10/15/2004EE 42 fall 2004 lecture 201 Lecture #20 Review Reminder: MIDTERM coming up next week (Monday October 18 th ) This week: Review and examples

10/15/2004EE 42 fall 2004 lecture 202 Midterm Monday, October 18, In class One page, one side of notes

10/15/2004EE 42 fall 2004 lecture 203 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.

10/15/2004EE 42 fall 2004 lecture 204 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

10/15/2004EE 42 fall 2004 lecture 205 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.

10/15/2004EE 42 fall 2004 lecture 206 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.

10/15/2004EE 42 fall 2004 lecture 207 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).

10/15/2004EE 42 fall 2004 lecture 208 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.

10/15/2004EE 42 fall 2004 lecture 209 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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