lecture09

# lecture09 - A B v = 0-i N R N A B v = 0-S i SC EE 205...

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EE 205 Coifman 9-1 Not all circuits are linear and things can get messy when its non-linear (remember problem 1-17?). Segregating the linear bits from the non-linear bits can make things simpler i A B + v L - S Source Interface Load If the source circuit in a two-terminal interface is linear, then the interface signals v and i do not change when the source circuit is replaced by its Th venin or Norton equivalent circuit. (Obviously, one could reverse the argument and apply it to a linear load and non-linear source)

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EE 205 Coifman 9-2 Th venin equivalent i A B L + - + v - v T R T Source Interface Load Norton equivalent i A B L + v - i N R N Source Interface Load
EE 205 Coifman 9-3 But how do you find i N and v T ? v T i = 0 A B + v OC - S i = 0 A B + - + v OC - v T R T i N i SC

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Unformatted text preview: A B + v = 0-i N R N A B + v = 0-S i SC EE 205 Coifman 9-4 I just wish this guy would do an example to put it all into perspective. .. Load +-B A 15 5 10 15 V Ω Ω Ω EE 205 Coifman 9-5 Prove it! i + v-S L + v TEST-S i TEST First, replace the load with a current source, i TEST EE 205 Coifman 9-6 Second, turn i TEST off, while leaving all sources within S on. .. + v TEST1-Sources ON i TEST OFF Third, turn i TEST on, but turn all sources within S off. .. + v TEST2-R EQ Sources OFF i TEST ON EE 205 Coifman 9-7 Would you buy a used example? +-B A 15 5 10 15 V i X + v TEST-i TEST Ω Ω Ω...
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## This note was uploaded on 07/17/2008 for the course EE 205 taught by Professor Coifman during the Spring '08 term at Ohio State.

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lecture09 - A B v = 0-i N R N A B v = 0-S i SC EE 205...

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