6.002x
CIRCUITS AND
ELECTRONICS
MOSFET SCS Model
MOSFET Amplifier
1 Review
n Dependent source in a circuit
+
–
a
+
v
aʹ′ –
b
i = f (v)
bʹ′
n Superposition with dependent sources. One approach:
- leave all dependent sources in;
- solve for one independent source at a time
- [section 3.5.1 of the text]
2
Review
Next, quick review of amp. Reading: Chapter 7.3–7.7
Vs
RL
vO
VCCS
vI
iD
iD =
K
(vI − 1)2
2
for vI ≥ 1V
+
–
= 0 otherwise
vO = VS − iD RL
K
(v I − 1)2
2
3
The i-v relation for a current source
i
v
Hold the thought
4
Key device Needed
Vs
vO
VCCS
vI
+
–
5
Key device Needed
Vs
VCCS
VCCS
B
A
D
vO
+
vI
+
–
i = f(v)
v
–
C
Let’s look at our old friend, the MOSFET …
6
Our old friend, the MOSFET
First, we sort of lied. The on-state behavior of the
MOSFET is quite a bit more complex than either the ideal
switch or the resistor model would have you believe.
7
Let’s Look at the MOSFET Behavior Graphically
D
G
S
8
Graphically
iDS
D
Demo
G
S
iDS
vDS
S MODEL
v+GS
–
+
vDS
–
iDS
vDS
SR MODEL
9
Graphically
iDS
D
Demo
v+GS
–
+
vDS
–
iDS
G
S
iDS
iDS
vGS ≥ VT
vGS ≥ VT
vGS < VT
S MODEL
vDS
vGS < VT
SR MODEL
vDS
vDS
10
Graphically
D
v+GS
–
iDS
vGS ≥ VT
S MODEL
vDS
vGS < VT
SR MODEL
Saturation
region vDS ≥ vGS
vGS 2
− VT
vGS ≥ VT vGS 3
...
vGS ≥ VT
e re
gion
iDS
Trio
d
iDS
vGS < VT
vDS = vGS − VT
vGS 1
vDS < vGS − VT
vDS
vGS < VT
Cutoff
region
vDS
G
+
vDS
–
iDS
S
when
vDS ≥ vGS − VT
Notice that
MOSFET
behaves like a
current source
11
MOSFET SCS Model
When vDS ≥ vGS − VT and vGS ≥ VT
the MOSFET is in its saturation region, and the switch current source
(SCS) model of the MOSFET is more accurate than the S or SR model
D
G
S
12
When to use which model
iDS
iDS
e re
gion
iDS
vGS ≥ VT
vDS
vGS < VT
Saturation
region
vGS2
vGS ≥ VT
vDS ≥ vGS − VT
vGS3
...
vGS < VT
Trio
d
vGS ≥ VT
vDS = vGS − VT
vGS1
vDS
vGS < VT
S MODEL
SR MODEL
for fun!
for digital designs
When to use each model in 6.002?
vDS
SCS MODEL
for analog designs
13
In More Advanced Courses…
Note: alternatively (in more advanced courses)
vDS ≥ vGS − VT
use SCS model
vDS < vGS − VT
use SR model
or, use SU (Switch Unified) Model (Section 7.8 of A&L) 14
Back to Amplifier
vI
VS
VS
AMP
vO
vI A vO
RL
vO
vI
iD
iD =
K
(vI − 1)2
2
To ensure the MOSFET operates as a VCCS, we must
operate it in its saturation region only. To do so, we
promise to adhere to the
“saturation discipline”
15
MOSFET Amplifier
VS
G
e re
gion
vO
K
2
= (vI − VT )
2
D
iDS
S
in saturation region
“Saturation discipline”
vDS ≥ vGS − VT
vGS 1
Saturation
region
vGS 2
vGS 3
...
vI
iDS
Trio
d
RL
vDS = vGS − VT
vGS ≥ VT
vGS < VT
vDS
Cutoff region
In other words, we will operate the amp circuit such that:
16
MOSFET Amplifier
VS
G
e re
gion
vO
K
2
= (vI − VT )
2
D
iDS
S
in saturation region
“Saturation discipline”
vDS ≥ vGS − VT
vGS 1
Saturation
region
vGS 2
vGS 3
...
