6.002
CIRCUITS AND
ELECTRONICS
Dependent Sources
and Amplifiers
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 2002: Lecture 8
Review
„
Nonlinear circuits — can use the
node method
„
Small signal trick resulted in linear
response
Today
„
Dependent sources
„
Amplifiers
Reading: Chapter 7.1, 7.2
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 2002: Lecture 8
Dependent sources
Seen previously
+
i
Resistor
Independent
Current source
+
i
v
–
R
v –
I
v
i=
R
i=I
2-terminal 1-port devices
New type of device: Dependent source
iI
i
O
+
control
port
vI
f ( vI )
–
+
vO
output
port
–
2-port device
E.g., Voltage Controlled Current Source
Current at output port is a function of voltage
at the input port
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 2002: Lecture 8
Dependent Sources: Examples
Example 1: Find V
+
R V
–
independent
current
source
I = I0
V = I0R
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 2002: Lecture 8
Dependent Sources: Examples
Example 2: Find V
voltage
controled
current
source
+
R V
–
K
I = f (V ) =
V
iI
+
+
R V
–
f (vI ) =
K
vI
iO
+
vI
vO
–
–
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 2002: Lecture 8
Dependent Sources: Examples
Example 2: Find V
voltage
controled
current
source
+
R V
–
K
I = f (V ) =
V
e.g. K = 10-3 Amp·Volt
R = 1kΩ
K
V = IR = R
V
or V 2 = KR
or V = KR
= 10 −3 ⋅ 10 3
= 1 Volt
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 2002: Lecture 8
Another dependent source example
RL
iIN
vI +
–
iD
+
+
vIN
vO
–
–
e.g.
VS +
–
iD = f (vIN )
iD = f (vIN )
K
2
= (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
Find vO as a function of vI .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 – Fall 2002: Lecture 8
Another dependent source example
VS
RL
iIN
vI +
–
iD
+
+
vIN
vO
–
–
iD = f (vIN )
e.g.
iD = f (vIN )
K
2
= (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
Find vO as a function of 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 2002: Lecture 8
Another dependent source example
VS
RL
vI
vI
+
–
vO
K
2
iD = (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
Find vO as a function of 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 2002: Lecture 8
Another dependent source example
VS
RL
vI
vI
+
–
vO
K
2
iD = (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
KVL
− VS + iD RL + vO = 0
vO = VS − iD RL
K
2
vO = VS − (vI − 1) RL
2
vO = VS
for vI ≥ 1
for vI < 1
Hold that thought
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 – Fall 2002: Lecture 8
Next, Amplifiers
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 – Fall 2002: Lecture 8
Why amplify?
Signal amplification key to both analog
and digital processing.
Analog:
AMP
IN
Input
Port
OUT
Output
Port
Besides the obvious advantages of being
heard farther away, amplification is key
to noise tolerance during communcation
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 2002: Lecture 8
Why amplify?
Amplification is key to noise tolerance
during communcation
No amplification
useful
signal
1 mV
e
nois
10 mV
huh?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 – Fall 2002: Lecture 8
Try amplification
e
nois
AMP
not bad!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 – Fall 2002: Lecture 8
Why amplify?
Digital:
Valid region
5V
5V
VIH IN
VIL
0V
5V
OUT
Digital System
IN
5V
VOL
OUT
V OH
VIH
VIL
0V
0V
VOH
t
V OL
0V
t
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 2002: Lecture 8
Why amplify?
Digital:
Static discipline requires amplification!
Minimum amplification needed:
VIH
VIL
VOH
VOL
VOH − VOL
VIH − VIL
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 2002: Lecture 8
An amplifier is a 3-ported device, actually
Power port
Input
port
iO
iI
+v
– I
Amplifier
+ v Output
– O port
We often don’t show the power port.
Also, for convenience we commonly observe
“the common ground discipline.”
In other words, all ports often share a
common reference point called “ground.”
POWER
IN
OUT
How do we build one?
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 2002: Lecture 8
Remember?
VS
RL
vI
vI
+
–
vO
K
2
iD = (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
KVL
− VS + iD RL + vO = 0
vO = VS − iD RL
K
2
vO = VS − (vI − 1) RL
2
vO = VS
for vI ≥ 1
for vI < 1
Claim: This is an amplifier
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 2002: Lecture 8
So, where’s the amplification?
Let’s look at the vO versus vI curve.
mA
e.g. VS = 10V , K = 2 2 , RL = 5 kΩ
V
K
2
vO = VS − RL (vI − 1)
2
2 −3
2
3
= 10 − ⋅10 ⋅ 5 ⋅ 10 (vI − 1)
2
vO = 10 − 5 (vI − 1)
vO
VS
2
ΔvO
1
ΔvO
>1
Δv I
ΔvI
vI
amplification
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
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6.002 – Fall 2002: Lecture 8
Plot vO versus vI
vO = 10 − 5 (vI − 1)
2
0.1 change
in vI
Demo
vI
vO
0.0
1.0
1.5
2.0
2.1
2.2
2.3
2.4
10.00
10.00
8.75
5.00
4.00
2.80
1.50
~ 0.00
1V change
in vO
Gain!
Measure vO .
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 2002: Lecture 8
One nit …
vO
What
happens
here?
1
vI
Mathematically,
K
2
vO = VS − RL (vI − 1)
2
So
is mathematically predicted behavior
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 2002: Lecture 8
One nit …
vO
K
2
vO = VS − RL (vI − 1)
2
What
happens
here?
vI
1
However, from
K
2
iD =
(vI − 1)
2
VS
for vI ≥ 1
RL
vO
VCCS
iD
For vO>0, VCCS consumes power: vO iD
For vO<0, VCCS must supply power!
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 2002: Lecture 8
If VCCS is a device that can source
power, then the mathematically
predicted behavior will be observed —
vO
K
2
i.e. vO = VS − RL (vI − 1)
2
vI
where vO goes -ve
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 2002: Lecture 8
If VCCS is a passive device,
then it cannot source power,
so vO cannot go -ve.
So, something must give!
Turns out, our model breaks down.
K
2
iD = (vI − 1)
2
will no longer be valid when vO ≤ 0 .
e.g. iD saturates (stops increasing)
and we observe:
Commonly
vO
1
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 2002: Lecture 8