4. single stage amplifier

4. single stage amplifier

1. Set 3: Single-Stage Amplifiers
1 SM
EECE488: Analog CMOS Integrated Circuit Design
3. Single-Stage Amplifiers
Shahriar Mirabbasi
Department of Electrical and Computer Engineering
University of British Columbia
[email protected]
Technical contributions of Pedram Lajevardiin revising the course notes are greatly acknowledged.

2. Set 3: Single-Stage Amplifiers
2 SM
Overview
1. Why Amplifiers?
2. Amplifier Characteristics
3. Amplifier Trade-offs
4. Single-stage Amplifiers
5. Common Source Amplifiers
1. Resistive Load
2. Diode-connected Load
3. Current Source Load
4. Triode Load
5. Source Degeneration

3. Set 3: Single-Stage Amplifiers
3 SM
Overview
6. Common-Drain (Source-Follower) Amplifiers
1. Resistive Load
2. Current Source Load
3. Voltage Division in Source Followers
7. Common-Gate Amplifiers
6. CascodeAmplifiers

4. Set 3: Single-Stage Amplifiers
4 SM
Reading Assignments
• Reading:
Chapter 3 of Razavi’s book
• In this set of slides we will study low-frequency small-signal behavior of
single-stage CMOS amplifiers. Although, we assume long-channel
MOS models (not a good assumption for deep submicron technologies)
the techniques discussed here help us to develop basic circuit intuition
and to better understand and predict the behavior of circuits.
Most of the figures in these lecture notes are © Design of Analog
CMOS Integrated Circuits, McGraw-Hill, 2001.

5. Set 3: Single-Stage Amplifiers
5 SM
Why Amplifiers?
• Amplifiers are essential building blocks of both analog and digital
systems.
• Amplifiers are needed for variety of reasons including:
– To amplify a weak analog signal for further processing
– To reduce the effects of noise of the next stage
– To provide a proper logical levels (in digital circuits)
• Amplifiers also play a crucial role in feedback systems
• We first look at the low-frequency performance of amplifiers. Therefore,
all capacitors in the small-signal model are ignored!

6. Set 3: Single-Stage Amplifiers
6 SM
Amplifier Characteristics -1
• Ideally we would like that the output of an amplifier be a linear
function of the input, i.e., the input times a constant gain:
x y
1
α
=
x
y
• In real world the input-output characteristics is typically a nonlinear
function:

7. Set 3: Single-Stage Amplifiers
7 SM
Amplifier Characteristics -2
• It is more convenient to use a linear approximation of a nonlinear
function.
• Use the tangent line to the curve at the given (operating) point.
x
y
• The larger the signal changes about the operating point, the worse the
approximation of the curve by its tangent line.
• This is why small-signal analysis is so popular!

8. Set 3: Single-Stage Amplifiers
8 SM
Amplifier Characteristics -3
• Let to get:
n
n
xx xx xx y) ( ) ( ) (
0
2
0 2 0 1 0
− + + − + − + ≈
α α α α
L
) (
0 1 0
xx y−
+
≈
α
α
x y xf y y∆
≈
−
=
−
=
∆
1 0 0
)(
α
α
• If x-x
0
=∆x is small, we can ignore the higher-order terms (hence the
name small-signal analysis) to get:
•α
0
is referred to as the operating (bias) point and α
1
is the small-signal
gain.
!
)(
0
n
xf
n
n
=
α
• A well-behaved nonlinear function in the vicinity of a given point can be
approximated by its corresponding Taylor series:
n
n
xx
n
xf
xx
xf
xx xf xf y) (
!
)(
) (
!2
)(''
) ()(' )(
0
0 2
0
0
0 0 0
−⋅ + + −⋅ + −⋅ + ≈L

9. Set 3: Single-Stage Amplifiers
9 SM
• In practice, when designing an amplifier, we need to optimize for some
performance parameters. Typically, these parameters trade
performance with each other, therefore, we need to choose an
acceptable compromise.
Amplifier Trade-offs

10. Set 3: Single-Stage Amplifiers
10 SM
Single-Stage Amplifiers
• We will examine the following types of amplifiers:
1. Common Source
2. Common Drain (Source Follower )
3. Common Gate
4. Cascode and Folded Cascode
• Each of these amplifiers have some advantages and some
disadvantages. Often, designers have to utilize a cascade combination
of these amplifiers to meet the design requirements.

11. Set 3: Single-Stage Amplifiers
11 SM
Common Source Basics -1
• In common-source amplifiers, the input is (somehow!) connected to the
gate and the output is (somehow!) taken from the drain.
• We can divide common source amplifiers into two groups:
1. Without source degeneration (no body
effect for the main transistor):
2. With source degeneration (have to take
body effect into account for the main
transistor):

12. Set 3: Single-Stage Amplifiers
12 SM
Common Source Basics -2
In a simple common source amplifier:
• gate voltage variations times g
m
gives the drain current
variations,
• drain current variations times the load gives the output voltage
variations.
• Therefore, one can expect the small-signal gain to be:
D m v
R g A⋅ =

13. Set 3: Single-Stage Amplifiers
13 SM
Common Source Basics -3
• Different types of loads can be used in an amplifier:
1. Resistive Load
2. Diode-connected Load
3. Current Source Load
4. Triode Load
• The following parameters of amplifiers are very important:
1. Small-signal gain
2. Voltage swing

14. Set 3: Single-Stage Amplifiers
14 SM
Resistive Load -1
• Let’s use a resistor as the load.
• The region of operation of M
1
depends on its size and the values of V
in
and R.
• We are interested in the small-signal gain and the headroom (which
determines the maximum voltage swing).
• We will calculate the gain using two different methods
1. Small-signal model
2. Large-signal analysis

