6.01: Introduction to EECS I
Lecture 9
April 5, 2011
6.01: Introduction to EECS I
Midterm Examination #2
Circuit Abstractions
Time:
Tuesday, April 12, 7:30 pm to 9:30 pm
Location:
Walker Memorial (if last name starts with A-M)
10-250 (if last name starts with N-Z)
Coverage:
Everything up to and including Design Lab 9.
You may refer to any printed materials that you bring to exam.
You may use a calculator.
You may not use a computer, phone, or music player.
No software lab in week 10.
Review sessions during office hours.
April 5, 2011
Last Time: Interaction of Circuit Elements
Last Time: Buffering with Op-Amps
Circuit design is complicated by interactions among the elements.
Adding an element changes voltages & currents throughout circuit.
Buffers can be used to simplify modular design.
1Ω
Example: closing a switch is equivalent to adding a new element.
12V
Io
2Ω
8V
8V
1Ω
2Ω
12V
Vo
This op-amp circuit produces an output voltage equal to its input
voltage (8V) while having no effect on the left part of the circuit.
There are also other useful ways to deal with element interactions.
Today: abstractions to characterize circuit interactions
Closing the switch decreases Vo and increases Io .
Systematic Changes
Check Yourself
Altering an element changes voltages and currents systematically.
3Ω
Io
Example: consider changes in Vo and Io when Ro is changed.
6Ω
90V
3Ω
90V
6Ω
Vo
Ro
Io
Vo
How many of the blue numbers are wrong?
Ro
Ro [Ω]
0
2
3
8
∞
1
Vo [V]
0
30
36
48
60
Io [A]
30
15
12
6
0
6.01: Introduction to EECS I
Lecture 9
April 5, 2011
Current-Voltage Relations
One-Ports
The current-voltage relation summarizes all possible behaviors of
circuit – regardless of what the circuit is connected to.
A “one-port” is a circuit that is regarded as a single, generalized
component.
3Ω
I
Io
Io
+
V
Vo
6Ω
90V
−
Vo
As with other components, a one-port has two terminals:
• current I enters “+” terminal and exits “−” terminal and
• produces voltage V across the terminals.
It allows us to think about circuit as a single element: a one-port.
I = −Io
+
V = Vo
−
Current-Voltage Relations
Current-Voltage Relations
Under what conditions is the current-voltage relation a straight line?
Current-voltage relations for resistors and sources are straight lines.
3Ω
I
Io
Io
I
V
R0
Vo
6Ω
90V
I
V
V0
V
I0
Vo
I
I
V
I
V
V = R0 I
V = V0
V
I = −I0
What happens when you combine different types of components?
Check Yourself
Parallel One-Ports
If the i-v curves for two one-ports are both straight lines, then the
i-v curve for the parallel combination is a straight line.
I
Graphical “proof”:
2Ω
+
5V −
I
B
V
I
C
V
V1
Ip
I1
+
V1
−
What is the corresponding current-voltage relation?
A
I1
V
I
D
V
I
+
I2
+
V2
−
I2
Vp
−
V2
Ip = I1 + I2
V p = V1 = V2
V
The “sum” of two straight lines is a straight line.
2
6.01: Introduction to EECS I
Lecture 9
April 5, 2011
Series One-Ports
Current-Voltage Relations
If the i-v curves for two one-ports are both straight lines, then the
i-v curve for the series combination is a straight line.
More generally, any combination of one-ports with straight i-v curves
will produce a one-port with a straight i-v curve.
Graphical “proof”:
Is
Example: if a one-port contains only resistors and sources, then the
current-voltage relation will be a straight line.
+
I1
+
V1
−
I2
+
V2
−
I1
Vs
3Ω
−
Vo
6Ω
90V
I2
Io
Io
Vo
Is = I1 = I2
V1
V2
Vs = V1 + V2
The “horizontal sum” of two straight lines is a straight line.
Linear Equations
Thevenin Equivalents
If the current-voltage relation is a straight line, then the relation
between element current and element voltage is a linear equation.
If the current-voltage relation is linear, the current-voltage relation
is the same as that for a voltage source in series with a resistor.
3Ω
I
Vo
6Ω
90V
Io
Io
I
Vo
V0
V
R
V
V0
1
R0
Linear equations have the form
V − V0
.
R
If I = 0, then V = V0 (the x-intercept of the plot).
ai Vi + bi Ii + ci = 0.
From the circuit, I =
√
(i.e., there are no Vi2 terms, no Ii terms, etc.)
V
V0
−
with V is the slope 1/R.
R
R
If all of the component equations are linear, then the i-v curve for the
one-port will also be linear (since the solution to a system of linear
components plus associated KVL and KCL equations is linear).
The rate of growth of I =
Norton Equivalents
Open-Circuit Voltage and Short-Circuit Current
If the current-voltage relation is linear, the current-voltage relation
is the same as that for a current source in parallel with a resistor.
If a one-port contains just resistors and current and voltage sources,
then its i-v relation is determined by two points.
I
First point: open-circuit voltage
I
V
1
R0
−I0
I
V
R
I
I0
1
R0
−I0
V0
V
V
From the circuit, V = (I + I0 )R.
If V = 0, then I = −I0 (the negative of the y-intercept of the plot).
Set I = 0 in the circuit. Then V = V0 = 1V.
The rate of growth of I = −I0 + V /R with V is the slope 1/R.
