EE 42 lecture 5

1. EE 42 lecture 5 • 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

2. Linear equations-> straight line • 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 I + V - V

3. Equivalent resistance for a linear passive circuit. • 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. I I + V - V

4. Why it must pass through the origin if there are no sources of power • 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 power! I I V + V -

5. Circuits with sources • 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 + V -

6. Models and measurements of circuits with sources • 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. • If we measure any two points, we can calculate any of these parameters.

7. Thevenin’s Theorm • 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 - V

8. Thevenin’s Theorm • 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 + V - I ~ + V - V

9. Thevenin’s Theorm • A large slope means a small value of R I R + V - I ~ + V - V

10. Norton’s Theorem • 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.

11. Norton’s Theorem • 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 I R + V - V

12. Taking Measurements • 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 .

13. Measuring Current • To measure current, insert the measuring instrument in serieswith 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 small resistance (like wire) when used as an ammeter. • It usually does not introduce serious error into your measurement, unless the circuit resistance is small. i DMM

14. Measuring Voltage • To measure voltage, insert the measuring instrument in parallelwith the device you are measuring. That is, put your measuring instrument across the measured voltage. • The measuring device will contribute a very large resistance (like air) when used as a voltmeter. • It usually does not introduce serious error into your measurement unless the circuit resistance is large. DMM + v -

15. Example • For the above circuit, what is i1? • Suppose i1 was measured using an ammeter with internal resistance 1 Ω. What would the meter read? 9 Ω 27 Ω 54 Ω 3 A i1 i2 i3

16. Measuring Resistance • To measure resistance, insert the measuring instrument in parallelwith the resistor you are measuring with nothing else attached. • The measuring device applies a voltage to the 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. DMM

17. Measurements to derive a Thevenin equivalent circuit • 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 • Turn off -> voltage sources to 0 volts (replace them with a short circuit), current sources to zero current (open)

18. Measurements to derive a Norton equivalent circuit • 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)

19. 9 Ω 27 Ω 54 Ω 3 A i1 i2 i3 9 Ω 18 Ω 3 A i1 Solution to Example: • 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, • Now, i1 = -3 A (18 Ω)/(10 Ω+18 Ω) = -1.93 A 1 Ω 10 Ω 27 Ω 54 Ω 3 A 18 Ω 3 A 9 Ω i1 i2 i1 i3