Unit 3.4 Negative Resistance

Unit 3.4 Negative Resistance - Using a Current Source… f...

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1 Unit 3: Other Circuit Analysis Techniques Unit 3.4 - Negative Resistance Consider the circuit below… 10V 1 10 Ω 50 Ω 10 Ω + V 1 - f No independent sources f To find Thévenin Equivalent, you must apply an independent source to the output terminals Add a test source--1A -20 = 1 20V - = 1 V = I V = R -20V = V , 6 V = ) 50 + 10 10 ( V = V 0 = 1 - 60 V + 10 ) 10V - (V test test test th test test test 1 test 1 test Solving 10V 1 10 Ω 50 Ω 10 Ω + V 1 - Itest=1A + Vtest - Solve for Vtest: The Thévenin Equivalent Circuit 10V 1 10 Ω 50 Ω 10 Ω + V 1 - Rth = -20 Ω How Can this Be? f Can only have negative resistance with dependent sources f The negative resistance describes the terminal behavior of the original circuit f When an independent source is attached to the terminals, the original dependent source supplies power 10 V source applied I = 10/(-20) = -.5A Power absorbed by 10V source= 5W Rth = -20 Ω 10V I= -.5A
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Unformatted text preview: Using a Current Source… f If we apply an independent current source, the original dependent source supplies power to it as well… V = 2(-20) = -40V Power supplied to independent source = 80W Rth = -20 Ω 2A + V = -40V-2 An Op-Amp Circuit Realization 1k Ω 9k Ω 50 Ω 10 Ω +-10 Ω 10V 1 10 Ω 50 Ω 10 Ω + V 1-The original circuit is a model for a circuit using an op-amp in a non-inverting configuration. Summary f A negative Thévenin resistance can be realized using dependent sources f These represent models for actual circuits f A negative Thévenin resistance means the dependent sources in the original circuit supply power to any independent sources connected to its terminals...
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This note was uploaded on 08/27/2009 for the course EE 16200 taught by Professor Williamneal during the Fall '08 term at University of Texas.

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Unit 3.4 Negative Resistance - Using a Current Source… f...

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