3
2Ω
2Ω
1V
6.01: Introduction to EECS I
Lecture 9
April 5, 2011
Open-Circuit Voltage and Short-Circuit Current
Open-Circuit Voltage and Short-Circuit Current
If a one-port contains just resistors and current and voltage sources,
then its i-v relation is determined by two points.
If a one-port contains just resistors and current and voltage sources,
then its i-v relation is determined by two points.
Second point: short-circuit current
The equivalent resistance is R0 =
I
1
R0
I
I
−I0
V
V0
Vo
.
Io
2Ω
2Ω
V
1V
1
R0
1
Set V = 0 (wire = short circuit). Then I = −I0 = − A.
2
R0 =
I
−I0
V0
V
2Ω
2Ω
V
Vo
1V
= 1 = 2Ω
Io
2A
Thevenin and Norton Equivalents
Thevenin Example
Equivalent circuits.
Find the Thevenin equivalent of this circuit.
I
I
I
− 12 A
1
2Ω
V
1V
2
V
10V
1Ω
3Ω
+
−
V
1
Open-circuit voltage (i.e., I = 0)
3
V0 = V =
× 10 = 7.5 V
3+1
I
1
2Ω
1V
I
− 12 A
V
1V
2
V
1/2
Thevenin Example
Thevenin Example
Find the Thevenin equivalent of this circuit.
This is the Thevenin equivalent.
I
10V
+
−
1Ω
3Ω
I
V
10V
Open-circuit voltage (i.e., I = 0)
3
V0 = V =
× 10 = 7.5 V
3+1
Short-circuit current (i.e., V = 0 → add a wire!)
10 V
= 10 A
I0 = −I =
1Ω
Equivalent resistance
V0
7.5 V
R0 =
=
= 0.75Ω
I0
10 A
+
−
1Ω
3Ω
I
V
+
−
7.5V
0.75Ω
V
These circuits are “equivalent” in the sense that their i-v curves are
the same.
I
7.5
−10
4
V
6.01: Introduction to EECS I
Lecture 9
Check Yourself
April 5, 2011
Conceptual Value of Equivalent Circuits
Equivalent circuits have conceptual value.
Determine the equivalent circuits viewed from 3 different ports.
10Ω
10Ω
10Ω
10A
5Ω
10Ω
10A
5Ω
A
4Ω
10Ω
B
10Ω
10A
Example: Will closing the switch increase or decrease I?
C
+20V
5Ω
How many of the following are wrong?
1.
2.
3.
A
20V
5A
4Ω
V0 =
I0 =
R0 =
B
40V
5A
6Ω
+
−
2Ω
4Ω
I
2Ω
2Ω
We could just solve two circuits questions – one with switch open
and one with switch closed – and compare currents.
C
60V
10A
6Ω
But this question can be answered without doing any calculations!
Conceptual Value of Equivalent Circuits
Superposition
Replace the parts to the left and right of I by equivalent circuits.
If a circuit contains only linear parts (resistors, current and voltage
sources), then any voltage (or current) can be computed as the sum
of those that result when each source is turned on one-at-a-time.
4Ω
+20V
+
−
2Ω
I
4Ω
2Ω
I
2Ω
R1
V0
R2
I0
Equivalent circuit to right of I depends on state of switch.
switch open
4Ω
I
+10V
+
−
switch closed
4Ω
I
2Ω
+10V
+
−
1Ω
Closing the switch decreases equivalent resistance to right of I.
Therefore, closing the switch increases I.
Superposition
Superposition
If a circuit contains only linear parts (resistors, current and voltage
sources), then any voltage (or current) can be computed as the sum
of those that result when each source is turned on one-at-a-time.
If a circuit contains only linear parts (resistors, current and voltage
sources), then any voltage (or current) can be computed as the sum
of those that result when each source is turned on one-at-a-time.
I
V0
R1
I
I0
R2
Turning off I0 (i.e., I0 = 0) is equivalent to an open circuit.
I1
V0
Then I1 =
R1
R2
R1
V0
R2
I0
Turning off V0 (i.e., V0 = 0) is equivalent to a short circuit.
I2
I0 = 0
V0 = 0
V0
.
R1 + R2
Then I2 = −
5
R2
I0 .
R1 + R2
R1
R2
I0
6.01: Introduction to EECS I
Lecture 9
Superposition
April 5, 2011
Check Yourself
If a circuit contains only linear parts (resistors, current and voltage
sources), then any voltage (or current) can be computed as the sum
of those that result when each source is turned on one-at-a-time.
Determine V using superposition.
I
R1
V0
R2
1.
2.
3.
4.
5.
Combining:
I1
V0
I = I1 + I2 =
R1
R2
I2
I0 = 0
1Ω
1V
V
1A
I0
V0 = 0
R1
R2
I0
1V
2V
0V
cannot determine
none of the above
V0
R2
−
I0 .
R1 + R2 R1 + R2
Check Yourself
Summary
Consequences of linearity.
Determine I using superposition.
If a one-port contains just linear elements (resistors, voltage sources,
and current sources) then
I
1Ω
1V
1.
2.
3.
4.
5.
• the current-voltage relation will be linear, and
1A
• it can be represented by a Thevenin or Norton equivalent circuit.
Linear one-ports can be characterized by two points on their i-v
curve (e.g., open-circuit voltage and short-circuit current).
1A
2A
0A
cannot determine
none of the above
Responses of multiple sources can be superposed to find electrical
responses of linear circuits.
6
MIT OpenCourseWare
http://ocw.mit.edu
6.01SC Introduction to Electrical Engineering and Computer Science
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.