Lecture #20 Review
Reminder:
MIDTERM coming up next week
(Monday October 18th)
This week: Review and examples
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EE 42 fall 2004 lecture 20
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Midterm
• Monday, October 18,
• In class
• One page, one side of notes
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EE 42 fall 2004 lecture 20
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Tips on solving for voltages and
currents
• Start by looking at what you are asked to find. Ask the
following questions:
• Is it a current or a voltage?
• What other quantity might help you find that voltage?
• Is it a resistor current or voltage? If so, can Ohm’s law
help?
• If looking for a voltage, can you find other voltages in a
closed loop (KVL)?
• If looking for current, can you find other currents and use
KCL?
• Think about where the current will flow, about whether
diodes will be forward or reversed biased.
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EE 42 fall 2004 lecture 20
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More methods
• You can always resort to nodal analysis.
• Writing seemingly random KVL and KCL
equations never hurts.
• Try to reduce parts of the circuit to a simpler
Norton or Thevenin equivalent.
• If dependent sources are in the circuit, try to
solve for the controlling voltage or current.
• Or, write an equation for it using Ohm’s law, KVL
or KCL.
• Look for short circuits and opens in the circuit,
due to reverse biased diodes, open switches
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EE 42 fall 2004 lecture 20
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Simple errors to avoid
• Remember that 0 V over a resistor implies
0 A current, and vice-versa.
• Remember that current sources generally
have nonzero voltage. Their voltage
adapts to satisfy KVL in the circuit.
• Remember that voltage sources generally
have nonzero current. Their current adapts
to satisfy KCL in the circuit.
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EE 42 fall 2004 lecture 20
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Voltage sources
and current sources
• The output voltage of a voltage source does not depend
on what is attached to the output. The voltage source
provides current to whatever is attached to the output, to
ensure that it carries the proper voltage.
• The output current of a current source does not depend
on what is attached to it. It will produce whatever
voltage is needed to get to that current.
• An irresistible force acting on an immovable object is
always due to an error in the model. For example, a
current source trying to force current through a reverse
biased diode, two voltage sources connected in parallel,
or two current sources connected in series.
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EE 42 fall 2004 lecture 20
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Tips on finding Thevenin/Norton
Equivalents:
• Remember that we are trying to identify the I-V line for
the circuit. We usually find two specific points on the line
(the x and y intercepts) but any two points will do.
• We match up the I-V line for the Thevenin/Norton circuit
with that of our more complex circuit. We do this when
we set:
• VTH = x-intercept
• IN = negative of y-intercept
• RTH = RN = 1/slope = VTH / IN
• To find the x-intercept, set the y value to zero. That
means set the current to zero. That’s why VTH is the
voltage drop from a to b when no current flows (open
circuit).
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• To find the y-intercept, set the x value to zero.
That means set the voltage to zero. That’s why
IN is the negative of the current from a to b when
no voltage drop is present (short circuit).
• The values of VTH and IN depend on the values
of the independent sources. The value of
• RTH (RN) depends on the values of resistors
and dependent sources. This means:
• RTH remains the same even if you set voltage
and current sources to 0 V (wire) and 0 A (air)
respectively. This can simplify RTH calculation.
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Equivalent resistance
of a passive circuit
• If there are no independent sources in the circuit, VTH
and IN are 0 V and 0 A respectively. Then you cannot
find RTH using VTH / IN since 0 / 0 is undefined. In this
case, you need to treat the circuit like an unknown
resistance:
• Apply a test voltage across the terminals.
• Measure the current that flows through the circuit from +
to -.
• Divide voltage by current to get RTH.
• You could also do this by applying a test current into the
device and finding the resulting voltage drop.
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Find Vo as a function of V1.
Assume the rails are +1
volt and -1 volt
V  104 (V  V )
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V  104 (V  V )
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