Lecture 15 - University of California, Berkeley
Download
Report
Transcript Lecture 15 - University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Lecture 15:
Small Signal Modeling
Prof. Niknejad
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Lecture Outline
Department of EECS
Review: Diffusion Revisited
BJT Small-Signal Model
Circuits!!!
Small Signal Modeling
Example: Simple MOS Amplifier
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Notation Review
iC f (vBE , vCE )
Large signal
IC iC f (VBE vBE ,VCE vCE )
Quiescent Point
(bias)
Q (VBE ,VCE )
IC ic f (VBE vbe ,VCE vce )
f
ic
vBE
transconductance
f
vbe
vCE
Q
vce
small signal
DC (bias)
small signal
(less messy!)
Q
Output conductance
Since we’re introducing a new (confusing) subject, let’s adopt some
consistent notation
Please point out any mistakes (that I will surely make!)
Once you get a feel for small-signal analysis, we can drop the notation
and things will be clear by context (yeah right! … good excuse)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Diffusion Revisited
Why is minority current profile a linear function?
Recall that the path through the Si crystal is a zig-zag series
of acceleration and deceleration (due to collisions)
Note that diffusion current density is controlled by width of
region (base width for BJT):
Half go left,
half go right
Density here fixed by potential (injection of carriers)
Physical interpretation: How many electrons (holes) have
enough energy to cross barrier? Boltzmann distribution give
this number.
Density fixed by
metal contact
Wp
Decreasing width increases current!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Diffusion Capacitance
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 0e kT n p 0 )
2
2
For a one-sided junction, the current is dominated
by these minority carriers:
qVD
qADn
ID
(n p 0 e kT n p 0 )
W p xdep , p
Dn
ID
2
Qn W x
p
dep , p
Department of EECS
Constant!
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Diffusion Capacitance (cont)
The proportionality constant has units of time
Qn Wp xdep , p
T
ID
Dn
Temperature
q Wp xdep , p
T
kT
n
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 T I D
Department of EECS
Qn
I
Cd
T
g d T
V
V
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
BJT Transconductance gm
The transconductance is analogous to diode
conductance
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Transconductance (cont)
Forward-active large-signal current:
iC I S e
vBE / Vth
(1 vCE VA )
• Differentiating and evaluating at Q = (VBE, VCE )
iC
vBE
Q
q
I S e qVBE / kT (1 VCE VA )
kT
iC
gm
vBE
Department of EECS
Q
qI C
kT
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
BJT Base Currents
Unlike MOSFET, there is a DC current into the
base terminal of a bipolar transistor:
I B IC F I S F eqVBE / 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
Department of EECS
gm
F
Q
iB
iC
iC
Q
vBE
Q
1
F
gm
vbe
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Small Signal Current Gain
0
iC
F
iB
iC 0 iB
ic 0ib
Since currents are linearly related, the derivative is a
constant (small signal = large signal)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Input Resistance rπ
r
1
iB
vBE
Q
1 iC
F vBE
r
Q
gm
F
F
gm
In practice, the DC current gain F and the small-signal
current gain o are both highly variable (+/- 25%)
Typical bias point: DC collector current = 100 A
r 100
25 mV
25 k
.1mA
Ri
Department of EECS
MOSFET
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
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 )
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Graphical Interpretation of ro
slope~1/ro
slope
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
BJT Small-Signal Model
ib r vbe
1
ic g m vbe vce
ro
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
BJT Capacitors
Emitter-base is a forward biased junction
depletion capacitance:
C j , BE 1.4C j , BE 0
Collector-base is a reverse biased junction
depletion capacitance
Due to minority charge injection into base, we have
to account for the diffusion capacitance as well
Cb F gm
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
BJT Cross Section
Core Transistor
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 Fall 2003, Lecture 15
Prof. A. Niknejad
Core BJT Model
Reverse biased junction
Base
Collector
g m v
Reverse biased junction &
Diffusion Capacitance
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
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Complete Small-Signal Model
“core” BJT
Reverse biased junctions
Real Resistance
(has noise)
External Parasitics
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Circuits!
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!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Modern ICs
Source: Intel Corporation
Used without permission
Source: Texas Instruments
Used without permission
First IC (TI, Jack Kilby, 1958): A couple of transistors
Modern IC: Intel Pentium 4 (55 million transistors, 3 GHz)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
A Simple Circuit: An MOS Amplifier
Input signal
RD
VDD
Supply “Rail”
vo
vs
vGS VGS vs
Department of EECS
VGS
I DS
Output signal
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Selecting the Output Bias Point
The bias voltage VGS is selected so that the output is
mid-rail (between VDD and ground)
For gain, the transistor is biased in saturation
Constraint on the DC drain current:
IR
VDD Vo VDD VDS
RD
RD
All the resistor current flows into transistor:
I R I DS , sat
Must ensure that this gives a self-consistent
solution (transistor is biased in saturation)
VDS VGS VT
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Finding the Input Bias Voltage
Ignoring the output impedance
I DS , sat
W
1
nCox (VGS VTn ) 2
L
2
Typical numbers: W = 40 m, L = 2 m,
RD = 25k, nCox = 100 A/V2, VTn = 1 V,
VDD = 5 V
I RD
VDD
W
1
I DS , sat nCox (VGS VTn ) 2
2 RD
L
2
5V
μA 1
100μA 20 100 2 (VGS 1) 2
50k
V 2
.1 (VGS 1) 2
Department of EECS
VGS 1.32
VGS VT .32 VDS 2.5
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Applying the Small-Signal Voltage
Approach 1. Just use vGS in the equation for the total
drain current iD and find vo
vGS VGS vs
vs vˆs cos t
vO VDD RD iDS
W 1
VDD RD nCox
(VGS vs VT ) 2
L 2
Note: Neglecting charge storage effects. Ignoring
device output impedance.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Solving for the Output Voltage vO
vO VDD RD iDS
vO VDD RD iDS
W 1
VDD RD nCox
(VGS vs VT ) 2
L 2
vs
W 1
2
VDD RD nCox
(VGS VT ) 1
L 2
V
V
GS
T
2
I DS
vs
vO VDD RD I DS 1
V
V
GS
T
2
VDD
2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Small-Signal Case
Linearize the output voltage for the s.s. case
Expand (1 + x)2 = 1 + 2x + x2 … last term can be
dropped when x << 1
2
2
v
2v
v
s
s
s
------------------------------------------------1
+
=
1
+
+
-------------------------
V GS – V Tn
V GS – V Tn V – V
GS
Tn
Neglect
2vs
vO VDD RD I DS 1
VGS VT
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Linearized Output Voltage
For this case, the total output voltage is:
2vs
VDD
vO VDD
1
2 VGS VT
vsVDD
VDD
vO
2 VGS VT
“DC”
Small-signal output
The small-signal output voltage:
vo
vsVDD
Av vs
VGS VT
Voltage gain
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
Plot of Output Waveform (Gain!)
Numbers: VDD / (VGS – VT) = 5/ 0.32 = 16
output
input
mV
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 15
Prof. A. Niknejad
There is a Better Way!
What’s missing: didn’t include device output
impedance or charge storage effects (must solve
non-linear differential equations…)
Approach 2. Do problem in two steps.
DC voltages and currents (ignore small signals
sources): set bias point of the MOSFET ... we had
to do this to pick VGS already
Substitute the small-signal model of the MOSFET
and the small-signal models of the other circuit
elements …
This constitutes small-signal analysis
Department of EECS
University of California, Berkeley