Lecture 20
Bipolar Junction Transistors (BJT): Part 4
Small Signal BJT Model
Reading:
Jaeger 13.5-13.6, Notes
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ECE 3040 - Dr. Alan Doolittle
Further Model Simplifications
((useful for circuit analysis)
y )
Ebers-Moll
Forward Active
Mode
Neglect
Small Terms
VEB

 VCB VT

 VEB VT

 VEB VT
I C   F I F 0  e
 1  I R 0  I C  I S e VT
 1  I C   F I F 0  e
 1  I R 0  e






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ECE 3040 - Dr. Alan Doolittle
Modeling the “Early Effect” (non-zero slopes in IV curves)
IC
•Base width changes due
to changes in the basecollector depletion width
with changes in VCB.
iB1< iB2< iB3
iB3 (theory)
iB2 (theory)
iB1 (theory)
•This changes
g T, which
changes IC, DC and BF
VCE
VA
Major BJT Circuit Relationships
iC  I S e
VEB
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VT
 iC  I S e
VEB
VT
 vCE 
1 

V
A 

 vCE 
iC
I S VEB VT

e
 F   FO 1 
 iB 
V


A 
F
FO

ECE 3040 - Dr. Alan Doolittle
Small Signal Model of a BJT
•Just as we did with a p-n diode, we can break the BJT up into
a large signal analysis and a small signal analysis and
“linearize” the non-linear behavior of the Ebers-Moll model.
•Small
S ll signal
i l Models
M d l are only
l useful
f l for
f Forward
F
d active
i mode
d
and thus, are derived under this condition. (Saturation and cutoff are
used for switches which involve very large voltage/current swings from the on to off states.)
•Small signal models are used to determine amplifier
characteristics (Example: “Gain” = Increase in the magnitude
of a signal at the output of a circuit relative to it
it’ss magnitude at
the input of the circuit).
•Warning:
Warning: Just like when a diode voltage exceeds a certain
value, the non-linear behavior of the diode leads to distortion
of the current/voltage curves (see previous lecture), if the
i t / t t exceedd certain
inputs/outputs
t i limits,
li it the
th full
f ll Ebers-Moll
Eb M ll model
d l
must be used.
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ECE 3040 - Dr. Alan Doolittle
Consider the BJT as a two-port Network
i1
+
V1
-
i2
Two Port
Network
General “y-parameter” Network
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+
V2
-
BJT “y-parameter” Network
i1=y11v1 + y12v2
ib=y11vbe + y12vce
i2=y21v1 + y22v2
ic=y21vbe + y22vce
ECE 3040 - Dr. Alan Doolittle
Consider the BJT as a two-port Network
ib=y11vbe + y12vce
ic=y21vbe + y22vce
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ECE 3040 - Dr. Alan Doolittle
Consider the BJT as a two-port Network
o is most
often taken
as a
constant, F
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ECE 3040 - Dr. Alan Doolittle
Alternative Representations
IC
 40 I C
Transconductance g m  y 21 
VT
 oVT  o
1


Input Resistance r 
y11
IC
gm
V A  VCE
1

Output Resistance ro 
y 22
IC
Y-parameter Model
Hybrid-pi Model
v1
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ECE 3040 - Dr. Alan Doolittle
Alternative Representations
g m vbe  g m r ib   o ib
Voltage Controlled Current
source version of Hybrid-pi
Hybrid pi
Model
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Current Controlled Current
source version of Hybrid-pi
Hybrid pi
Model
ECE 3040 - Dr. Alan Doolittle
Single Transistor Amplifier Analysis: Summary of Procedure
Important!
Steps to Analyze a Transistor Amplifier
1.) Determine DC operating point and
calculate small signal parameters (see
next page)
2.) Convert to the AC only model.
•DC Voltage sources are shorts to ground
•DC Current sources are open circuits
•Large capacitors are short circuits
•Large inductors are open circuits
3.) Use a Thevenin circuit (sometimes a
Norton) where necessary. Ideally the
base should be a single resistor + a single
source. Do not confuse this with the DC
Thevenin you did in step 1.
4.) Replace transistor with small signal
model
5.) Simplify the circuit as much as
necessary
necessary.
6.) Calculate the small signal parameters
(r, gm, ro etc…) and then gains etc…
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Step 1
Step
2
Step
3
Step
4
Step
5
ECE 3040 - Dr. Alan Doolittle
Single Transistor Amplifier Analysis
Step 1 detail
Important!
DC Bias Point
Thevenin
Ib
RTH
Vbe
Ie
3V=IERE+Vbe+IBRTH
3V=IIC((o+1)/
3V
1)/o) Re+0.7V+I
0.7V IBRTH
3V=IB o((o+1)/o) Re+0.7V+IBRTH
3V= IB(100+1)1300+0.7+ IB7500
IB=16.6 uA, IC= IB o=1.66 mA, IE=(o+1) c/ o=1.67 mA
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ECE 3040 - Dr. Alan Doolittle
Single Transistor Amplifier Analysis
Calculate small signal parameters
Step 6 detail
Transconductance
g m  y 21 
IC
 40 I C  0 . 0664 S
VT
 V

