Download Emitter Current Bias Quick Review

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?
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?
2009 Kenneth R. Laker, updated 25Sep10 KRL
1
ESE319 Introduction to Microelectronics
BJT Biasing Cont. & Small
Signal Model
Bias Design Example using “1/3, 1/3, 1/3 Rule”
● Small Signal BJT Models
● Small Signal Analysis
●
2009 Kenneth R. Laker, updated 25Sep10 KRL
2
ESE319 Introduction to Microelectronics
Emitter Feedback Bias Design
I1
IC
IB
IE
Voltage bias circuit
2009 Kenneth R. Laker, updated 25Sep10 KRL
Single power supply version
3
ESE319 Introduction to Microelectronics
Thevinen Equivalent
Use Thevinen's theorem:
R2
V Th=V B =
V CC
R1R 2
R1
Let: =
R2
1
V B=
V CC
1
R1 R 2
R1
R B=
=
R1 R 2 1
2009 Kenneth R. Laker, updated 25Sep10 KRL
Let's specify VB and RB, and
solve for ρ then R1 and R2:
VB
1
=
1 V CC
V CC
=
−1
VB
R1 =1 R B
R1
R 2=

4
ESE319 Introduction to Microelectronics
“1/3, 1/3, 1/3 Rule” Bias Procedure:
5.Or choose
V CC
V CC
RT =R1R 2 =
=
I 1 I C /10
2009 Kenneth R. Laker, updated 25Sep10 KRL
1. Bias so that VCC is split equally across
RC, VCE (or VCB), and RE (or VB).
2. Select desired collector current.
3. Assume IE = IC to determine RC
and RE.
4. Add 0.7 V to VRe = VCC/3 to find VB.
Assume base current through RB is
negligible; hence V B≈V BG
5. Choose RB approximately equal to
 R E /10 . (Use lowest value of 
6. Finally, compute ρ then R1 and R2.
5
ESE319 Introduction to Microelectronics
Example
50 ≤  ≤ 100
V CC =12V
V CC
V R E =V RC =
=4 V
3
I C =1 mA
2009 Kenneth R. Laker, updated 25Sep10 KRL
Then:
4
RC = R E = −3 =4⋅103 =4 k 
10
V CC
V B=
0.7V 4 V 0.7 V =4.7 V
3
min 1 R E 50⋅4000
R B=
=
=20 k 
10
10
For a single power supply:
V CC
12
=
−1=
−1=1.55
VB
4.7
R1 =1 R B =2.55⋅20=51 k 
R1 51
R 2= =
=32.9 k 
 1.55
6
ESE319 Introduction to Microelectronics
Completed Bias Design
Electronics Workbench
simulation results
Δ < ±10%
4.7 V
1 mA
RT =R 1R 2 =83.9 k 
VB
IB
⇒ I 1=
=0.056 mA
RT
10
2009 Kenneth R. Laker, updated 25Sep10 KRL
Our design differs from the simulation because we neglected
the base current.
There is no point in including the
base current, since we will build
the circuit using resistors that
come only in standard sizes and
with 5% tolerances attached to
their values. The closest available resistors in the RCA Lab
are 47 kΩ, 33 kΩ, and 3.3 kΩ.
7
ESE319 Introduction to Microelectronics
BJT Small Signal Models
Conceptually, the signal we wish
to amplify is connected in series
with the bias source and is of quite
small amplitude.
We will linearize the signal analysis
to simplify our mathematics – to
avoid having to deal with the
nonlinear exponential collector
characteristic.
2009 Kenneth R. Laker, updated 25Sep10 KRL
8
ESE319 Introduction to Microelectronics
BJT Small Signal Models
2009 Kenneth R. Laker, updated 25Sep10 KRL
9
ESE319 Introduction to Microelectronics
Linearization Process
Split the total base-emitter voltage and total collector current into bias and signal components:
v BE
VT
iC =I C i c = I S e = I S e
V BE v be
VT
V BE
VT
v be
VT
=IS e e
IC
Identify and substitute the bias current into the model:
i C =I C e
v be
VT
Expand the exponential in a Taylor series:
∞
f  x=∑
n=0
2009 Kenneth R. Laker, updated 25Sep10 KRL
n
x
 n
f 0
n!
10
ESE319 Introduction to Microelectronics
Analysis Using Collector Current Model
v be
VT
i C =I C e
v
iC
V
=e
IC
Useful constants:
be
log 10
log 10 2=0.301
T
log 10 e=0.4343
  