15. Set 3: Single-Stage Amplifiers
15 SM
Resistive Load -2
Gain – Method 1: Small-Signal Model• This is assuming that the transistor is in saturation, and channel length
modulation is ignored.
• The current through R
D
:
• Output Voltage:
• Small-signal Gain:
IN m D
vg i
⋅
=
D IN m D D OUT
R
vg
R
i v⋅
⋅
−
=
⋅
−
=
D m
IN
OUT
v
R g
v
v
A⋅ −= =

16. Set 3: Single-Stage Amplifiers
16 SM
Resistive Load -3
Gain – Method 2: Large-Signal Analysis• If V
IN
<V
TH
, M1 is off, and V
OUT
= V
DD
= V
DS
.
m D TH IN ox n D
IN
OUT
v
TH IN ox n D dd D D dd OUT
g R V V
L
W
C R
V
V
A
V V
L
W
C R V i R V V
⋅ −= − ⋅ ⋅ ⋅⋅ −=
∂
∂
=
− ⋅ ⋅ ⋅⋅⋅ − =⋅ − =
) (
) (
2
1
2
µ
µ
• As V
IN
becomes slightly larger than V
TH
, M
1
turns on and goes into
saturation (V
DS
≈V
DD
> V
GS
-V
TH
≈0).
0=
∂
∂
=
=⋅ − =
IN
OUT
v
DD D D DD OUT
V
V
A
V i R V V
• As V
IN
increases, V
DS
decreases, and M
1
goes into triode when
V
IN
-V
TH
= V
OUT
. We can find the value of V
IN
that makes M
1
switch
its region of operation.
) ( ) (
2
1
2
TH IN TH IN ox n D dd D D dd OUT
V V V V
L
W
C R V i R V V− = − ⋅ ⋅ ⋅⋅⋅ − =⋅ − =
µ

17. Set 3: Single-Stage Amplifiers
17 SM
Resistive Load -4
Gain – Method 2: Large-Signal Analysis (Continued)
• As V
IN
increases, V
DS
decreases, and M
1
goes into triode.
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
∂
∂
⋅ − +
∂
∂
⋅ − ⋅ ⋅ ⋅ ⋅ −=
∂
∂
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
− ⋅ − ⋅ ⋅ ⋅ ⋅ − =⋅ − =
IN
OUT
OUT OUT
IN
OUT
TH IN ox n D
IN
OUT
OUT
OUT TH IN ox n D DD D D DD OUT
V
V
V V
V
V
V V
L
W
C R
V
V
V
V V V
L
W
C R V i R V V
) (
2
) (
2
µ
µ
• We can find A
v
from above. It will depend on both V
IN
and V
OUT
.
• If V
IN
increases further, M
1
goes into deep triode if V
OUT
<< 2(V
IN
-V
TH
).
D ON
ON
DD
ON
D
DD
TH IN ox n D
DD
OUT
OUT TH IN ox n D DD D D DD OUT
R R
R
V
R
R
V
V V
L
W
C R
V
V
V V V
L
W
C R V i R V V
+
⋅ =
⋅ +
=
− ⋅ ⋅ ⋅ ⋅ +
=
⋅ − ⋅ ⋅ ⋅ ⋅ − =⋅ − =
1
1 ) ( 1
) (
µ
µ

18. Set 3: Single-Stage Amplifiers
18 SM
Resistive Load -5
Example: Sketch the drain current and g
m
of M
1
as a function of V
IN
.
•g
m
depends on V
IN
, so if V
IN
changes by a large amount the small-
signal approximation will not be valid anymore.
• In order to have a linear amplifier, we don’t want gain to depend on
parameters like g
m
which depend on the input signal.

19. Set 3: Single-Stage Amplifiers
19 SM
Resistive Load -6
• Gain of common-source amplifier:
• To increase the gain:
1. Increase g
m
by increasing W or V
IN
(DC portion or bias). Either way,
I
D
increases (more power) and V
RD
increases, which limits the
voltage swing.
2. Increase R
D
and keep I
D
constant (g
m
and power remain constant).
But, V
RD
increases which limits the voltage swing.
3. Increase R
D
and reduce I
D
so V
RD
remains constant.
¾If I
D
is reduced by decreasing W, the gain will not change.
¾If I
D
is reduced by decreasing V
IN
(bias), the gain will increase.
Since R
D
is increased, the bandwidth becomes smaller (why?).
• Notice the trade-offs between gain, bandwidth, and voltage swings.
eff
RD
D
RD
ox n
D
RD
TH IN ox n D m v
V
V
I
V
L
W
C
I
V
V V
L
W
C R g A
⋅−
= ⋅ −= ⋅ − −= ⋅ −=
2
2 ) (
µ µ

20. Set 3: Single-Stage Amplifiers
20 SM
Resistive Load -7
• Now let’s consider the simple common-source circuit with channel
length modulation taken into account.
• Channel length modulation becomes more important as R
D
increases
(in the next slide we will see why!).
• Again, we will calculate the gain in two different methods
1. Small-signal Model
2. Large Signal Analysis

21. Set 3: Single-Stage Amplifiers
21 SM
Resistive Load -8
Gain – Method 1: Small-Signal Model
• This is assuming that the transistor is in saturation.
• The current through R
D
:
• Output Voltage:
• Small-signal Gain:
IN m D
v
g
i
⋅
=
(
)
(
)
o D IN m o D D OUT
rR vg rR i v⋅
⋅
−
=
⋅
−
=
()
o D m
IN
OUT
v
rR g
v
v
A
⋅ −= =