1
 o T  o  1506 
y 11
IC
gm
Input Resistance r 
Output Resistance ro 
V  V CE
V
1
 A
 A  45 . 2 K 
y 22
IC
IC
Vout
880
VTH=0.88 VS
RTH
vout   g m vbe RL
Important!
Vbe
and
RL= RC| R3| ro
r
gmVbe
vbe  vTh
r
RTh  r
and
vTh  0.88v S

 r

vout  vout  vbe  vth 
     g m RL 
0.88
 
vS
 vbe  vth  v S 
 RTh  r 
 1506 
Av   0.0664 45,200 || 4300 || 100,000 
0.88
 880  1506 
Av  139 V / V
ECE 3040 - Dr. Alan Doolittle
Av  Voltage Gain 
For Extra Examples see:
Jaeger section 13.6, and
pages 627-630 (top of 630)
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Important!
Supplement Added to describe more details of the
Solution of this Problem
Bipolar Junction Transistors (BJT): Part 5
D t il off Amplifier
Details
A lifi Analysis
A l i
Reading:
Jaeger 13.5-13.6, Notes
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Detailed Example: Single Transistor Amplifier Analysis
Important!
Notes on slides 14-25 were prepared by a previous student.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Step 1: Determine DC Operating Point
Important!
Remove the Capacitors
Because the
impedance of a
capacitor is Z =
1/(jωC), capacitors
have infinite
impedance or are
open circuits in DC
(ω = 0).
Inductors (not present
in this circuit) have an
impedance Z = jωL,
jωL
and are shorts in DC.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Step 1: Determine DC Operating Point
Determine the DC Thevenin Equivalent
Important!
Replace all connections to the transistor with their Thevenin equivalents.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Step 1: Determine DC Operating Point
Important!
Calculate Small Signal Parameters
Identify the type of transistor (npn in this example) and
draw the base, collector, and emitter currents in their
proper direction
di i andd their
h i corresponding
di voltage
l
polarities.
Applying KVL to the controlling loop
((loopp 1):
)
VTHB – IBRTHB – VBE – IERE = 0
+
+
IC
-
+
Applying
pp y g KCL to the transistor:
IE = IB + IC
Because IC = βIB,
IE = IB + IC = IB + βIB = IB(1+β)
IB
VBE
+
IE
1
-
Substituting for IE in the loop equation:
VTHB – IBRTHB – VBE – IB(1+β)RE = 0
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ECE 3040 - Dr. Alan Doolittle
Step 1: Determine DC Operating Point
Important!
Plug in the Numbers
VTHB – IBRTHB – VBE – IB(1+β)RE = 0
VTHB – VBE – IB(RTHB + (1+ β)RE) = 0
VTHB = 12R1/(R1+R2) = 3 V
RTHB = R1 || R2 = 7.5 kΩ
Assume VBE = 0.7 V
Assume β for this particular transistor is
given to be 100.
3 – 0.7
0 7 – IB(7500 + (1+100)*1300) = 0
IB = 16.6 μA
IC = βIB = 1.66 mA
IE = IB + IC = 1.676
1 676 mA
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+
IC
-
+
+
IB
VBE
+
IE
1
-
ECE 3040 - Dr. Alan Doolittle
Step 1: Determine DC Operating Point
Important!
Check Assumptions: Forward Active?
VC = 12 – ICRC = 12 - (1.66 mA)(4300) = 4.86 V
VE = IERE = (1.67
(1 67 mA)(1300) = 22.18
18 V
VB = VTHB – IBRTHB = 3 – (16.6μA)(7500) = 2.88 V
+
IC
-
Check:
For an npn transistor in forward active:
VC > VB
VC
+
-
+
IB
4.86 V > 2.88 V
VB – VE = VBE = 0.7 V
2.88 V – 2.18 V = 0.7 V
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VB
VBE
-
VE
+
IE
1
-
ECE 3040 - Dr. Alan Doolittle
Step 2: Convert to AC-Only Model
Important!
Short the Capacitors and DC Current Sources
