BJT Biasing Cont. & Small Signal Model - Penn...
Transcript of BJT Biasing Cont. & Small Signal Model - Penn...
ESE319 Introduction to Microelectronics
1Kenneth R. Laker, updated 18Sep13 KRL
BJT Biasing Cont. & Small Signal Model
● Bias Design Example using “1/3, 1/3, 1/3 Rule”● Small Signal BJT Models● Small Signal Analysis
ESE319 Introduction to Microelectronics
2Kenneth R. Laker, updated 18Sep13 KRL
Emitter Feedback Bias Design
Two power supply version Single power supply version
I1 I
C
IE
IB
VB
RC
RE
RB
VCC
RC
RE
R1
R2
V B=R2
R1R2V CC
RB=R1 R2
R1R2R1=RB
V CC
V B
R2=R1
V B
V CC−V B
Specify VCC
, VB and RB, and solve for R
1 and R
2:
ESE319 Introduction to Microelectronics
3Kenneth R. Laker, updated 18Sep13 KRL
“1/3, 1/3, 1/3 Rule” Bias Procedure:1. Bias so that VCC is split equally acrossRC, VCE , and RE .2. Specify the desired collector current.3. Assume IE = IC to determine RC & RE.4. Add 0.7 V to V
RE = VCC/3 to find VB.
Assume base current through RB isnegligible; hence 5. Choose RB approximately equal to . (β
min is lowest value of β )
6. Finally compute R1 and R2, and check that I
B << I
1 << I
C
V B≈V BEV RE
min1RE /105.Or choose RT=R1R2=
V CC
I 1=
V CC
I C /10
VB
RC
RE
RB V
CC
Check RB << (β
min +1)R
E
ESE319 Introduction to Microelectronics
4Kenneth R. Laker, updated 18Sep13 KRL
Example
V CC=12V
V RE=V RC
=V CC
3=4 V
I C=1 mA
Then:RC=RE=
410−3=4⋅103=4 k
V B=V CC
30.7 V=4V 0.7V =4.7 V
RB=min1RE
10≈
50⋅400010
=20 k
For a single power supply:50≤ ≤150
RB
RE
RC
VCC
VB
R1=RB V CC
V B=20 k 12V
4.7V=51 k
R2=R1
V B
V CC−V B=51 k 4.7V
7.3V=32.9 k
RT=R1R2=83.9 k⇒ I 1=0.14 mA
ESE319 Introduction to Microelectronics
5Kenneth R. Laker, updated 18Sep13 KRL
Completed Bias DesignWhy does our design differ from the simu-lation?
Electronics Workbench simulation results
4.7 V1 mA
RT=R1R2=83.9 k
Δ < ±10%
RC
R1 R
C
RE
R2
VCC
ESE319 Introduction to Microelectronics
6Kenneth R. Laker, updated 18Sep13 KRL
Completed Bias DesignOur design differs from the simulation be-cause we neglected the base current.
There is no point in including the base current, since we will build the circuit us-ing resistors that come only in standard sizes and with 5% tolerances attached to their values. The closest available resist-ors in the Detkin Lab are 47 kΩ, 33 kΩ, and 3.3 kΩ (or 4.7 kΩ).
Voltage & currents are specified as ranges (V
min ≤ V ≤ V
max), not absolute values.
Electronics Workbench simulation results
4.7 V1 mA
RT=R1R2=83.9 k
Δ < ±10%
RC
R1 R
C
RE
R2
VCC
ESE319 Introduction to Microelectronics
7Kenneth R. Laker, updated 18Sep13 KRL
Completed Bias DesignOur design differs from the simulation be-cause we neglected the base current.
There is no point in including the base current, since we will build the circuit us-ing resistors that come only in standard sizes and with 5% tolerances attached to their values. The closest available resist-ors in the Detkin Lab are 47 kΩ, 33 kΩ, and 3.3 kΩ (or 4.7 kΩ).
Voltage & currents are specified as ranges (V
min ≤ V ≤ V
max), not absolute values.
Electronics Workbench simulation results
4.7 V1 mA
RT=R1R2=83.9 k
Δ < ±10%
RC
R1 R
C
RE
R2
VCC
ESE319 Introduction to Microelectronics
8Kenneth R. Laker, updated 18Sep13 KRL
BJT Small Signal Models
Conceptually, the signal we wish to amplify is connected in series with the bias source and is of small amplitude.
We will linearize the signal analysisto simplify our mathematics – to avoid having to deal with thenonlinear exponential collector current
iC− I S evBE
V T
VCC
VB
RE
RCR
B
ESE319 Introduction to Microelectronics
9Kenneth R. Laker, updated 18Sep13 KRL
BJT Small Signal Models
ESE319 Introduction to Microelectronics
10Kenneth R. Laker, updated 18Sep13 KRL
Linearization Process
iC= I Cic= I S evBE
V T = I S eV BEvbe
V T = I S eV BE
V T evbe
V T
Split the total base-emitter voltage and total collector cur-rent into bias and signal components:
iC= I C evbe
V T
Identify and substitute the bias current into the model:
Expand the exponential in a Taylor series:
f x =∑n=0
∞
f n 0 xn
n!
