6.002
CIRCUITS AND
ELECTRONICS
Capacitors
and First-Order Systems
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 12
Motivation
Demo
5V
5V
B
C
A
5V
0V
5
A
0
5
Expect this, right?
But observe this!
B
0
5
Expected
Observed
C
0
Reading:
Chapters 9 & 10
Delay!
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 12
The Capacitor
D
n-channel MOSFET
symbol
G
S
drain
gate
m+
e+
t +
a+
l +
+
n
o
x
i
d
e
source
s
i
l n-channel
p
i MOSFET
n-channel c
o
n
n
D
G
CGS
S
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 12
Ideal Linear Capacitor
+
+ A
++++
-----
E
d
EA
d
obeys DMD!
total charge on
capacitor
= +q − q = 0
C=
i
C
q
+
v
–
q = C v
coulombs farads volts
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 12
Ideal Linear Capacitor
i
q
C
q = C v
+
v
–
dq
i=
dt
d (Cv )
=
dt
dv
=C
dt
⎡ E = 1 Cv 2 ⎤
⎢⎣
⎥⎦
2
A capacitor is an energy storage device
Æ memory device Æ history matters!
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 12
Analyzing an RC circuit
Thévenin Equivalent:
vI (t ) +
–
R
C
+
vC (t )
–
Apply node method:
vC − vI
dvC
+C
=0
R
dt
dvC
+ vC = vI
RC
dt
t ≥ t0
vC (t0 ) given
units
of time
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 12
Let’s do an example:
+
v I (t )
R
+
–
C
vC (t )
–
vI (t ) = VI
vC (0 ) = V0 given
dvC
RC
+ vC = VI
dt
X
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 12
Example…
vI (t ) = VI
vC (0 ) = V0 given
RC
dvC
+ vC = VI
dt
X
vC (t ) = vCH (t ) + vCP (t )
total homogeneous
particular
Method of homogeneous and particular
solutions:
1
Find the particular solution.
2 Find the homogeneous solution.
3 The total solution is the sum of
the particular and homogeneous
solutions.
Use the initial conditions to solve
for the remaining constants.
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 12
1 Particular solution
dvCP
+ vCP = VI
dt
RC
vCP = VI
RC
works
dVI
+ VI = VI
dt
0
In general, use trial and error.
vCP : any solution that satisfies the
original equation X
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 12
2 Homogeneous solution
dvCH
RC
+ vCH = 0
dt
Y
vCH : solution to the homogeneous
equation Y
(set drive to zero)
vCH = A e st
assume solution
of this form. A,
s?
dA e st
+ A e st = 0
RC
dt
R CA s e st + A e st = 0
Discard trivial A = 0 solution,
Characteristic equation
R C s +1 = 0
s= −
or
1
RC
vCH = Ae
−t
RC
RC
called time
constant
τ
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 12
3 Total solution
vC = vCP + vCH
vC = VI + A e
−t
RC
Find remaining unknown from initial
conditions:
at t = 0
Given,
vC = V0
so,
V0 = VI + A
or
A = V0 − VI
thus
vC = VI + (V0 − VI ) e
also
iC = C
−t
RC
dvC
(V − VI )
e
=− 0
dt
R
−t
RC
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 12
vC = VI + (V0 − VI ) e
−t
RC
vC
VI
V0
0
t
RC
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 12
Examples
vC
vC
5V
5V
5 + 5e
−t
RC
5e
t
0V
VO = 0V
VI = 5V
5
0
−t
RC
t
0V
VO = 5V
VI = 0V
5
0
τ = RC
Remember
B
demo
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 12