•
•
•
•
DC voltage sources are
shorts
h t (no
( voltage
lt
drop/gain through a
short circuit).
DC current sources are
open (no current flow
through an open
circuit).
Large capacitors are
shorts (if C is large,
1/jωC is small).
Large inductors are
open (if L is large, jωL
is large).
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ECE 3040 - Dr. Alan Doolittle
Step 2: Convert to AC-Only Model
Important!
(Optional) Simplify Before Thevenizing
r2
30 kΩ
rs
2 kΩ
vs
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rc
4 3 kΩ
4.3
rs
rL
r1
10 kΩ
100
kΩ
rc||rL
4.12 kΩ
2 kΩ
vs
r1||r2
7.5 kΩ
ECE 3040 - Dr. Alan Doolittle
Step 3: Thevenize the AC-Only Model
Important!
rthC = rc||rL
rthB = rs||r
|| 1||r
|| 2
rs
rc||rL
4.12 kΩ
2 kΩ
vs
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r1||r2
7.5 kΩ
4 12 kΩ
4.12
vthC = 0 V
1.58 kΩ
vthB = vs *
(r1||r2)/(rs+[r1||r2])
rthE = 0 Ω
vthE = 0 V
ECE 3040 - Dr. Alan Doolittle
Step 4: Replace Transistor With Small Signal Model
Important!
rthC = rc||rL
4.12 kΩ
rthB
hB = rs||r1||r2
vthC
hC = 0 V
1.58 kΩ
rthE = 0 Ω
vthE = 0 V
vthB = vs *
[(r1||r2)/(rs+r1||r2)]
rthB
After replacing the transistor, apply Ohm’s
Law: V = IR to find vout.
ro and rthC are in parallel, so that Ohm’s
Law becomes: vout = -IR = -(gmvBE)(ro||rthC)
Because rthC = rc||rL
vout = -(gmvBE)(ro||rc||rL)
B
C
+
vthB
vBE
E
rπ
gmvBE
ro
rthC
vout/vBE = -gm(ro||rc||rL) is the
+ gain from transitor input (v )
BE
to transistor/circuit output vout)
vout
-
E
TRANSISTOR
EXTERNAL
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ECE 3040 - Dr. Alan Doolittle
Important!
Step 5: Calculate Gain and Small Signal Parameters
rthB
B
C
+
vBE
vthB
E
TRANSISTOR
EXTERNAL
+
rπ
gmvBE
ro
rthC vout
-
As previously determined:
vthB/vs = (r1||r2)/([r1||r2] + rs)
Applying a voltage divider:
vBE/vthB = rπ/(rπ+rthB)
E
G i = vout/v
Gain
/ s = (v
( thB/v
/ s)(v
)( BE/v
/ thB)(v
)( out/v
/ BE)
Gain
G
i factor:
f t
vout/vBE = -gm(ro||rc||rL)
Because calculating the DC operating point was done first, we have equations for gm,
rπ, and ro in terms of previously calculated DC currents and voltages.
Plugging in the numbers:
Gain = vout/vs = -139 V/V
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ECE 3040 - Dr. Alan Doolittle
Interpretation/Analysis of Results
Gain = vout/vs = (vthB/vs)(vBE/vthB)(vout/vBE) = -139 V/V
Important!
vthB/vs = (r1||r2)/([r1||r2] + rs)
vBE/vthB = rπ/(rπ+rthB)
Both terms are loss
factors, i.e. they can
never be greater than 1
in magnitude and thus
cause the gain to
decrease.
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This term is the gain factor
and is responsible for
amplifying the signal.
signal
vout/vBE = -gm(ro||rc||rL)
The AC input signal has
been amplified 139 times in
magnitude. The negative
sign
i indicates
i di
there
h has
h been
b
a phase shift of 180°.
ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Base Charging Capacitance (Diffusion Capacitance)
In active mode when the emitter-base is forward
biased, the capacitance of the emitter-base junction
is dominated byy the diffusion capacitance
p
((not
depletion capacitance).
Recall for a diode we started out by saying:
Sum up all the
minority carrier
charges
h
on either
ith
side of the junction
C Diffusion 

