Lecture 18 Transistor Amplifiers (I) Common-Source Amplifier

Lecture 18
Amplificateurs à Transistor (I)
Common-Source Amplifier
• Amplifier fundamentals
Outline
• Common-source amplifier
• Common-source amplifier with current-source
supply
6.012 Spring 2004 Lecture 18 1

Amplifier Fundamentals
• Source resistance R S is associated only with small
signal sources
• Choose I D = I SUP ---> DC output current
– I OUT
= 0
– V OUT
= 0
Input
sources
Voltage Input
Intrinsic
Amplifier
V +
Load
v s
V BIAS
+
−
+
−
R S
v IN
= V BIAS
+ v s
Current Input i IN
= I BIAS
+ i s
i s
R S
I BIAS
Input
I SUP
Supply
Current
I SUP
Active
Device
i D
= f(input)
i OUT
= i d
i D
+
v OUT
−
R L
V −
6.012 Spring 2004 Lecture 18 2

2. Common-Source Amplifier:
Consider the following circuit:
V + =V DD
signal source
R S
v s
V BIAS
R D
V - =V SS
i R
i D
+
v OUT
-
signal
load
R L
• Consider intrinsic voltage amplifier - no loading
•R S = 0
•R L ---> ∞
•V GS = V BIAS -V SS
• V BIAS , R D and W/L of MOSFET selected to bias
transistor in saturation and obtain desired output
bias point (i.e. V OUT = 0).
Watch notation: v OUT (t)=V OUT +v out (t)
6.012 Spring 2004 Lecture 18 3

Load line view of amplifier:
load line
I R =I D
V SS
V DD -V SS
R D
V BIAS -V ss = V DD -V SS
V GG -V SS =V DD -V SS
V BIAS -V ss
V GG -V SS
0
V DD
V BIAS -V ss = V T
V GG -V SS =V T
V OUT
Transfer characteristics of amplifier:
V OUT
V DD
Want:
V SS
0 V T
V DD -V SS
• Bias point calculation;
• Limits to signal swing
• Small-signal gain;
V GG -V SS
V BIAS -V ss
• Frequency response [in a few days]
6.012 Spring 2004 Lecture 18 4

Bias point: choice of V BIAS , W/L, and R D to keep
transistor in saturation and to get proper quiescent V OUT .
Assume MOSFET is in saturation:
I D = W 2L µ nC ox ( V BIAS − V SS − V T ) 2
If we select V OUT =0:
I D = I R = W 2 L µ nC ox
Then:
I R = V DD − V OUT
R D
( V BIAS − V SS − V T ) 2 = V DD
R D
V BIAS =
2I D
W
+ V SS + V T
L µ nC ox
Equation that allows us to compute needed V BIAS given
R D and W/L.
6.012 Spring 2004 Lecture 18 5

Signal swing:
V DD
R D
signal source
+
R S
v OUT
v s
V BIAS
-
V SS
• Upswing: limited by MOSFET going into cut-off.
v out,max = V DD
• Downswing: limited by MOSFET leaving saturation.
or
V DS,sat = V GS − V T =
2I D
W
L µ nC ox
v out ,min − V SS = V BIAS − V SS − V T =
Then:
v out,min = V BIAS − V T
2I D
W
L µ nC ox
6.012 Spring 2004 Lecture 18 6

Generic view of the effect of loading on
small-signal operation
Two-port network view of small-signal equivalent
circuit model of a voltage amplifier:
R in is input resistance
R out is output resistance
A vo is unloaded voltage gain
R s
R out
v s
+
-
+
+
+
v in R in Avo v in v out
-
-
-
R L
input
loading
unloaded circuit
output
loading
Voltage divider at input:
Voltage divider at output:
v in = R in
v out = R L
v s
R in + R s
A vo v in
R out + R L
Loaded voltage gain:
v out
v s
=
R in
R in + R S
A vo
R L
R L + R out
6.012 Spring 2004 Lecture 18 7

Small-signal voltage gain A vo : draw small-signal
equivalent circuit model: Remove R L and R S
D
G
+ +
+
v t v gs gm v gs
v out
- -
S
r o
R D
-
+
v t
+
-
gm v t (r o //R D )
v out = −g m v t ( r o // R D )
Then unloaded voltage gain:
A vo = v out
v t
=−g m ( r o // R D )
v out
-
6.012 Spring 2004 Lecture 18 8

Input Resistance
• Calculation of input resistance, R in :
– Load amplifier with R L
– Apply test voltage (or current) at input, measure test
current (or voltage).
For common-source amplifier:
i t
+
+
v t v gs
g m v gs (r o //R D ) R L
-
-
i t = 0 ⇒ R in = v t
i t
=∞
No effect of loading at input.
6.012 Spring 2004 Lecture 18 9