22. Set 3: Single-Stage Amplifiers
22 SM
Resistive Load -9
Gain – Method 2: Large-Signal Analysis
()
()
()
()
o D m
D o
m D o
o
D
m D
D D
m D
TH IN ox n D
OUT TH IN ox n D
v
IN
OUT
TH IN OUT TH IN ox n D
IN
OUT
OUT TH IN ox n D DD D D DD OUT
r R g
R r
g Rr
r
R
g R
I R
g R
V V
L
W
C R
V V V
L
W
C R
A
V
V
V V V V V
L
W
C R
V
V
V V V
L
W
C R V I R V V
⋅ −=
+
⋅ ⋅ −
=
⋅ +
⋅ −
=
λ⋅ ⋅ +
⋅ −
=
λ⋅ − ⋅ ⋅ ⋅µ⋅ ⋅ +
⋅λ+⋅ − ⋅ ⋅ ⋅µ⋅ −
=
⎥
⎦
⎤
⎢
⎣
⎡
∂
∂
⋅λ⋅ − ⋅ + ⋅λ+ ⋅ − ⋅ ⋅ ⋅µ⋅ −=
∂
∂
⋅λ+⋅ − ⋅ ⋅ ⋅µ⋅⋅ − = ⋅ − =
1
1
1
) (
2
1
1
1) (
) (
2
1
1) (
1 ) (
2
1
2
2
• As V
IN
becomes slightly larger than V
TH
, M
1
turns on and goes into
saturation (V
DS
≈V
DD
> V
GS
-V
TH
≈0).

23. Set 3: Single-Stage Amplifiers
23 SM
Resistive Load -10
Example:
• Assuming M
1
is biased in active region, what is the
small-signal gain of the following circuit?
()
o m o m
IN
OUT
v
rg r g
v
v
A
⋅ −= ∞⋅ −= =
•I
1
is a current source and ideally has an infinite impedance.
• This is the maximum gain of this amplifier (why?), and is known as the
intrinsic gain.
• How can V
IN
change if I
1
is constant?
()
OUT TH IN ox n D
V V V
L
W
C I
⋅ +⋅ − ⋅ ⋅ ⋅⋅ =
λ µ
1) (
2
1
2
• Here we have to take channel-length modulation into account. As V
IN
changes, V
OUT
also changes to keep I
1
constant.

24. Set 3: Single-Stage Amplifiers
24 SM
Diode Connected Load -1
• Often, it is difficult to fabricate tightly controlled or reasonable size
resistors on chip. So, it is desirable to replace the load resistor with a
MOS device.
• Recall the diode connected devices:
Body Effect R
X
(when λ≠0) R
X
(when λ=0)
NO
YES
m
o X
g
r R
1
=
m
X
g
R
1
=
mb m
o X
g g
r R
+
=
1
mb m
X
g g
R
+
=1

25. Set 3: Single-Stage Amplifiers
25 SM
Diode Connected Load -2
• Now consider the common-source amplifier with two types of diode
connected loads:
1. PMOS diode connected load:
(No body effect)
2. NMOS diode connected load:
(Body effect has to be taken into account)

26. Set 3: Single-Stage Amplifiers
26 SM
Diode Connected Load -3
PMOS Diode Connected Load:
• Note that this is a common source configuration
with M
2
being the load. We have:
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅ −= ⋅ −= =
1 2
2
1 1 1
1
o o
m
m o X m
IN
OUT
v
rr
g
g rR g
v
v
A
• Ignoring the channel length modulation (r
o1
=r
o2
=∞), we can write:
1 1
2 2
2 2
2
1 1
1
2
1
2
1
2
2
1
1
2
1
2
1
2
2
2
2
1
TH GS
TH SG
TH SG
D
TH GS
D
m
m
v
p
n
D ox p
D ox n
m
m
m
m v
V V
V V
V V
I
V V
I
g
g
A
L
W
L
W
I
L
W
C
I
L
W
C
g
g
g
g A
−
−
−=
−
⋅
−
⋅
−= −=
⎟
⎠
⎞
⎜
⎝
⎛
⋅µ
⎟
⎠
⎞
⎜
⎝
⎛
⋅µ
−=
⋅⎟
⎠
⎞
⎜
⎝
⎛
⋅ ⋅µ
⋅⎟
⎠
⎞
⎜
⎝
⎛
⋅ ⋅µ
−= −=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∞∞ ⋅ −=

27. Set 3: Single-Stage Amplifiers
27 SM
Diode Connected Load -4
NMOS Diode Connected Load:
• Again, note that this is a common source configuration
with M
2
being the load. We have:
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⋅ −= ⋅ −= =
1 2
2 2
1 1 1
1
o o
mb m
m o X m
IN
OUT
v
rr
g g
g rR g
v
v
A
• Ignoring the channel length modulation (r
o1
=r
o2
=∞), we can write:
TH GS
TH GS
v
m
m
mb m
m
mb m
m v
V V
V V
L
W
L
W
A
g
g
g g
g
g g
g A
−
−
⋅
+
−=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+
−=
+⋅
−=
+
−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∞∞
+
⋅ −=
1
2
2
1
2
1
2 2
1
2 2
1
1
1
1
1
) 1(
1
η η
η

28. Set 3: Single-Stage Amplifiers
28 SM
Diode Connected Load -5
• For a diode connected load we observe that (to the first order
approximation):
1. The amplifier gain is not a function of the bias current. So,
the change in the input and output levels does not affect the
gain, and the amplifier becomes more linear.
2. The amplifier gain is not a function of the input signal
(amplifier becomes more linear).
3. The amplifier gain is a weak function (square root) of the
transistor sizes. So, we have to change the dimensions by a
considerable amount so as to increase the gain.

