EECS 105 Spring 2004, Lecture 22
Lecture 21: BJTs
(Bipolar Junction Transistors)
Prof J. S. Smith
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Context
In Friday’s lecture, we discussed BJTs
(Bipolar Junction Transistors)
Today we will find large signal models
for the bipolar junction transistor,
and start exploring how to use
transistors to make amplifiers and
other analog devices
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Reading
Today’s lecture will finish chapter 7, Bipolar Junction
Transistors (BJT’s)
Then, we will start looking at amplifiers, chapter 8 in
the text.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Lecture Outline
z
z
z
z
BJT Physics (7.2)
BJT Ebers-Moll Equations (7.3)
BJT Large-Signal Models
BJT Small-Signal Models
Next: Circuits
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Currents in the BJT
z
z
A BJT is ordinarily designed so that the minority
carrier injection into the base is far larger than the
minority carrier injection into the emitter.
It is also ordinarily designed such that almost all
the minority carriers injected into the base make it
all the way across to the collector
Department of EECS
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Current controlled
z
So the current is determined by the minority current
across the emitter-base junction
IC ≈ I S e
z
qVBE
kT
But since the majority of the minority current goes
right through the base to the collector:
IC ≈ − I E
z
And so the amount of current that must be supplied
by the base is small compared to the current
controlled:
I C >> I B
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT operating modes
z
Forward active
–
–
z
Saturation
–
z
Emitter-Base forward biased
Base-Collector reverse biased
Both junctions are forward biased
Reverse active
–
–
–
Emitter-Base reverse biased
Base-Collector forward biased
Transistor operation is poor in this direction, becauseβ is
low: lighter doping of the layer designed to be the
collector means that there is a lot of minority carrier
injection out of the Base.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Collector Characteristics (IB)
Saturation Region
(Low Output Resistance)
Breakdown
Linear Increase
Reverse Active
(poor Transistor)
Forward Active
Region
(Very High Output Resistance)
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
The origin of current gain in BJT’s
z
z
z
The majority of the minority carriers injected from the
emitter go across the base to the collector and are swept out
by the electric field in the depletion region of the collectorbase junction.
The base contact doesn’t have to supply that current to
maintain the voltage of the base—the voltage which is
causing the current in the first place.
The current which does have to be supplied by the base
contact comes from two main sources:
–
–
z
Recombination in the base (can often neglect in Silicon)
Injection of minority carriers into the emitter
If we find the ratio of the current to the current that must be
supplied by the base, that will give us the current gain β.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Diffusion Currents
Minority holes
in emitter-they
recombine at
the contact
Minority
electrons
in the base
Base-collector
depletion extracts
the minority
carriers from the
base
The minority carriers injected into the base have a concentration
gradient, and thus a current. Since emitter doping is higher, this
current is much larger than the current due to the minority
carriers injected from the base to the emitter. This is the source
of BJT current gain.
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Diffusion Revisited
z
z
z
Why is minority current profile a linear function?
The diffusion current is proportional to the
gradient×diffusion constant. Since current is
constant→gradient is constant
Note that diffusion current density is controlled by width of
region (base width for BJT):
Density here fixed by potential (injection of carriers)
It is proportional to the number of majority carriers on
The other side of the barrier, and is exponential with the
(lowered) barrier height.
Density at the contact is
equal to the equilibrium
value (strong G/R)
Wp
z
Decreasing width increases current!
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT Currents
Collector current is nearly identical to the (magnitude)
of the emitter current … define
α F = .999
I C = −α F I E
Kirchhoff:
− I E = IC + I B
DC Current Gain:
I C = −α F I E = α F ( I B + I C )
IC =
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αF
IB = βF IB
1−αF
βF =
αF
.999
=
= 999
1 − α F .001
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Origin of αF
Base-emitter junction: some reverse injection of holes
into the emitter Æ base current isn’t zero
Some electrons lost due to
recombination
E
B
C
β F ≈ 100
Typical: α F ≈ .99
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Collector Current
Diffusion of electrons across base results in
J
diff
n
⎛ qDn n pB 0 ⎞ qVkTBE
= qDn
=⎜
⎟e
dx ⎝ WB ⎠
dn p
⎛ qDn n pB 0 AE ⎞
IS = ⎜
⎟
WB
⎝
⎠
IC = I S e
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qVBE
kT
University of California, Berkeley
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Base Current
In silicon, recombination of carriers in the base
can usually be neglected, so the base current is
mostly due to minority injection into the emitter.