Output Resistance
• Calculation of output resistance, R out :
– Load amplifier with R S
– Apply test voltage (or current) at output, measure test
current (or voltage).
– Set input source equal zero
For common-source amplifier:
i t
R S
+
v gs
-
g m v gs (r o //R D )
+
vt
-
v gs = 0 ⇒ g m v gs = 0 ⇒ v t = i t ( r o // R D )
R out = v t
i t
= r o // R D
6.012 Spring 2004 Lecture 18 10

Two-port network view of common-source amplifier
Voltage Amplifier
R s
R out
v s
+
-
+
+
+
v in R in Avo v in v out
-
-
-
R L
input
loading
Intrinsic circuit
output
loading
v out
v s
v out
v s
=
R in
R in + R S
A vo
( )
=−g m r o // R D
R L
R L
R L + R out
( )
R L + r o // R D
=−g m r o // R D // R L
6.012 Spring 2004 Lecture 18 11

Current Source Supply
I—V characteristics of current source:
i SUP
+
v SUP
i SUP
I SUP
1
r oc
_
v SUP
Equivalent circuit models :
i SUP
+
v SUP
I SUP
roc
r oc
_
large-signal model
small-signal model
• i SUP = 0 for v SUP ≤ 0
• i SUP = I SUP + v SUP / r oc for v SUP > 0
• High small-signal resistance r oc.
6.012 Spring 2004 Lecture 18 12

3. Common-source amplifier with currentsource
supply
V DD
i SUP
signal source
R S
i D
+
signal
load
v s
v OUT
R L
V BIAS
-
Loadline View
i SUP =I D
load line
V SS
V BIAS -V SS =V DD -V SS
I SUP
V SS
V BIAS -V SS
0
V DD
V BIAS -V SS =V T
V OUT
6.012 Spring 2004 Lecture 18 13

Use PMOS for current source supply
V DD
v s
V BIAS
V B
signal source
R S
V SS
i D
i SUP
v OUT
Bias point: Assume both transistors in saturation
V OUT = 0
I SUP =−I Dp =
I SUP = I Dn =
⎛
⎜
⎝
⎛
⎜
⎝
W ⎞
⎟
2L⎠
W ⎞
⎟
2L⎠
µ p C ox ( V DD − V B + V Tp ) 2
p
µ n C ox ( V BIAS − V SS − V Tn ) 2
n
V BIAS =
⎛
⎜
⎝
W
L
2I SUP
+ V
⎞
SS + V T
⎟ µ
⎠ n C ox
n
6.012 Spring 2004 Lecture 18 14

Signal swing:
V DD
V B
i SUP
signal source
R S i D
v
v OUT
s
V BIAS
V SS
• Upswing: limited by PMOS leaving saturation.
V SD,sat = V SG + V Tp =
⎛
⎜
⎝
W
L
2I SUP
⎞
⎟
⎠
µ p C ox
p
V DD − v out,max = V DD − V B + V Tp
v out,max = V B − V Tp
• Downswing: limited by NMOS leaving saturation.
• Same result as with resistive supply current.
v out,min = V BIAS − V T
6.012 Spring 2004 Lecture 18 15

3. Common-source amplifier with currentsource
supply (contd.)
Current source characterized by high output resistance:
r oc . Significantly higher than amplifier with resistive supply.
p-channel MOSFET: r oc = 1/λI Dp
V DD
V B
i SUP
signal source
R S i D
v s
V BIAS
V SS
• Voltage gain: A vo = -g m (r o //r oc ).
v OUT
• Input resistance :R in = ∞
• Output resistance: R out = r o //r oc .
6.012 Spring 2004 Lecture 18 16

Relationship between circuit figures of merit and
device parameters
Remember:
W
g m = 2I D
L µ nC ox
1
r o ≈ ∝ L
λ n I D I D
Then:
Device*
Parameters
Circuit Parameters
|A vo | R in R out
g m (r o //r oc ) ∝ ro //r oc
I SUP ↑ ↓ - ↓
W ↑ ↑ - -
µ n C ox ↑` ↑ - -
L ↑ ↑ - ↑
* adjustments are made to V BIAS so that none of
the other parameters change
CS amplifier with current source supply is a good
voltage amplifier (R in high and |A vo | high), but R out high
too ⇒ voltage gain degraded if R L

What did we learn today?
Summary of Key Concepts
for CS amplifier
• Bias Calculations
• Signal Swing
• Small Signal Circuit Parameters
– Voltage Gain - A VO
– Transconductance - G m
– Input Resistance - R in
– Output Resistance - R out
• Relationship between small signal circuit and
device parameters
6.012 Spring 2004 Lecture 18 18