29. Set 3: Single-Stage Amplifiers
29 SM
Diode Connected Load -6
4. The gain of the amplifier is reduced when body effect
should be considered.
5. We want M
1
to be in saturation, and M
2
to be on (M
2
cannot
be in triode (why?)):
6. The voltage swing is constrained by both the required
overdrive voltages and the threshold voltage of the diode
connected device.
7. A high amplifier gain leads to a high overdrive voltage for
the diode connected device which limits the voltage swing.
2 1 1 1
:2 , :1
TH DD OUT eff TH GS OUT
V V V M V V V V M
− < = − >

30. Set 3: Single-Stage Amplifiers
30 SM
Diode Connected Load -6
Example:
• Find the gain of the following circuit if M1 is biased in saturation and
I
s
=0.75I
1
.
()
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅ −=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∞ ⋅ −= ⋅ −= =
1 2
2
1 1 2
2
1 1 1
1 1
o o
m
m o o
m
m o Is X m
IN
OUT
v
rr
g
g r r
g
g rrR g
v
v
A
• Ignoring the channel length modulation (r
o1
=r
o2
=∞) we get:
1 1
2 2
2
1
2 2
2
1 1
1
2
2
1
1
2
1
2
1
4 2
2
2
2
2
1
TH GS
TH SG
p
n
v
TH SG
D
TH GS
D
D ox p
D ox n
m
m
m
m v
V V
V V
L
W
L
W
A
V V
I
V V
I
I
L
W
C
I
L
W
C
g
g
g
g A
−
−
⋅−=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅−=
−
⋅
−
⋅
−=
⋅⎟
⎠
⎞
⎜
⎝
⎛
⋅ ⋅
⋅⎟
⎠
⎞
⎜
⎝
⎛
⋅ ⋅
−= −=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∞∞ ⋅ −=
µ
µ
µ
µ

31. Set 3: Single-Stage Amplifiers
31 SM
Diode Connected Load -7
Example (Continued):
• We observe for this example that:
1. For fixed transistor sizes, using the current source increases the
gain by a factor of 2.
2. For fixed overdrive voltages, using the current source increases
the gain by a factor of 4.
3. For a given gain, using the current source allows us to make the
diode connected load 4 times smaller.
4. For a given gain, using the current source allows us to make the
overdrive voltage of the diode connected load 4 times smaller.
This increases the headroom for voltage swing.

32. Set 3: Single-Stage Amplifiers
32 SM
Current Source Load -1
• Note that current source M
2
is the load.
•
• Recall that the output impedance of M
2
seen from V
out
:
()
()
1 2 1 1 1
2
o o m o X m
IN
OUT
v
o
X
X
X
rr g rR g
v
v
A
r
i
v
R
⋅ −= ⋅ −= =
= =
• For large gain at given power, we want large
r
o
and
Increase L and W keeping the aspect ratio constant (so r
o
increases
and I
D
remains constant). However, this approach increases the
capacitance of the output node.
• We want M
2
to be in saturation so
W
L
L
W
L
I
r
D
o
2
1
1 1
=
⋅
∝
⋅λ
=
2 2 2 2eff DD OUT eff TH SG OUT DD SD
V V V V V V V V V− < → = − > − =

33. Set 3: Single-Stage Amplifiers
33 SM
Current Source Load -2
• We also want M
1
to be in saturation:
1 1 1 1
eff OUT eff TH GS OUT DS
V V V V V V V
> →
=
−
>
=
• Thus, we want V
eff1
and V
eff2
to be small, so that there is more
headroom for output voltage swing. For a constant I
D
, we can increase
W
1
and W
2
to reduce V
eff1
and V
eff2
.
• The intrinsic gain of this amplifier is:
• In general, we have:
o m v
r
g
A
⋅
−
=
L A
WL
r
L
W
g
v o m
∝ → ∝ ∝
2
,
• But since current in this case is roughly constant:
LW A L
I
r
L
W
I
L
W
C g
v
D
o D ox n m
∝ → ∝
⋅
= ∝ ⋅ ⋅ ⋅ =
λ
µ
1
, 2

34. Set 3: Single-Stage Amplifiers
34 SM
Triode Load
• We recognize that this is a common source
configuration with M2 being the load. Recall that if
M
2
is in deep triode, i.e., V
SD
<<2(V
SG
-|V
TH
|), it
behaves like a resistor.
(
)
()
()
()
1 2 1 2
,
1 1
: 2
o ON m v
TH b dd ox p TH SG ox p
ON
TH SG SD
r R g A
V V V
L
W
C V V
L
W
C
R
V V VIf
⋅ −=
− − ⋅ ⋅ ⋅
=
− ⋅ ⋅ ⋅
=
− <<
µ µ
•V
b
should be low enough to make sure that M
2
is in deep triode region
and usually requires additional complexity to be precisely generated.
•R
ON2
depends on µ
p
, C
ox
, and V
TH
which in turn depend on the
technology being used.
• In general, this amplifier with triode load is difficult to design and use!
• However, compared to diode-connected load, triode load consumes less
headroom:
DD OUT eff TH GS OUT
V V M V V V V M≈ = − >: , :
2 1 1 1

35. Set 3: Single-Stage Amplifiers
35 SM
Source Degeneration -1
• The following circuit shows a common source configuration
with a degeneration resistor in the source.
• We will show that this configuration makes the common
source amplifier more linear.
• We will use two methods to derive the gain of this circuit:
1. Small-signal Model
2. Using the following Lemma
Lemma:
In linear systems, the voltage gain is equal to –G
m
R
out
.