Diffusion of holes across emitter results in
J
diff
p
⎞
dpnE ⎛ qD p pnE 0 ⎞ ⎛ qVkTBE
= − qD p
=⎜
−
1
e
⎜
⎟
⎟
dx ⎝ WE ⎠ ⎝
⎠
⎞
⎛ qD p pnE 0 AE ⎞ ⎛ qVkTBE
− 1⎟
IB = ⎜
⎟⎜e
WE
⎝
⎠⎝
⎠
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Current Gain
⎛ qDn n pBo AE ⎞
⎜
⎟ ⎛
WB
IC ⎝
⎠ = Dn ⎞ ⎛ n pB 0 ⎞ ⎛ WE ⎞
βF = =
⎜
⎟⎜
⎟⎜
⎟
I B ⎛ qD p pnEo AE ⎞ ⎝⎜ D p ⎠⎟ ⎝ pnE 0 ⎠ ⎝ WB ⎠
⎜
⎟
WE
⎝
⎠
Minimize base width
ni2
N A, B
N D,E
⎛ n pB 0 ⎞
⎜
⎟= 2 =
ni
N A, B
⎝ pnE 0 ⎠
N D,E
Maximize doping in emitter
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Simple NPN BJT model
z
A simple model for a NPN BJT:
C
I B (t ) →
B
βiB (t )
+
VBE (t )
IB
−
E
Real diode, not
an ideal diode
C
+
+
VCE
VBE
−
−IE
−
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Ebers-Moll Equations
Exp. 6: measure E-M parameters
Derivation: Write emitter and collector currents in terms
of internal currents at two junctions
(
)
(
)
I E = − I ES eVBE / Vth − 1 + α R I CS eVBC / Vth − 1
(
)
(
)
I C = α F I ES eVBE / Vth − 1 − I CS eVBC / Vth − 1
α F I ES = α R I CS
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Ebers-Moll Equivalent Circuit
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University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Parasitic capacitances
z
z
z
To model devices adequately at high frequencies,
we need to account for the charge that we must
move in or out of the devices.
In the FET, this is clearly a capacitance, but in a
BJT the majority of the stored charge is in the form
of minority carriers which are diffusing across the
device in forward operation, but aren’t there when
the transistor is not conducting, so obviously they
must be extracted, or allowed to diffuse away.
This stored charge can be modeled as a capacitance
in small signal models.
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Diffusion Capacitance
z
The total minority carrier charge for a one-sided
junction is (area of triangle)
qV
D
1
1
Qn = qA ⋅ bh2 = qA ⋅ (W − xdep , p )(n p 0 e kT − n p 0 )
2
2
z
For a one-sided junction, the current is dominated
by these minority carriers:
ID =
qVD
qADn
( n p 0 e kT − n p 0 )
W p − xdep , p
Dn
ID
=
2
Qn (W − x
p
dep , p )
Constant!