iC
v be
=
log 10 e
IC
VT
Let iC be 2x IC, i.e. i C / I C =2:
v be
0.301=
0.4343
VT
0.301
v be=
0.025≈0.017V
0.4343
 
2009 Kenneth R. Laker, updated 25Sep10 KRL
11
ESE319 Introduction to Microelectronics
Linearization Using Taylor Series
∞
f  x=∑ f
Expand in a Taylor series:
 n
n=0
∞
For the exponential function: e =∑
x
v be
VT
∞
e =∑
n=0
or:
v be
VT
n=0
 
1 v be
n! V T
xn
0
n!
xn
n!
n
   
2
3
v be 1 v be
1 v be
e ≈1 


VT 2 VT
6 VT
2009 Kenneth R. Laker, updated 25Sep10 KRL
12
ESE319 Introduction to Microelectronics
Linearization Continued
Recall: vbe of about 17 mV causes a 2x change in collector current.
Let's expand in Taylor series for this value of vbe.
v be 0.017
=
=0.68
V T 0.025
v
1
1
V
2
3
e =10.68 0.68  0.68 
2
6
be
T
v be
VT
e ≈10.680.23120.0524=1.9444
Compare with:
e 0.68=1.973
Four term expansion is accurate to about 1.5%, two term expansion is
only accurate to about 15%, i.e.
v be≈0.17 V is not such a small signal
2009 Kenneth R. Laker, updated 25Sep10 KRL
13
ESE319 Introduction to Microelectronics
Quick Small Signal Model Review
v BE
VT
v be
VT
i C =I S e = I C e
What is iC, IC, vBE, vbe?
   
v be
VT
2
3
v be 1 v be
1 v be
e ≈1 


VT 2 VT
6 VT
Under what condition(s) can one justifiably approximate the above infinite series as
v
v be
V
e ≈1
VT
Why is this important?
be
T
2009 Kenneth R. Laker, updated 25Sep10 KRL
14
ESE319 Introduction to Microelectronics
Small Signal Model
Using the grouping into bias and signal voltages/currents:
i C =I C e
v be
VT
And using the first two terms of the Taylor series expansion:
iC ≈ I C


IC
1
1
v be = I C 
v be = I C i c
VT
VT
We define transconductance and incremental (or ac) current as:
d iC
g m =

d v BE i
C
=I C
IC
=
VT
i c =g m v be
bias current
2009 Kenneth R. Laker, updated 25Sep10 KRL
15
ESE319 Introduction to Microelectronics
Incremental (small-signal) BJT Model
1
1
i B = I B i b = i C =  I C i c 


 
1
1
1 IC
i B= i C = I C 
v be


 VT
Define the incremental base current and base resistance:
 
1
1 IC
1
i b = i c=
v be = v be

 VT
r
bias current
VT 
r  =
=
I C gm
2009 Kenneth R. Laker, updated 25Sep10 KRL
where I C =V T g m
16
ESE319 Introduction to Microelectronics
Equivalent Models
Equations (1):
i c =g m v be
v be
i b=
r
i e =i b i c
Equations (2):
Equations (3):
i c = i b
ie
i b=
1
i c =g m v be
ic
i b=

i e =i b i c
v be v be
i e =1i b =1 =
r r e
IC V T
i c =g m r  i b =

i b=i b
VT IC
r
 VT
r e=
=
1 1 I C
Choose the model that simplifies the circuit analysis.
2009 Kenneth R. Laker, updated 25Sep10 KRL
17
ESE319 Introduction to Microelectronics
Equivalent Circuits
ib
b
Equations (1): i
c
rrπ
g m v be
e
Equations (2):
c
b
ib
r0
i b
ie
v ce
i c = g m v be
r0
IC
VA
g m=
r 0=
VT
IC
v be
i b=
r
VT
VT 
r = =
=
IB
IC gm
i e =i b i c
ie
ic
r0
Equations (3):
c
b
ic