36. Set 3: Single-Stage Amplifiers
36 SM
Source Degeneration -2
Gain – Method 1: Small Signal Model
() ()
S D S mb m O
D O m
IN
OUT
v
D IN m
O
S
O
D
S mb S m OUT
O
S
D
OUT
OUT
S
D
OUT
mb S
D
OUT
IN m
D
OUT
S
D
OUT
S OUT BS S
D
OUT
IN S OUT IN
D
OUT
OUT
O
S OUT OUT
BS mb m OUT
R R R g g r
Rr g
v
v
A
R v g
r
R
r
R
R g R g v
r
R
R
v
v
R
R
v
g R
R
v
v g
R
v
R
R
v
R i v R
R
v
v R i v v
R
v
i
r
R i v
v g v g i
+ + ⋅ + +⋅
⋅⋅ −
= =
⋅ ⋅ −=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + ⋅ + ⋅ +⋅
⋅ +
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅ ⋅ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅ + ⋅ =
−
⋅ = ⋅ −= ⋅ + = ⋅ − =
−
=
⋅ −
+ ⋅ +⋅ =
1
1
,
,
1
1

37. Set 3: Single-Stage Amplifiers
37 SM
Source Degeneration -3
Gain – Method 2: Lemma
• The Lemma states that in linear systems, the voltage gain is
equal to –G
m
R
out
. So we need to find G
m
and R
out
.
1. G
m
:
Recall that the equivalent transconductance of the above
Circuit is:
()
()
S S mb m
m
S S mb S m
m
m
R R g g r
r g
R R g R g r r
r g
v
i
G
O
O
O O
O
IN
OUT
+ ⋅ + +
⋅
=
+ ⋅ + ⋅ ⋅ +
⋅
= =
] 1[

38. Set 3: Single-Stage Amplifiers
38 SM
Source Degeneration -4
Gain – Method 2: Lemma (Continued)
1. R
OUT
:
We use the following small signal model to derive the small
signal output impedance of this amplifier:
()
()() ()
()()
() ()
() ()
D O S mb m S
D O S mb m S
D X OUT
O S mb m S O S mb S m S
X
X
X
O S X mb S X m X S X
O BS mb m X S X X
S X BS S X
R r R g g R
R r R g g R
R R R
r R g g R r R g R g R
i
v
R
r Ri g Ri g i Ri
r v g v g i Ri v
Ri v Ri v
+ ⋅ ⋅ + + +
⋅ ⋅ ⋅ + + +
= =
⋅ ⋅ + + + =⋅ ⋅ + ⋅ + + = =
⋅ ⋅ −⋅ − ⋅ −⋅ − + ⋅ =
⋅ ⋅ −⋅ − + ⋅
=
⋅ −= ⋅ −=
) (1
) (1
) (1 1
,
1
1
• Since typically r
O
>>R
S
:
(
)
(
)
(
)
O S mb m O S mb m O S mb m S X
r R g g r R g g r R g g R R⋅ ⋅ + =⋅ ⋅ + + =⋅ ⋅ + + + =) (1 ) (1

39. Set 3: Single-Stage Amplifiers
39 SM
Source Degeneration -5
Gain – Method 2: Lemma (Continued)
()
(
)
(
)
()
()
D O S mb m O
D O m
D S O mb O m O
D S O mb O m O
S S mb S m O O
O m
OUT m v
R r R g g r
Rr g
R R r g r g r
R R r g r g r
R R g R g r r
r g
R G A
+⋅ ⋅ + + +
⋅⋅ −
=
+ ⋅ ⋅ +⋅ + +
⋅ ⋅ ⋅ +⋅ + +
⋅
+ ⋅ + ⋅ ⋅ +
⋅
−= ⋅ −=
) (1
1
1
() ()
()
D O S mb m S
D O S mb m S
OUT
R r R g g R
R r R g g R
R
+⋅ ⋅ + + +
⋅ ⋅ ⋅ + + +
=
) (1
) (1
()
S S mb m O
O m
m
R R g g r
r g
G
+ ⋅ + ⋅+
⋅
=
) 1(

40. Set 3: Single-Stage Amplifiers
40 SM
Source Degeneration -6
• If we ignore body effect and channel-length modulation:
Method 1 – Small-signal Model:
S m
D m
OUT m v
R g
R
g
R G A
⋅ +
⋅
−
= ⋅ −=
1() ()
()
()
(
)
()
D
S mb m
D S mb m
D S O mb O m O
D S O mb O m O
OUT
R
R g g
R R
g
g
R R r g rg r
R R r
g
r
g
r
R
mb
g
or
=
⋅ + +
⋅
⋅
+
+
=
+ ⋅ ⋅ +⋅ + +
⋅
⋅
⋅
+
⋅
+ +
=
→
∞→
1
1
1
1
lim
0
()
()
S m
m
S mb m
m
S S mb S m O O
O m
m
R g
g
R g g
g
R R g R g r r
r
g
G
mb
g
or
⋅ +
=
⋅ + +
=
+ ⋅ + ⋅ ⋅ +
⋅
=
→
∞→
1 1
lim
0
S m
D m
IN
OUT
v
S m
IN GS
S GS m IN GS D GS m OUT
R g
R g
v
v
A
R g
v v
R v
g
v v R v
g
v
⋅ +
⋅ −
= = →
⋅ +
⋅ =
⋅
⋅
−
=
⋅ ⋅ −=
1 1
1
,
Method 2 – Taking limits:

41. Set 3: Single-Stage Amplifiers
41 SM
Source Degeneration -7
Obtaining G
m
and R
out
directly assuming
λ
=
γ
=0:
1. G
m
:
S m
m
IN
D
m
S m
IN GS
S GS m IN GS GS m D
R g
g
v
i
G
R g
v v
R v
g
v v v
g
i
⋅ +
= = →
⋅ +
⋅ =
⋅
⋅
−
=
⋅ =
1 1
1
,
2. R
OUT
:
D
X
X
OUT
D
X
GS m
D
X
X
GS S GS m GS
R
i
v
R
R
v
vg
R
v
i
v R vg v
= =
= ⋅ + =
=
→ ⋅ ⋅ −=0
S m
D m
OUT m v
R g
R
g
R G A
⋅ +
⋅
−
= ⋅ −=
1

42. Set 3: Single-Stage Amplifiers
42 SM
Source Degeneration -8
• If we ignore body effect and channel-length modulation:
S m
D m
v D OUT
S m
m
m
R g
R
g
A R R
R g
g
G
⋅ +
⋅
−
= → =
⋅ +
=
1
,
1
• We Notice that as R
S
increases G
m
becomes less dependent on g
m
:
S S m
m
m
R R g
g
G
s
R
s
R
1
1
lim lim=
⋅ +
=
∞→ ∞→
• That is for large R
S
:
OUT S IN
S IN
OUT
m
iR v
R v
i
G⋅ ≈ → ≈ =
1
• Therefore, the amplifier becomes more linear when R
S
is large enough.
Intuitively, an increase in v
IN
tend to increase I
D
, however, the voltage
drop across R
S
also increases. This makes the amplifier less sensitive
to input changes, and makes I
D
smoother!
• The linearization is achieved at the cost of losing gain and voltage
headroom.

43. Set 3: Single-Stage Amplifiers
43 SM
Source Degeneration -9
• We can manipulate the gain equation so the numerator is the
resistance seen at the drain node, and the denominator is the
resistance in the source path.
• The following are I
D
and g
m
of a transistor without R
S
.
S
m
D
S m
D m
v
R
g
R
R g
R
g
A
+
−
=
⋅ +
⋅
−
=
1 1
•I
D
and g
m
of a transistor considering R
S
are:
• When I
D
is small such that R
S
g
m
<<1, G
m
≈g
m
.
• When I
D
is large such that R
S
g
m
>>1, G
m
≈1/R
S
.

44. Set 3: Single-Stage Amplifiers
44 SM
Alternative Method to Find the Output-Resistance of a
Degenerated Common-Source Amplifier

45. Set 3: Single-Stage Amplifiers
45 SM
Why Buffers?
• Common Source amplifiers needed a large load impedance to provide
a large gain.
• If the load is small but we need a large gain (can you think of an
example?) a bufferis used.
• Source-follower (common-drain) amplifiers can be used as buffers.
Ideal Buffer:
1. R
IN
=∞: the input current is zero; it doesn’t load the previous stage.
2. R
OUT
=0: No voltage drop at the output; behaves like a voltage source.
1 ,0 ,
=
=
∞
=
V OUT IN
A
R R

46. Set 3: Single-Stage Amplifiers
46 SM
Resistive Load -1
• We will examine the Source follower amplifier with two different loads:
1. Resistive Load
2. Current Source Load
–Resistive Load:
• As shown below the output (source voltage) will follow the input (gate
voltage). We will analyze the following circuit using large-signal and
small-signal analysis.

47. Set 3: Single-Stage Amplifiers
47 SM
Resistive Load -2
Large Signal Analysis:
• The relationship between V
IN
and V
OUT
is:
()( )
()( )
S OUT DD TH OUT IN ox n OUT
S DS TH GS ox n D S OUT
R V V V V V
L
W
C V
R V V V
L
W
C I R V
⋅ ⋅− ⋅ +⋅ − − =
⋅ ⋅ +⋅ − =⋅ =
λ λ µ
λ µ
1
2
1
1
2
1
2
2
(
)
IN
OUT
IN
OUT
SB F IN
OUT
OUT
TH
IN
TH
OUT SB F SB F TH TH
V
V
V
V
V V
V
V
V
V
V
V V V V V
∂
∂
⋅=
∂
∂
⋅
+Φ⋅
=
∂
∂
⋅
∂
∂
=
∂
∂
= Φ⋅ − +Φ⋅ ⋅+ =
η
γ
γ
22
, 2 2
0
()
()
()()
IN
OUT
S TH GS ox n
S DS
IN
TH
IN
OUT
TH OUT IN ox n
IN
OUT
V
V
R V V
L
W
C
R V
V
V
V
V
V V V
L
W
C
V
V
∂
∂
⋅ −⋅ ⋅ − +
⋅ ⋅ +⋅
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
−⋅ − − =
∂
∂
λ µ
λ µ
2
2
1
1 1
• Differentiate with respect to V
IN
:
• Need to find the derivative of V TH
with respect to V
IN
:

48. Set 3: Single-Stage Amplifiers
48 SM
Resistive Load -3
()
S
o
mb m
S m
o
S
S mb S m
S m
IN
OUT
V
S m S D S m S m
IN
OUT
R
r
g g
R g
r
R
R g R g
R g
V
V
A
R g RI R g R g
V
V
⋅
⎟
⎟
⎠⎞
⎜
⎜
⎝⎛
+ + +
⋅
=
+ ⋅ + ⋅ +
⋅
=
∂
∂
=
⋅ = ⋅ ⋅ + ⋅⋅ + ⋅ +⋅
∂
∂
1
1 1
1
λ η
Large Signal Analysis (Continued):
• The small signal gain can be found:
• If channel-length modulation is ignored (r
o
=∞) we get:
()
S mb m
S m
IN
OUT
V
R g g
R
g
V
V
A
⋅ + +
⋅
=
∂
∂
=
1