I C
ESE319 Introduction to Microelectronics
11Kenneth R. Laker, updated 18Sep13 KRL
Analysis Using Collector Current Model
iC= I C evbe
V T
iC
I C=e
vbe
V T
log10 iC
I C = vbe
V T log10 e
log102=0.301
Useful constants:
log10e=0.4343
Let iC = IC + i
c be 2x I
C, i.e.
0.301= vbe
V T 0.4343
vbe=0.301
0.43430.025≈0.017 V
iC / I C=2
ESE319 Introduction to Microelectronics
12Kenneth R. Laker, updated 18Sep13 KRL
Linearization Using Taylor Series
For the exponential func-tion:
f x =ex=∑n=0
∞ xn
n!
evbe
V T=∑n=0
∞ 1n ! vbe
V T n
or:
evbe
V T=1vbe
V T
12 vbe
V T 2
16 vbe
V T 3
≈1vbe
V T
Expand in a Taylor series: f x =∑n=0
∞
f n 0 xn
n!
when 12
vbe
V T2
≪1
ESE319 Introduction to Microelectronics
13Kenneth R. Laker, updated 18Sep13 KRL
Linearization Continued Recall: vbe of about 17 mV causes a 2x change in collector current i
C = I
C + i
c.
Let's expand in Taylor series for this value of vbe.vbe
V T= 0.017
0.025=0.68
evbe
V T=10.68 12 0.6821
6 0.683
evbe
V T≈10.680.23120.0524=1.9444
Compare with: e0.68=1.973
Four term expansion is accurate to about 1.5%, two term expansion is only accurate to about 15%, i.e.
vbe≈0.017V is not sucha small signal
ESE319 Introduction to Microelectronics
14Kenneth R. Laker, updated 18Sep13 KRL
Small Signal Model
And using the first two terms of the Taylor series expansion:iC= I Cic= I C e
vbe
V T
Using the grouping into bias and signal voltages/currents:
iC≈ I C 1 1V T
vbe= I CI C
V Tvbe=I Cic
We define transconductance and incremental (or ac) current as:
ic=gm vbegm=d iC
d vBEiC= I C
=I C
V T
bias current
ESE319 Introduction to Microelectronics
15Kenneth R. Laker, updated 18Sep13 KRL
Small Signal - Graphical Rep-
resentation
iC
ESE319 Introduction to Microelectronics
16Kenneth R. Laker, updated 18Sep13 KRL
Incremental (small-signal) BJT ModeliB= I Bib=
1
iC=1 I Cic
iB=1
I C1 I C
V T vbe
Define the incremental base current and base resistance:
ib=1 ic=
1 I C
V T vbe=1r
vbe
r=V T
I C=
gmbias current
I B ib
ESE319 Introduction to Microelectronics
17Kenneth R. Laker, updated 18Sep13 KRL
Equivalent ModelsEquations (3):
ic=gm vbe
ie=ibic
ib=ic
ib=
vbe
r
Equations (1):ic=gm vbe
ie=ibic
Equations (2):
ic= ib
ib=ie
1ie=1 ib=1
vbe
r=
vbe
re
Choose the model that simplifies the circuit analysis.
re=r
1=
1
V T
I Cic=gm r ib=
I C
V T
V T
I Cib= ib
ESE319 Introduction to Microelectronics
18Kenneth R. Laker, updated 18Sep13 KRL
Equivalent CircuitsEquations (3):
b
e
cic
g m vbe
ie=ibic
ib=ic
Equations (2):b
e
c
re
ib
ic= i bvce
r0
ib=ie
1
ie=vbe
re
Equations (1):
ib=vbe
r
ic=gm vbevce
r0
ie=ibic
rπ
b c
e
r r0
gm=I C
V Tr0=
V A
I C
r=V T
I B=
V T
I C=
g m
r0
re=V T
I E=
gm
=r
1
g m vbe r0
ic=gm vbevce
r0
icib
ie
ib ic
ie
ib ic
ie
ESE319 Introduction to Microelectronics
19Kenneth R. Laker, updated 18Sep13 KRL
Quick Small Signal Model Review
What is iC, I
C, v
BE, v
be?
Under what condition(s) can one justifiably approximate the above infinite series as
iC= I S evBE
V T =I C evbe
V T
evbe
V T≈1vbe
V T
12 vbe
V T 2
16 vbe
V T 3
evbe
V T≈1vbe
V T
Why is this important?
