Download Lecture 5

EE 42 lecture 5
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Equivalent resistance of a passive linear circuit
Thevenin’s Theorem for a single source linear
circuit
Norton’s Theorem for a single source linear
circuit
Measurements
Nodal analysis
EECS 42 fall 2004 lecture 5
9/10/2004
Linear equations-> straight line
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We know it is a straight line because the circuit
gives us a set of linear equations, and so they
have a linear solution.
I
+
V
-
EECS 42 fall 2004 lecture 5
9/10/2004
I
V
Equivalent resistance for a linear passive
circuit.
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I
If we have a circuit which includes only wires
and resistors, the relationship between the
voltage and current are related by a straight line,
through the origin.
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V
-
EECS 42 fall 2004 lecture 5
9/10/2004
I
V
Why it must pass through the origin if
there are no sources of power
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The function of current vs voltage must pass
through the origin, otherwise it would pass
through a quadrant where we could extract
power—and we said there were no sources of
I
power!
I
+
V
-
EECS 42 fall 2004 lecture 5
9/10/2004
V
Circuits with sources
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If we have any circuit which includes DC sources
(voltage or current) and linear resistances, then
the I-V curve is still linear, but it does not have to
pass through the origin.
I
I
+
V
EECS 42 fall 2004 lecture 5
9/10/2004
V
Models and measurements of circuits
with sources
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Since the I-V curve is a straight line, we can
characterize it with two parameters, for example:
 The
current intercept and slope.
 The voltage intercept and slope.
 Both the current and voltage intercept.
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If we measure any two points, we can calculate
any of these parameters.
EECS 42 fall 2004 lecture 5
9/10/2004
Thevenin’s Theorm
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We can make an equivalent circuit for any circuit
which includes only resistors and linear sources
with a single voltage source and a resistor. This
translates the voltage intercept and slope
equation into a circuit.
I
I
+
V
EECS 42 fall 2004 lecture 5
9/10/2004
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Thevenin’s Theorm
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The point where the current is equal to zero
gives us the voltage put out by the voltage
source in the equivalent circuit.
(1/slope) gives us the resistance R of the
resistor in the equivalent circuit.
I
R
I
+
V
EECS 42 fall 2004 lecture 5
9/10/2004
~
V
+
V
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Thevenin’s Theorm
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A large slope means a small value of R
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R
I
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V
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EECS 42 fall 2004 lecture 5
9/10/2004
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+
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Norton’s Theorem
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We can also make a different equivalent circuit,
for any circuit which contains only linear sources
and resistors, with a current source
The needed value of current from the current
source is the amount of current put out by the
actual circuit when the voltage is zero (the
current intercept)
This model is called a Norton equivalent circuit.
EECS 42 fall 2004 lecture 5
9/10/2004
Norton’s Theorem
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The point where the voltage is equal to zero
gives us the current put out by the voltage
source in the equivalent circuit.
(1/slope) gives us the resistance R of the
resistor in the equivalent circuit.
I
I
+
V
EECS 42 fall 2004 lecture 5
9/10/2004
I
V
R
Taking Measurements
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To measure voltage, we use a two-terminal
device called a voltmeter.
To measure current, we use a two-terminal
device called a ammeter.
To measure resistance, we use a two-terminal
device called a ohmmeter.
A multimeter can be setup to function as any of
these three devices.
In lab, you use a DMM to take measurements,
which is short for digital multimeter .
EECS 42 fall 2004 lecture 5
9/10/2004
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Measuring Current
To measure current, insert the measuring
instrument in series with the device you are
measuring. That is, put your measuring instrument
in the path of the current flow.
The measuring device
will contribute a very
i
small resistance (like wire)
when used as an ammeter.
It usually does not
introduce serious error into
DMM
your measurement, unless
the circuit resistance is small.
EECS 42 fall 2004 lecture 5
9/10/2004
Measuring Voltage
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To measure voltage, insert the measuring
instrument in parallel with the device you are
measuring. That is, put your measuring instrument
across the measured voltage.
The measuring device
DMM
will contribute a very
large resistance (like air)
when used as a voltmeter.
+ v It usually does not
introduce serious error into
your measurement unless
the circuit resistance is large.
EECS 42 fall 2004 lecture 5
9/10/2004
Example
3A
9Ω
27 Ω
i1
54 Ω
i2
i3
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For the above circuit, what is i1?
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Suppose i1 was measured using an ammeter with
internal resistance 1 Ω. What would the meter read?
EECS 42 fall 2004 lecture 5
9/10/2004
Measuring Resistance
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To measure resistance, insert the measuring
instrument in parallel with the resistor you are
measuring with nothing else attached.
The measuring device
applies a voltage to the
DMM
resistance and measures
the current, then uses Ohm’s
law to determine resistance.
It is important to adjust the settings of the meter
for the approximate size (Ω or MΩ) of the
resistance being measured so appropriate
voltage is applied to get a reasonable current.
EECS 42 fall 2004 lecture 5
9/10/2004
Measurements to derive a Thevenin
equivalent circuit
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Measure the open circuit voltage.
The voltage for the voltage source in the Tevenin model
is this open circuit voltage.
Measure the short circuit current.
The resistance to put into the Thevenin model is
R=Voc/Iss
Alternatively, you can turn off all of the sources within the
black box and measure its resistance
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Turn off -> voltage sources to 0 volts (replace them with a short
circuit), current sources to zero current (open)
EECS 42 fall 2004 lecture 5
9/10/2004
Measurements to derive a Norton
equivalent circuit
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Measure the short circuit current
The current for the current source in the Norton model is
this open circuit voltage.
Measure the open circuit voltage.
The resistance to put into the Norton model is R=Voc/Iss
Alternatively, you can turn off all of the sources within the
black box and measure its resistance

Turn off -> voltage sources to 0 volts (replace them with a short
circuit), current sources to zero current (open)
EECS 42 fall 2004 lecture 5
9/10/2004
Solution to Example:
3A
9Ω
27 Ω
i1
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i2
3A
9Ω
18 Ω
i1
i3
By current division, i1 = -3 A (18 Ω)/(9 Ω+18 Ω) = -2 A
When the ammeter is placed in series with the 9 Ω,
1Ω
3A
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54 Ω
9Ω
27 Ω
i1
54 Ω
i2
3A
10 Ω
i3
Now,
i
=
-3
A
(18
Ω)/(10
Ω+18
Ω)
=
-1.93
A
1
EECS 42 fall 2004 lecture 5
9/10/2004
18 Ω
i1