49. Set 3: Single-Stage Amplifiers
49 SM
Resistive Load -4
Small Signal Analysis:
• We get the following small signal model:
()
()() ()
()
()
1
1
1
, ,
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+ + ⋅
⋅
=
+ ⋅ +⋅ +⋅
⋅ ⋅
=
⋅ ⋅ +⋅ ⋅ + +
⋅ ⋅
= =
⋅⋅ ⋅ = ⋅ ⋅ +⋅ ⋅ + + ⋅
+
⋅
⋅ −⋅ + − ⋅ =
−
=
−
=
⋅
⋅
+ ⋅ =
mb m
O
S
S m
v
O O mb O m S
O S m
O S mb O S m O S
O S m
IN
OUT
v
IN O S m O S mb O S m O S OUT
O S
O S
OUT mb OUT IN m OUT
OUT BS OUT IN GS O S BS mb GS m OUT
g g
r
R
R g
A
r r g rg R
rR g
rR g rR g r R
rR g
v
v
A
vrR g rR g rR g r R v
r R
rR
v g v v g v
v v v v v rR v g vg v

50. Set 3: Single-Stage Amplifiers
50 SM
Resistive Load -5
• Graph of the gain of a source-follower amplifier:1. M
1
never enters the triode region as long as V
IN
<V
DD
.
2. Gain is zero if V
IN
is less than V
TH
(because g
m
is 0).
3. As V
IN
increases, g
m
increases and the gain becomes:
4. As V
OUT
increases, ηdecreases, and therefore, the maximum gain
increases.
5. Even if R
S
=∞, the gain is less than one:
6. Gain depends heavily on the DC level of the input (nonlinear amplifier).
η
+
=
+
≈
1
1
mb m
m
v
g g
g
A
1
1
<
+ +
≈
o
mb m
m
V
r
g g
g
A

51. Set 3: Single-Stage Amplifiers
51 SM
Current Source Load
• In a source follower with a resistive load, the
drain current depends on the DC level of V
IN
,
which makes the amplifier highly nonlinear.
• To avoid this problem, we can use a current
source as the load.
• The output resistance is:
2
1 1
1 1 1 2 1
1 1
1 1
1 1
,
1 1
o
mb m
o I M OUT o I
mb m
o M
r
g g
r R R R r R
g g
r R= = → = =
• If channel length modulation is ignored (r
o1
=r
o2
=∞) :
1 1 1 1 1 1
1 1 1 1 1
mb m mb m mb m
OUT
g g g g g g
R
+
= =∞ ∞=
• Note that the body effect reduces the output impedance of the source
follower amplifiers.

52. Set 3: Single-Stage Amplifiers
52 SM
Voltage Division in Source Followers -1
• When Calculating output resistance seen at the source of M
1
,
i.e., R
M1
,
,
we force v
IN
to zero and find the output impedance:
1 1
1 1
1 1
mb m
o M
g g
r R=
• However, if we were to find the gain of the amplifier, we would not
suppress v
IN
.
• Here, we would like to find an equivalent circuit of M
1
, from which we can
find the gain.
• Consider the small-signal model of M
1
:

53. Set 3: Single-Stage Amplifiers
53 SM
Voltage Division in Source Followers -2
• For small-signal analysis v
BS
= v
DS
, so g
mb
v
BS
dependant current source
can be replaced by a resistor (1/g
mb
) between source and drain.
• Note that, when looking at the circuit from the source terminal, we can
replace the g
m
v
GS
dependant current source with a resistor (of value
1/g
m
) between source and gate.
• Simplified circuit:

54. Set 3: Single-Stage Amplifiers
54 SM
Voltage Division in Source Followers -3
Example:
• Find the gain of a source follower amplifier with a resistive
load.
• We draw the small signal model of this amplifier as shown
below to get:
m mb
O S
mb
O S
IN
OUT
v IN
m mb
O S
mb
O S
OUT
g g
rR
g
rR
v
v
A v
g g
rR
g
rR
v
1 1
1
1 1
1
+
= = → ⋅
+
=
m O S mb O S O S
m O S
m mb O S O S
O S
mb O S O S
O S
m
mb
O S
mb
O S
v
grR grR r R
grR
g grR r R
rR
grR r R
rR
g
g
r R
g
r R
A
⋅⋅ + ⋅⋅ + +
⋅⋅
=
+
⋅⋅ + +
⋅
⋅⋅ + +
⋅
=
+
+ +
+ +
=
1 1
1 1
1
1 1
1
• We can show that this is equal to what we obtained before:

55. Set 3: Single-Stage Amplifiers
55 SM
Voltage Division in Source Followers -4
Example:
• Find the gain of a source follower amplifier with a current
source load.
• Small-signal model of this amplifier is:
1 1
1 2
1
1 2
1 1
1 2
1
1 2
1 1
1
1 1
1
m mb
O O
mb
O O
IN
OUT
v IN
m mb
O O
mb
O O
OUT
g g
rr
g
rr
v
v
A v
g g
rr
g
rr
v
+
= = → ⋅
+
=
• If we ignore channel length modulation:
1 1
1
1 1
1
1 1
1
1 1
1
m mb
mb
IN
OUT
v IN
m mb
mb
OUT
g g
g
v
v
A v
g g
g
v
+
= = → ⋅
+
=

