
1
1R
2=
R1
determineR
2
I1
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ESE319 Introduction to Microelectronics
162009 Kenneth R. Laker, update 23Sep11
A Rule of Thumb for Single Supply Biasing1. ChooseR
T= R
1+ R
2so that I
1
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172009 Kenneth R. Laker, update 23Sep11
IC
R1
R2
RC
RE
VCCI
E
VCB+
-
VB
+
-
+
-
VRc
Three design goals so far
RB1REVBVBE
Constraint: VCC=VRcVCBVBTRADEOFF Increase V
B=> Reduce V
Rc+ V
CB
VRc
too large => potential for saturation to reduce op-
erating range (i.e. vC < vB).V
Rctoo small => potential for cutoff to reduce operat-
ing range (i.e. iC
-> small or zero).
NEED A COMPROMISE!
An Unavoidable Design Tradeoff
I1IC also I1IB
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ESE319 Introduction to Microelectronics
182009 Kenneth R. Laker, update 23Sep11
Another Useful Rule of Thumb1/3, 1/3, 1/3 Rule
VB=VCC
3VRc=IC RC=
VCC
3VCB=VCCVRcVB=
VCC
3
VCC=VRcVCBVB
RC=VRc
IC=
VCC
3IC
RE=VBVBE
IE=
VCC
3 0.7V
IE
VCCR
2
R1R
2
=VB=VCC
3
Design Equations
where
IC
R1
R2
RC
RE
VCCIE
VCB+
-
VB
+
-
+
-
VRc
VB=V
BEI
EREI
1R
2
VBE=0.7V
VB
>> VBE
VCE=VRcVReVCC
3
VRe=IE RE
or
I1=
VCC
RT=
IC
10
R2=RT V
B
VCC=13
RT
R1=RT1
VB
VCC=
2
3RT
RT
= R1
+ R2
I1
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ESE319 Introduction to Microelectronics
192009 Kenneth R. Laker, update 23Sep11
Biasing for Operating Point Stability
Quick ReviewWhat is the condition on the value ofV
B?
What is the condition on the value ofRE?
What is the purpose forR1
&R2?
What happened toRB?
What is the constraint on the value ofI1?
What is the 1/3, 1/3, 1/3 Rule?
IC
R1
R2
RC
RE
VCCI
E
VCB+
-
VB
+
-
+
-
VRc
I1
IB
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ESE319 Introduction to Microelectronics
202009 Kenneth R. Laker, update 23Sep11
Constant Emitter Current BiasThe current mirroris used to create a current source:
1. A BJT collector is the cur-rent source:
IC=ISe
VBE
VT
IREF
I
2. A diode-connected transistor sets thecurrent.
3. ChooseRref
for the desired
current:
Rref=VCC0.7
IREFVCC
0.7
IC1
VBE2=VBE1
IO
IREF=IC1IB1IB2IC1IE1
IREF=VCCVBE1
Rref
RrefVCC IC1
IB1IB2
VCE1=VBE1
4. IfQ1 = Q2 thenIC2
= IC1
=> IOIrefmatched
IO=IC2
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ESE319 Introduction to Microelectronics
212009 Kenneth R. Laker, update 23Sep11
Constant Emitter Current
IfQ1
and Q2
have the same saturation current:
Now: VBE1
= VBE2
IS1=IS2
And the transistors are at the same temperature: T1=T2
The two collector currents set primarily by Rref
are
equal, as long as Q2 is not saturated.
IO=IC2=IREFVCC0.7
Rref
VCE1
VCE2
, the Early Effect needs to be included in simulations.
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ESE319 Introduction to Microelectronics
222009 Kenneth R. Laker, update 23Sep11
Constant Emitter Current Early Voltage
IC2=IO
IB2
IB1
IC1
Assume Q1 = Q2
IO=IC2=ISeVBE/VT1
VCE2
VA
Early Effect
SinceIB
is not effected by VA
, i.e.
IB2= ISeVBE/VT= IC2
F
For Q2:
For Q1:IB1
= IB2
IREF=IC1IB1IB2=IC12 IB
VBE2=VBE1=VCE1=VBE
IREF=ISeVBE/VT1
VCE1
VA2
IS
e
VBE
VT
F=1VCE2
VAwhere
ISeVBE/VT1
VBE
VA2
IS
e
VBE
VT=
Solving for IS eVBE/VT
IS eVBE/VT=
IREF
1VBE
VA2
IO=IC2=IREF
1VCE2
V A
1VBE
VA
2
(1) (2)
Sub (2) into (1)
RrefIREF
VCC
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232009 Kenneth R. Laker, update 23Sep11
Constant Emitter Current Early Voltage Cont.