ib
g m v be
e
re
e
v ce
i c =i b 
r0
ie
i b=
1
v be
i e=
re
r
VT 
r e=
= =
I E g m 1
2009 Kenneth R. Laker, updated 25Sep10 KRL
ic
ie
i c = g m v be 
i b=
v ce
r0
ic

i e =i b i c
18
c
r0
ESE319 Introduction to Microelectronics
Small Signal Circuit Analysis
The small signal model will replace the large signal model
and be used for (approximate) signal analysis once the
transistor is biased.
we may include Rs with Rb
b
c
rr
π
ib
gmvbe
ie
e
Transistor
ic
r
vc
Biased circuit
(and all bypass caps are approx short circuits)
2009 Kenneth R. Laker, updated 25Sep10 KRL
19
ESE319 Introduction to Microelectronics
Combining AC and DC Sources
RS
vin
R1
R2
RC
vO
RE
RS
VCC
vin
<=>
AC Source with low output
impedance upsets bias.
2009 Kenneth R. Laker, updated 25Sep10 KRL
RC
RB
RE
vO
VCC
VB
Thevenin equivalent at base.
20
ESE319 Introduction to Microelectronics
Combining AC and DC Sources Cont.
ISSUES:
1. Signal source vin shorts out VB.
2. DC bias on vin interferes with op point design.
2009 Kenneth R. Laker, updated 25Sep10 KRL
21
ESE319 Introduction to Microelectronics
“Blocking Capacitor” Remedy
RS
vin
R1
R2
RC
RE
vO
VCC
What is the purpose of “blocking capacitor” Cin?
How does one determine the value of Cin?
2009 Kenneth R. Laker, updated 25Sep10 KRL
22
ESE319 Introduction to Microelectronics
Biased Circuit Small Signal Analysis
v sig =i b  R B r  ie R E
ib
r
ie
ic
vc
v sig =i b  R B r  i b 1 R E
v sig
i b=
R B r  1 R E
r
v be =i b r =
v sig
R B r 1 R E
Let: =100 and I C =1 mA
VT
0.025
r  =
=100
=2.5 k 
IC
0.001
I C 0.001 A
Note: g m= V = 0.025 V =0.04 S =40 mS
T
2009 Kenneth R. Laker, updated 25Sep10 KRL
S = siemen
23
ESE319 Introduction to Microelectronics
Small Signal Analysis - Continued
v be=i b r =
ib
r
ie
ic
vc
i c =g m v be =
r
R B r 1 R E
gm r
R B r 1 R E
v sig
v sig
− RC
v c =−RC i c =
v sig
R B r  1 R E
IC V T
g m r =

=
VT IC
≈100
2009 Kenneth R. Laker, updated 25Sep10 KRL
− RC
1 R E ≫ R B r  ⇒ v c≈
v sig
1 R E
vc
RC
Av =
≈−
v sig
RE
24
ESE319 Introduction to Microelectronics
Multisim Model of “Bias Design”
vc
RC
Av =
≈− =−1
v sig
RE
1. RE needs to be large to achieve good op. pt. Stability!
2. Consequence: |Av| is low.
2009 Kenneth R. Laker, updated 25Sep10 KRL
25
ESE319 Introduction to Microelectronics
Multisim Input-output Plot
vc
RC
Av =
≈− =−1
v sig
RE
2009 Kenneth R. Laker, updated 25Sep10 KRL
26
ESE319 Introduction to Microelectronics
Conclusions
Conservative 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?
2009 Kenneth R. Laker, updated 25Sep10 KRL
27
ESE319 Introduction to Microelectronics
Conclusion Cont.
How does one obtain operating point stability, and
simultaneously achieve a respectable voltage gain?
“Blocking
Capacitor”
C byp
2009 Kenneth R. Laker, updated 25Sep10 KRL
“Bypass
Capacitor”
28