1. 1 ECE 221 Electric Circuit Analysis I Chapter 13 Thévenin & Norton Equivalent Herbert G. Mayer, PSU Status 11/30/2014 For use at Changchun University of Technology CCUT

2. 2 Syllabus Motivation Motivation Thought Experiment Thought Experiment Purpose Purpose Thévenin Problem Thévenin Problem Norton Equivalent Norton Equivalent A Sample A Sample References References

3. 3 Motivation When working with real electric sources, such as typical household power supplies, the actual circuit behind the plugs is unknown It should be unknown, is it is unreasonable to expect users to know what exactly is behind the plug! We only know 1.) that there is a source of constant voltage at terminals a and b with a complex but finite resistance internally And 2.) that the current supplied depends on external load, up to some practical limits When the limit is exceeded, a fuse flips and we lose the power source 

4. 4 Motivation It is desirable to understand the physical limits of such a CVS Once we know the limits, we can model the whole complex CVS source with an equivalent, but simpler model é One such model is the Thévenin equivalent, an imaginary CVS with identical behavior éé Named after 19 th century French telegraph engineer Léon Charles Thévenin, 1857-1926 é Such a Thévenin equivalent source consists of CVS with V Th Volt and an internal resistance of R Th Ω in series

5. 5 Thought Experiment

6. 6 é We find the Thévenin parameters by the following experiment: 1. Leaving the circuit open at terminals a and b, allows us to measure the voltage V Th with no load, i.e. load with resistance R L = ∞ Measuring the voltage yields V Th 2. Short-circuiting the terminals a and b allows us to measure the maximum load current i SC Measuring the current yields i SC 3. Thus we can compute R Th Computing yields R Th = V Th / i SC

7. 7 Summary é The Thévenin Equivalent is a simple, CVS electric circuit with a serial resistor, named R Th This model is electrically equivalent to any arbitrary circuit, whose key parameters, idle voltage and short-circuit current are known, as measured above Such a circuit’s equivalent resistance is: R Th = V Th / i SC é A sample Thévenin transformation follows, taken from [1], p. 113-115

8. 8 Sample Thévenin Problem

9. 9 The following circuit has various internal sources and resistances, 2 external plugs a and b Goal is to model the exact behavior of this circuit via the R Th and i SC equivalents First measure idle voltage V Th at the plugs, here called V ab, then measure the short circuit current i SC And compute a resulting equivalent resistance R Th With plugs open, no current flows in 4 Ω resistor We compute v1, parallel to the CCV and the 20 Ω resistor And with open plugs, the 4 Ω R might as well be missing

10. 10 Sample Thévenin Problem

11. 11 Thévenin Problem: Open Plugs (v1 - 25)/5+ V1/20-3=0 4*v1 -100 -60 + v1=0 5*v1=160 v1= v ab = v Th = 32 V

12. 12 Thévenin Problem: Short-Circuit Now the plugs a and b are short-circuited A real current flows through the 4 Ω R, now parallel to the CCS We compute v2, the voltage along the 3 A CCS Once v2 is known all other units, specifically i SC can be computed

13. 13 Sample Thévenin Problem

14. 14 Thévenin Problem: Short-Circuit V2/20 + (v2-25)/5 - 3 + v2/4=0 v2 + 4*v2 + 5*v2=160 10*v2=160 v2 =16 V i SC =16/4=4 A R Th = V Th /4= 32/4 =8 Ω

15. 15 Norton Equivalent We have covered transformations of CVS to equivalent CCS é The Thévenin transformation yields a CVS with R in series Hence we know we can combine the 2 methods and generate a CCS equivalent to any circuit by just measuring its idle voltage and short-circuit current The equivalent Norton current i N is 4 A, with the 8 Ω resistor in parallel: i N = i Th = 32/8=4 A

16. 16 A Sample VL CVS = 32 * 8 / ( 8 + 8 )= 16 V iL CVS = 32 / ( 8 + 8 )= 2 A iL CCS = 4 * 8 / ( 8 + 8 )= 2 A VL CCS = ½ * 4 * 8= 16 V

17. 17 References 1. 1.Electric Circuits, James W. Nielsson and Susan A. Riedel, Pearson Education Inc, publishing as as Prentice Hall, © 2015, ISBN- 13: 978-0-13-376003-3