IO=IC2=IREF
1VCE2
V A
1VBE
VA
2
=>IO
IREF=
1VCE2
VA
1VBE
VA
2
Let VA= => Early effect is negligibleIO
IREF=
1
12
1
IO=IREF=>
Let VA = finite and VA = 50 V, VBE= 0.7 V, =100
IO
IREF= fVCE2=
10.02VCE21.034
=0.9710.02VCE2
If also =
IREF
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ESE319 Introduction to Microelectronics
242009 Kenneth R. Laker, update 23Sep11
We thus can use a current mirrorto provide stable control of tran-sistor collector current.R
refsets
the emitter and collector currentsand the collector-ground voltageforQ
amp.
vin
is the ac input voltage source.
RB
can be any reasonable value
this is not voltage biasing!
Rref
Rref
RB
VB
VCC
vin
Iref
iE
Qamp
Q1
Q2
iEvin
VCC
RBRC
BJT Emitter Current Source Bias
RC
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ESE319 Introduction to Microelectronics
252009 Kenneth R. Laker, update 23Sep11
Summary Two practical methods for achieving stable
bias for a BJT are:
Use a dc voltage source in the base with afeedback resistance in the emitter circuit. Place a dc current source directly in the emitter
circuit.
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ESE319 Introduction to Microelectronics
262009 Kenneth R. Laker, update 23Sep11
Emitter-Feedback Bias Design
1. Use single supply for base bias and collector sources.2. Use theI
C/10 rule for the currentI
1through the base bias
network (R1 andR2).3-1. Try less negative feedback using a smaller emitterresistorR
Esaving more of the V
CCsupply voltage for the
RC
voltage drop.
or
3-2. Use the 1/3, 1/3, 1/3 Rule.
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272009 Kenneth R. Laker, update 23Sep11
I1=
IC
10
Try 3-1 Emitter-Feedback Bias Design
VCC=12V
Complete the bias design giventhe following design values:
IC=1mA VC=6V
It follows:
RC=VRc
IC=VCCVC
IC=6k
I1= IC10=0.1mAR1R2=VCCI
1
= 1210
4=120 k
VC
=100RC
RE
R1
R2
VCC
VRc=6V
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282009 Kenneth R. Laker, update 23Sep11
3-1 Emitter-Feedback Bias - Cont.
Let's choose a small feedback voltage, sayV
Re= 1 V. Ignoring the base current:
RE=VRe
IE1V
IC=1 k
Then the voltage across R2
is 1.7 V
R2=
1.7V
104
A=17 k
R1=120k17k=103 k
RB=R1R214.5 k1
101RE10 k
I1
= 0.1 mA
IC=1mA
R1R2=120
k
R1
RE
RB1
101RERecall:
VB=1.7V
RC
VCC
RC
= 6kPrevious
slide
R2
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292009 Kenneth R. Laker, update 23Sep11
3-2 Emitter-Feedback 1/3, 1/3, 1/3 Rule Bias
VRc=VB=VCC
3=4V
RC=VRc
IC=
4V
103
A=4 k
RE= 4V0.7VIC=3.3 k
Then the voltage acrossR2
is VB
= 4 V
R2=
4V
104
A=40 k
4
Let's choose
R1=120k40k=80 kRB
1
101RERecall:
RB=R1R226.7 k1
101RE33 k
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ESE319 Introduction to Microelectronics
302009 Kenneth R. Laker, update 23Sep11
RECALL: Bias Stability Condition Argument
IB
IC VB=IBRBVBE1RE IB
IB=VBVBE
RB1RE
IE=1IB
IC=IB=
RB1REVBVBE
IC=IB
1
VBVBE
RE=
VBVBE
RE
RB1REif
RC
RE
RB
VB
VCC
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312009 Kenneth R. Laker, update 23Sep11
IC=IB=
RB1REVBVBE
RC
RE
RB
VB
RE
IC
IB
VCC= 12VV
B= 1.7 V
VBE
= 0.7 V
IC
= 1 mA
= 100
RC
= 6k
RE
= 1 k
RB
= 14.5 k
= 100; IC
= 0.873 mA
= 200; IC= 0.932 mA = ; I
C= 1 mA
= 100 & RB
= 0 ; IC
= 0.99 mA
3-1 Emitter-Feedback Sensitivity
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322009 Kenneth R. Laker, update 23Sep11
3-1 Emitter-Feedback IS
Sensitivity
//Rule of thumb BJT bias sensitivityBeta=100;VsubT=0.025;VB=1.7;Rb=14.5;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
Sensitivity toIS
//3-1 Bias Scheme with Re=1 K
Re=1;IC=
RB1REVBVBE
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332009 Kenneth R. Laker, update 23Sep11
3-1 Scilab Plot (Zoomed)
IS=1011
IS=1018
IC0.4mA
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ESE319 Introduction to Microelectronics
342009 Kenneth R. Laker, update 23Sep11
IS=1011
IS=1018
IC0.1mARB0
or
RB1RE
Compare to IC
vs IS
Results Plot with Ideal
Emitter Feedback (1/3,1/3,1/3 Rule)
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352009 Kenneth R. Laker, update 23Sep11
Conclusion
1/3, 1/3, 1/3 Rule Provides a Good Compromise BaseVoltage Bias Scheme