Department of EECS
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Diffusion Capacitance (cont)
z
The proportionality constant has units of time
τT =
Qn
=
ID
τT =
Temperature
z
q
kT
(Wp − xdep, p )
(W
Dn
p − xdep , p )
µn
2
2
Distance across
P-type base
Diffusion Coefficient
Mobility
The physical interpretation is that this is the transit
time for the minority carriers to cross the p-type
region. Since the capacitance is related to charge:
Qn = τ T I D
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Cd =
∂Qn
∂I
= τT
= g dτ T
∂V
∂V
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT Transconductance gm
z
The transconductance is analogous to diode
conductance
Department of EECS
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Transconductance (cont)
z
Forward-active large-signal current:
iC = I S evBE / Vth (1 + vCE VA )
Q here means quiescent
point not charge
• Differentiating and evaluating at Q = (VBE, VCE )
∂iC
∂vBE
=
Q
q
I S e qVBE / kT (1 + VCE VA )
kT
gm =
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∂iC
∂vBE
=
Q
qI C
kT
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Notation Review
Large signal
iC = f (vBE , vCE )
I C + ∆iC = f (VBE + ∆vBE , VCE + ∆vCE )
Quiescent Point
(bias)
small signal
DC (bias)
I C + ic = f (VBE + vbe , VCE + vce )
Q = (VBE , VCE )
∂f
ic ≈
∂vBE
∂f
vbe +
∂vCE
Q
small signal
(less messy!)
vce
Q
transconductance
Output conductance
Remember, the point of a small signal model is to
produce a set of equations which relate the small
variations in currents and voltages to each other
linearly → and to create a linear equivalent circuit
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT Base Currents
Unlike a MOSFET, there is a DC current into the
base terminal of a bipolar transistor:
I B = I C β F = ( I S β F ) e qVBE / kT (1 + VCE VA )
To find the change in base current due to change
in base-emitter voltage:
ib =
∂iB
∂vBE
∂iB
vbe
∂vBE
Q
ib =
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gm
βF
=
Q
∂iB
∂iC
∂iC
Q
∂vBE
=
Q
1
βF
gm
vbe
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Small Signal Current Gain
β0 =
∆iC
= βF
∆iB
∆iC = β 0 ∆iB
ic = β 0ib
z
Since currents are linearly related, the derivative is a
constant (small signal = large signal)
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Input Resistance rπ
( rπ )
−1
=
∂iB
∂vBE
=
Q
rπ =
z
z
1 ∂iC
β F ∂vBE
=
Q
gm
βF
βF
gm
In practice, the DC current gain βF and the small-signal
current gain βo are both highly variable (+/- 25%)
Typical bias point: DC collector current = 100 µA
rπ = 100
25 mV
= 25 kΩ
.1mA
Ri = ∞ Ω
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Output Resistance ro
Why does current increase slightly with increasing vCE?
Collector (n)
WB
Base (p)
Emitter (n+)
Answer: Base width modulation (similar to CLM for MOS)
Model: Math is a mess, so introduce the Early voltage
iC = I S e vBE / Vth (1 + vCE V A )
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Graphical Interpretation of ro
slope~1/ro
slope
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT Small-Signal Model
ib = rπ vbe
ic = g m vbe +
1
vce
ro
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT Parasitic Capacitors
z
Emitter-base is a forward biased junction Æ
depletion capacitance:
C j , BE ≈ 1.4C j , BE 0
z
z
Collector-base is a reverse biased junction Æ
depletion capacitance
Due to minority charge injection into base, we have
to account for the diffusion capacitance as well
Cb = τ F g m
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
BJT Cross Section
Core Transistor
z
z
z
External Parasitic
Core transistor is the vertical region under the
emitter contact
Everything else is “parasitic” or unwanted
Lateral BJT structure is also possible
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Core BJT Model
Reverse biased junction
Base
Collector
g m vπ
Reverse biased junction &
Diffusion Capacitance
z
z
z
Fictional Resistance
(no noise)
Emitter
Given an ideal BJT structure, we can model most of the
action with the above circuit
For low frequencies, we can forget the capacitors
Capacitors are non-linear! MOS gate & overlap caps are
linear
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Complete Small-Signal Model
Real Resistance
(has noise)
“core” BJT
Reverse biased junctions
External Parasitics
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
Circuits!
z
When the inventors of the bipolar transistor first
got a working device, the first thing they did was to
build an audio amplifier to prove that the transistor
was actually working!
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EECS 105 Spring 2004, Lecture 22
Prof. J. S. Smith
A Simple Circuit: An MOS Amplifier
Input signal
Supply “Rail”
RD
VDD
vo
vs
vGS = VGS + vs
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VGS
I DS
Output signal
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