QD  qA

0
Neglect charge
injected from the
base into
to the
t e
emitter due to p+
emitter in pnp
dQD
dv D'
dQD dt
dt dv D'
 vD VT
  x Lp
 vD VT
  x Ln




p no  e
 1 e
dx  qA n po  e
 1e
dx
0




'
'
Excess charge
h
stored
d iis due
d almost
l
entirely
i l to the
h
charge injected from the emitter.
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ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Base Charging Capacitance (Diffusion Capacitance)
•The
Th BJT acts
t like
lik a very efficient
ffi i t “siphon”:
“ i h ” As
A majority
j it carriers
i
from the emitter are injected into the base and become “excess
minority carriers”, the Collector “siphons them” out of the base.
•We can view the collector current as the amount of excess charge
in the base collected by the collector per unit time.
•Thus,
us, we ccan eexpress
p ess thee charge
c ge due to
o thee eexcess
cess hole
oe
concentration in the base as:
Q B  iC  F
or the excess charge in the base depends on the magnitude of
current flowing and the “forward” base transport time, F, the
average time the carriers spend in the base.
•It can be shown (see Pierret section 12.2.2)
12 2 2) that:
W2
F 
2 DB
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where,
W  Base Quasi  neutral region width
DB  Minority carrier diffusion coefficient
ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Base Charging Capacitance (Diffusion Capacitance)
Thus, the diffusion capacitance is,
2

QB
W  iC


CB 
Q  po int  
v BE
 2 DB  v BE
IC
CB   F
  F gm
VT
Q  po int
The upper operational frequency of the transistor is limited by
1
the forward base transport time:
f 
2 F
Note the similarity to the Diode Diffusion capacitance we
found previously:
C Diffusion  g d  t
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where  t

p

no

L p  n po Ln qA
IS
is the transit time
ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Base Charging Capacitance (Total Capacitance)
In active mode for small forward biases the depletion
capacitance of the base-emitter junction can contribute to the
total capacitance
C jEo
C jE 
1
V EB
Vbi for emitter base
where,
C jEo  zero bias depletion capaci tan ce
Vbi for emitter base  built in voltage for the E  B junction
Thus, the total emitter-base capacitance is:
C  C B  C jE
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ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Base Charging Capacitance (Depletion Capacitance)
In active mode when the collector-base is reverse biased, the
capacitance of the collector-base junction is dominated by the
depletion capacitance (not diffusion capacitance).
capacitance)
C o
C 
1
VCB
Vbi for collector base
where
h ,
C o  zero bias depletion capaci tan ce
Vbi for collector base  built in voltage for the B  C junction
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ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Collector to Substrate Capacitance (Depletion Capacitance)
In some integrated circuit BJTs (lateral BJTs in particular) the
device has a capacitance to the substrate wafer it is fabricated
in This results from a “buried”
in.
buried reverse biased junction.
junction Thus,
Thus
the collector-substrate junction is reverse biased and the
capacitance of the collector-substrate junction is dominated by
Emitter
the depletion
p
capacitance
p
((not diffusion capacitance).
p
)
C CS
C CS 
1
VCS
Vbi for collector  substrate
where,
p
n-base
p-collector
n-substrate
C CS  zero bias depletion capaci tan ce
Vbi for collector  substrate  built in voltage for the C  substrate junction
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ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Parasitic Resistances
•rb = base resistance between metal interconnect
and B- E junction
•rrc = parasitic collector resistance
•rex = emitter resistance due to polysilicon contact
•These resistance's can be included in SPICE
sim lations but
simulations,
b t are usually
s all ignored in hand
calculations.
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ECE 3040 - Dr. Alan Doolittle
Completing the Small Signal Model of the BJT
Complete Small Signal Model
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ECE 3040 - Dr. Alan Doolittle