vI
iDS
Trio
d
RL
vDS = vGS − VT
vGS ≥ VT
vGS < VT
vDS
Cutoff region
In other words, we will operate the amp circuit such that
17
Let’s analyze the circuit
First, replace the MOSFET
with its SCS model
VS
RL
vI G
vO
D
S
iDS =
K
(vI − VT )2
2
in saturation
region
Two methods:
1. Analytical
2. Graphical
18
Let’s analyze the circuit
Find vO vs vI
VS
RL
(vO = vDS in our example)
1 Analytical method:
vO vs vI
A
vO
vGS = vI G
+
+ v
– I
–
D
iDS =
for
K
(vI − VT )2
2
vO ≥ vI − VT
S
19
From A : iDS =
iDS
K
(vI − VT )2
2
iDS
iDS =
Saturation
region
K 2
vO
2
vGS2
vGS3
...
Graphical method
Trio
regiode
n
2
vDS = vGS − VT
vGS < VT
vDS
From B : vO = VS − iDS RL
0
vDS
20
2
Graphical method
A :
vO vS vI
K 2
K
(vI − VT )2 , for iDS ≤ vO
2
2
V
v
= S − O
RL RL
iDS =
B : iDS
iDS
iDS ≤
K 2
vO
2
Remember, superimpose
“loadline due to rest of
circuit (B)” over device
relation (A)
Constraints
A and B
must be met
A
vI = vGS
VS
vO
21
2
Graphical method
A :
B :
K
2
iDS = (vI − VT ) ,
2
V
v
iDS = S − O
RL RL
iDS
VS
RL
iDS ≤
B
vO vS vI
for iDS
K 2
≤ vO
2
Constraints
A and B
must be met
Then, given VI, we
can find VO, IDS
K 2
vO
2
A
vI = vGS
VS
vO
22
Large Signal Analysis
of Amplifier (under “saturation discipline”)
1
vO versus vI
2
Valid input operating range and
valid output operating range (to
guarantee saturation region
operation)
23
Large Signal Analysis
1
vO versus vI
Demo
24
A Curious Aside
vO
VS
K
2
VS − (vI − VT ) RL
2
VS
Cutoff
region
vO = vI − VT
gets into
triode region
VT
vI
25
Large Signal Analysis
2
VS
RL
What are the valid operating ranges
under the saturation discipline?
vI G
iDS
D
vO
S
vI
Our
Constraints
vO
26
Large Signal Analysis
RL
What are the valid operating ranges
under the saturation discipline?
iDS
VS
RL
iDS ≤
vI
G
D
vO
S
K 2
vO
2
VS vO
−
RL RL
vI
K
2
iDS = (vI − VT )
2
iDS =
Z
2
VS
VS
vO
Our
Constraints
vI ≥ VT
vO ≥ vI − VT
K 2
iDS ≤ vO
2
vI=VT
vO=VS and iDS=0
27
Large Signal Analysis
2
What are the valid operating ranges
under the saturation discipline?
iDS
VS
RL
vO =
Our
Constraints
− 1 + 1 + 2 KRLVS
KRL
iDS ≤
K 2
vO
2
vI ≥ VT
V v
iDS = S − O
vI
RL RL
K
2
iDS = (vI − VT )
2
VS
vO
vO ≥ vI − VT
K 2
iDS ≤ vO
2
vI=VT
vO=VS and iDS=0
28
Large Signal Analysis Summary
1
2
vO versus vI
Valid operating ranges under the
saturation discipline?
Valid input range:
We will look at
another way of
obtaining (2)
shortly
Corresponding output range:
29
Another Way of Obtaining (2) Valid Op Range
vO
Vs
5V
VT
vI
1V
30 Another Way of Obtaining (2) Valid Op Range
vO
~ 5V VS
~ 1V
VT
1V
vI
~ 2V
31
Another Way of Obtaining (2) Valid Op Range
vO
2
~ 5V VS
vO > vI − VT
K (vI − VT )
vO = VS −
RL
2
vO = vI − VT
vO = vI − VT
~ 1V
vO = vI
vO < vI − VT
VT
1V
~ 2V
32