REVIEW - Thevenin

REVIEW - Thevenin - ABOUT THEVENIN & NORTON EQUIVALENT...

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1 Any linear circuit can be replaced by a Thevenin equivalent circuit which is specified by three parameters, the Thevenin voltage, the short circuit current, and the Thevenin resistance. Linear circuits are described by a straight line i-v relationship: The short circuit current occurs when v=0. The open circuit occurs when i=0. The slope of the line is ! 1 R TH where R TH is the Thevenin resistance. PAGE THEVENIN EQUIV FROM MEASUREMENTS 2-5 THEVENIN EQUIV FOR INDEPENDENT SOURCES 6-7 THEVENIN EQUIV FOR DEPENDENT SOURCES 8-10 THEVENIN EQUIV USING TEST SOURCE 11-12 MAXIMUM POWER TRANSFER 13-15
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2 THEVENIN EQUIV CIRCUIT FROM MEASUREMENTS (HW1, S08) 3. You measure the i-v characteristics of the current source shown below. (a) Plot i s versus v s . SOLUTION: This element does not follow the passive sign convention so I reversed the sign and plotted the following data. This can be readily plotted as shown below. The circles represent the measured data. (b) Construct a circuit model using a current source and a resistor that will go inside the box labeled CCS that is valid for 0 v s 30 based upon the i-v characteristic you plotted in (a). SOLUTION: This is asking us for a circuit model corresponding to the linear data. Thevenin is only valid for linear data. The circuit model being asked for is a current source in parallel with a resistor as shown below. i d (mA) v d (V) -40 0 -35 10 -30 20 -25 30 -18 40 -8 50 0 55 + - v d i d 80 70 60 50 40 30 20 10 -40 -20 i d (mA) v d (V)
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3 i d (mA) v d (V) + - I R We can find the resistor by computing the slope of the i-v curve for 0 vs<30 volts. As you can see this is the linear portion of the curve. R = ! V ! I = 30 " 0 " 25 mA " " 40 mA ( ) = 30 15 mA = 2000 # We find the current source I by considering one of the ends of the curve. We can get the end by making either the current zero or the voltage zero. Consider how we can make the voltage zero. This happens when we short the output of the circuit model as shown below. i d (mA) 0 volts + - I R In this circuit we now have v d =0. Our equivalent circuit says that the current i d should be equal to –I since all the current will now go through the short. The – sign comes from the direction of i d . The table of measurements tells us that i d =-40mA when v d =0 so we conclude I=+40mA. Our circuit model is now i d (mA) v d (V) + - 40mA 2000 (c) Assume you connect a 30k Ω resistor to your circuit model (i.e., between the + and – terminals shown above). What is the current i s going through the 30k Ω resistor?
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4 In this part of the problem we connect a resistor across the output of our circuit model and want to determine the current going through the 30k Ω resistor as shown below. i
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REVIEW - Thevenin - ABOUT THEVENIN &amp; NORTON EQUIVALENT...

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