56. Set 3: Single-Stage Amplifiers
56 SM
Voltage Division in Source Followers -5
Example:
• Find the gain of a source follower amplifier with a resistive
load and biased with a current source.
• Small-signal model of this amplifier is:
1 1
1 2
1
1 2
1 1
1 2
1
1 2
1 1
1
1 1
1
m mb
O L O
mb
O L O
IN
OUT
v IN
m mb
O L O
mb
O L O
OUT
g g
rRr
g
rRr
v
v
A v
g g
rRr
g
rRr
v
+
= = → ⋅
+
=

57. Set 3: Single-Stage Amplifiers
57 SM
Voltage Division in Source Followers -6
Example:
• Find the gain of a source follower amplifier with a resistive
load.
• Small-signal model of this amplifier is:
1 2 2 1
1 2
2 2 1
1 2
1 2 2 1
1 2
2 2 1
1 2
1 1 1 1
1 1 1
1 1 1 1
1 1 1
m mb m mb
O O
mb m mb
O O
IN
OUT
v IN
m mb m mb
O O
mb m mb
O O
OUT
g g g g
rr
g g g
rr
v
v
A v
g g g g
rr
g g g
rr
v
+
= = → ⋅
+
=

58. Set 3: Single-Stage Amplifiers
58 SM
Advantages and Disadvantages -1
1. Source followers have typically small output impedance.
2. Source followers have large input impedance.
3. Source followers have poor driving capabilities..
4. Source followers are nonlinear. This nonlinearity is caused by:
¾Variable bias current which can be resolved if we use a current
source to bias the source follower.
¾Body effect; i.e., dependence of V
TH
on the source (output)
voltage. This can be resolved for PMOS devices, because each
PMOS transistor can have a separate n-well. However, because
of low mobility, PMOS devices have higher output impedance.
(In more advanced technologies, NMOS in a separate p-well,
can be implemented that potentially has no body effect)
¾Dependence of r
o
on V
DS
in submicron devices.

59. Set 3: Single-Stage Amplifiers
59 SM
Advantages and Disadvantages -2
5. Source followers have voltage headroom limitations due to level shift.
Consider this circuit (a common source followed by a source follower):
• If we only consider the common source stage, V
X
>V
GS1
-V
TH1
.
• If we only consider the source follower stage, V
X
>V
GS3
-V
TH3
+ V
GS2
.
• Therefore, adding the source follower will reduce the allowable voltage
swing at node X.
• The DC value of V
OUT
is V
GS2
lower than the DC value of V
X
.

60. Set 3: Single-Stage Amplifiers
60 SM
Common-Gate
A
v
=
(
g
m
+
g
mb
)R
D
=
g
m
(1
+
η
)R
D

61. Set 3: Single-Stage Amplifiers
61 SM
Common-Gate
A
v
=
(
g
m
+
g
mb
)r
o
+1
r
o
+(g
m
+g
mb
)r
o
R
S
+R
S
+R
D
R
D

62. Set 3: Single-Stage Amplifiers
62 SM
Common-Gate
A
v
=
(
g
m
+
g
mb
)r
o
+1
r
o
+(g
m
+g
mb
)r
o
R
S
+R
S
+R
D
R
D ) || )( ( :0
D o mb m v S
Rr g g A R
+
≈
=
for

63. Set 3: Single-Stage Amplifiers
63 SM
Common-Gate Input Impedance
o mb m
o
o mb m
D
o mb m
o D
in
r g g
r
r g g
R
r g g
r R
R
) (1 ) (1 ) (1+ +
+
+ +
=
+ +
+
=
)
1
||
1
|| (
) (1
mb m
o
o mb m
D
in
g g
r
r g g
R
R+
+ +
=

64. Set 3: Single-Stage Amplifiers
64 SM
Common-Gate Input Impedance
• Input impedance of common-gate stage is relatively low only if
R
D
is small
• Example: Find the input impedance of the following circuit.

65. Set 3: Single-Stage Amplifiers
65 SM
Example
• Calculate the voltage gain of the following circuit:
o mb m v
r g g A) (1
+
+
=

66. Set 3: Single-Stage Amplifiers
66 SM
Common-Gate Output Impedance
D S o S mb m out
R R rR g g R||} ]) (1{[
+
+
+
=

67. Set 3: Single-Stage Amplifiers
67 SM
Example
• Compare the gain of the following two circuits ( λ=
γ
=0 and 50Ωtransmission
lines!)

68. Set 3: Single-Stage Amplifiers
68 SM
Cascode Stage
D oo mb m
D o o o mb m out
R rr g g
R r rr g g R
||] ) [(
||} ]) (1{[
2 1 2 2
1 2 1 2 2
+ ≈
+ + + =
A
V
≈g
m1
{[r
o1
r
o2
(g
m2
+
g
mb2
)]||R
D
]}
• Cascade of a common-source stage and a common-gate stage is called a “cascode” stage.

69. Set 3: Single-Stage Amplifiers
69 SM
Cascode Stage
A
V
≈
g
m
1
[(r
o
1r
o
2
g
m
2
)
||
(r
o
3r
o
4
g
m
3
)]

70. Set 3: Single-Stage Amplifiers
70 SM
Output Impedance Comparison

71. Set 3: Single-Stage Amplifiers
71 SM
Shielding Property

72. Set 3: Single-Stage Amplifiers
72 SM
Board Notes

73. Set 3: Single-Stage Amplifiers
73 SM
Triple Cascode
• What is the output resistance of this circuit?
• Problem?

74. Set 3: Single-Stage Amplifiers
74 SM
Folded Cascode

75. Set 3: Single-Stage Amplifiers
75 SM
Output Impedance of a Folded Cascode
2 3 1 2 2 2
) || ]( ) (1[
o o o o mb m out
r r r r g g R
+
+
+
=