ESE319 Introduction to Microelectronics
20Kenneth R. Laker, updated 18Sep13 KRL
DC & AC Circuit Decomposition
RE
RCR
b
VCC
VB
RE
RCR
b
VCC
VB
RE
RCR
b
DC AC
DOES Rb = R
B = R
1\\R
2 ? ASSUMPTION: ?
ESE319 Introduction to Microelectronics
21Kenneth R. Laker, updated 18Sep13 KRL
DC & AC Circuit Decomposition
RE
RCR
b
VCC
VB
RE
RCR
b
VCC
VB
RE
RCR
b
DC AC
NO Rb ≠ R
B = R
1\\R
2
ASSUMPTION: signal v
s is pure AC!
ac source vs has
series input resistance
Rb=RBRs≈RB
ESE319 Introduction to Microelectronics
22Kenneth R. Laker, updated 18Sep13 KRL
Quick Review
“?” “?” “?”
“?”
“?”
What are each of the “?” parameters?
ESE319 Introduction to Microelectronics
23Kenneth R. Laker, updated 18Sep13 KRL
Quick Review
gm=I C
V T
r0=V A
I C
r=V T
I B=
V T
I C=
g m
re=V T
I E=
gm
=r
1
ESE319 Introduction to Microelectronics
24Kenneth R. Laker, updated 18Sep13 KRL
Small Signal Circuit AnalysisThe small signal model will replace the large signal model and be used for (approximate) signal analysis once the transistor is biased.
small signal BJT model
AC small signal circuit (Bias circuit set to “zero” (using superposition)
gmvberπ
b c
e
r rib
ie
ic
vc
RC
RE
RB
actual circuit, AC + DC
Assume for time being that Rb = R
B = R
1||R
2
rib
ie
ic
vc
RC
RE
RB
Rb
RC
REV
BV
B
vs
vs
Rb=RBRs≈RB
ESE319 Introduction to Microelectronics
25Kenneth R. Laker, updated 18Sep13 KRL
Biased Circuit Small Signal Analysis
r=V T
I C=100 0.025
0.001=2.5k
vs=ibRBrie RE
vs=ibRBrib1RE
ib=vs
Rbr1RE
vbe=ib r=r
RBr1REvs
rib
ie
gm=I C
V T=
0.001 A0.025V
=0.04 S=40 mSNote: S = siemen
Let: =100 and I C=1mA
ic
vc
20 k Ohm
e
cb
RC
RE
RB
vs
ESE319 Introduction to Microelectronics
26Kenneth R. Laker, updated 18Sep13 KRL
Small Signal Analysis - Continued
vbe=ib r=r
RBr1REv s
ic=gm v be=gm r
RBr1REv s
gm r=I C
V T
V T
I C=
vc=−RC ic=−RC
RBr1REvs
1RE≫RBr⇒ vc≈−RC
1REvs
Av=vc
vs≈−
RC
RE
=100
rib
ie
vc
ic
20 k OhmR
C
RE
RB
vs
ESE319 Introduction to Microelectronics
27Kenneth R. Laker, updated 18Sep13 KRL
Multisim Model of “Bias Design”
Av=vc
vs≈−
RC
RE=−1
1. RE needs to be large to achieve good op. pt. Stability!
2. Consequence: |Av| is low.
20 k Ohm
RC
RE
RB
VCC
VB
vs
vc
ESE319 Introduction to Microelectronics
28Kenneth R. Laker, updated 18Sep13 KRL
Multisim Input-output Plot
Av=vc
vs≈−
RC
RE=−1
ESE319 Introduction to Microelectronics
29Kenneth R. Laker, updated 18Sep13 KRL
Combining AC and DC Sources
vsv
s
vO
Rs
R1
R2
RC
RE
RC
RE
RB
VB
vO V
CCVCC
<=>
AC Source with low outputimpedance upsets bias.
Thevenin equivalent at base.
Rs=50
ESE319 Introduction to Microelectronics
What, if anything, is wrong with this amplifier design?
If there is something(s) wrong, how can it (they) be remedied?
ESE319 Introduction to Microelectronics
What, if anything, is wrong with this amplifier design?1. V
B is effectively shorted out by v
s.
2. DC on vs corrupts the op. pt. (i.e. V
C & I
C).
3. The voltage gain is - RC/RE.
ESE319 Introduction to Microelectronics
If there is something(s) wrong, how can it (they) be remedied?
1. Insert blocking capacitor Cin in series with R
S.
2. Insert bypass capacitor Cbyp
in parallel with all or part of RE.
ESE319 Introduction to Microelectronics
32Kenneth R. Laker, updated 18Sep13 KRL
ConclusionsConservative 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?
ESE319 Introduction to Microelectronics
33Kenneth R. Laker, updated 18Sep13 KRL
Conclusion Cont.How does one obtain operating point stability, and simultaneously achieve a respectable voltage gain?
Cbyp
“Bypass Capacitor”
“Blocking Capacitor”
vs