6.002
CIRCUITS AND
ELECTRONICS
MOSFET Amplifier
Large Signal Analysis
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Review
„
Amp constructed using dependent source
control a
a′
port
„
DS
b output
b ′ port
Dependent source in a circuit
+
–
a +
b
v
i = f (v )
a′ –
b′
„
Superposition with dependent sources:
one way tleave all dependent sources in;
solve for one independent source at a
time [section 3.5.1 of the text]
„
Next, quick review of amp …
Reading: Chapter 7.3–7.7
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Amp review
VS
RL
vO
VCCS
vI
K
2
iD = (vI − 1)
2
+
–
for vI ≥ 1V
= 0 otherwise
vO = VS − iD RL
K
(vI − 1)2
2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Key device Needed:
v
A
B
i = f (v )
voltage controlled
current source
C
Let’s look at our old friend, the MOSFET …
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Key device Needed:
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.
D
G
vGS < VT
D
S
S
?
G
vGS ≥ VT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Graphically
Demo
iDS
+
vGS
–
iDS
egio
n
iDS
vGS ≥ VT
vGS < VT
vGS < VT
vDS
S MODEL
vDS = vGS − VT
vGS 1
Saturation
region
vGS 2
vGS3
...
vGS ≥ VT
Trio
de r
iDS
v+DS
–
vDS
SR MODEL
vGS < VT Cutoff
vDS
region
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Graphically
iDS
+
vGS
–
iDS
iDS
egio
n
S MODEL
vGS 2
vGS3
...
vGS < VT
Saturation
region
Trio
de r
vGS ≥ VT
vDS
vDS = vGS − VT
vGS 1
iDS
vGS ≥ VT
vGS < VT
v+DS
–
vDS
SR MODEL
vGS < VT
vDS
when
vDS ≥ vGS − VT
Notice that
MOSFET
behaves like a
current source
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
MOSFET SCS Model
When
vDS ≥ 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
vGS < VT
S
D
D
G
S
vGS
G
≥ VT
iDS = f (vGS )
K
2
= (vGS − VT )
2
S
when
vDS ≥ vGS − VT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Reconciling the models…
iDS
iDS
vGS ≥ VT
vGS < VT
vDS
S MODEL
for fun!
vGS < VT
Saturation
region
vGS 2
vGS3
...
vGS ≥ VT
Trio
de r
egio
n
iDS
vDS = vGS − VT
vGS 1
vDS
SR MODEL
for digital
designs
vGS < VT
vDS
SCS MODEL
for analog
designs
When to use each model in 6.002?
Note: alternatively (in more advanced courses)
vDS ≥ vGS − VT
vDS < vGS − VT
use SCS model
use SR model
or, use SU Model (Section 7.8 of A&L)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Back to Amplifier
VS
vI
AMP
vO
VS
RL
vI
G
D
S
vO
K
2
iDS = (vI − VT )
2
in saturation
region
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”
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
MOSFET Amplifier
VS
RL
vI
G
D
S
vO
K
2
iDS = (vI − VT )
2
in saturation
region
To ensure the MOSFET operates as a VCCS,
we must operate it in its saturation region
only. We promise to adhere to the
“saturation discipline.”
In other words, we will operate the amp
circuit such that
vGS ≥ VT and vDS ≥ vGS – VT
vO ≥ vI – vT
at all times.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Let’s analyze the circuit
First, replace the MOSFET with its
SCS model.
VS
RL
vO
G
vGS = vI
+
–
+
vI
–
D
iDS
S
K
2
= (vI − VT )
2
A
for vO ≥ vI − VT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Let’s analyze the circuit
VS
RL
vO
G
vGS = vI
+
–
+
vI
–
D
iDS =
K
(vI − VT )2
2
A
for vO ≥ vI − VT
S
(vO = vDS in our example)
1
Analytical method: vO vs vI
vO = VS − iDS RL
B
K
2
or vO = VS − (vI − VT ) RL for vI ≥ VT
2
vO ≥ vI − VT
vO = VS
vI < VT
(MOSFET turns off)
for
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Graphical method vO vs vI
K
2
From A : iDS = (vI − VT ) ,
2
vO ≥ vI − VT
2
for
⇓
2iDS
vO ≥
K
⇓
iDS
B : iDS
K 2
≤ vO
2
VS v0
=
−
RL RL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
2
Graphical method vO vs vI
K 2
K
2
A : iDS = (vI − VT ) , for iDS ≤ vO
2
2
VS vO
=
−
i
B : DS
RL RL
iDS
VS
RL
iDS
B
Lo
ad
K 2
≤ vO
2
A
li n
e
vI
= vGS
VS
Constraints
A
and
B
vO
must be met
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
2
Graphical method vO vs vI
iDS
VS
RL
iDS ≤
K 2
vO
2
A
B
vI
VI
I DS
VO
VS
vO
Constraints A and B must be met.
Then, given VI, we can find VO, IDS .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
of Amplifier
(under “saturation discipline”)
1
vO versus vI
2
Valid input operating range and
valid output operating range
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
vO versus vI
1
vO
K
2
VS − (vI − VT ) RL
2
VS
VT
vO = vI − VT
gets into
triode region
vI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
What are valid operating ranges
under the saturation discipline?
vI ≥ VT
Our
K 2
iDS ≤ vO
Constraints vO ≥ v I − VT
2
2
iDS
VS
RL
iDS ≤
K 2
vO
2
K
(vI − VT )2
2
vI
VS vO
iDS =
−
RL RL
iDS =
VS
?
vO
vI = VT
vO = VS and iDS = 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
2
Large Signal Analysis
What are valid operating ranges
under the saturation discipline?
iDS
iDS
K 2
≤ vO
2
K
2
iDS = (vI − VT )
2
vI
V
v
iDS = S − O
RL RL
vO
− 1 + 1 + 2 KRLVS
vI = VT +
KRL
− 1 + 1 + 2 KRLVS
vO =
KRL
VS vO
iDS =
−
RL RL
vI = VT
vO = VS and iDS = 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
Summary
1
vO versus vI
vO = VS −
2
K
(vI − VT )2 RL
2
Valid operating ranges under the
saturation discipline?
Valid input range:
vI : VT
to
− 1 + 1 + 2 KRLVS
VT +
KRL
corresponding output range:
vO : VS to
− 1 + 1 + 2 KRLVS
KRL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9