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ECE 3301 General Electrical Engineering Section 12 Source Transformation 1 Practical Voltage Source • Consider the practical model of a voltage source consisting of an ideal voltage source in series with a resistance. RS a VS b 2 Open-circuit voltage • The open-circuit voltage at terminals a-b is found by applying Kirchhoff’s Voltage Law. IS RS a VS VRS VOC = VS – VRS VOC VOC = VS – ISRS b IS = 0, so VOC = VS 3 Short-circuit Current • The short-circuit current at terminals a-b is found by inserting a short-circuit link and applying Ohm’s Law. RS a VS ISC VS ISC = R S b 4 Practical Voltage Source • Consider a practical voltage source with a variable load resistance. RS a I0 VS RL V0 b 5 Practical Voltage Source • The voltage across the load resistor is found by applying the voltage divider formula. RS a I0 VS RL V0 VSRL V0 = R + R L S b 6 Practical Voltage Source • As RL 0 (short-circuit), V0 0 • One cannot maintain a voltage across a short-circuit. RS a I0 VS RL V0 VSRL V0 = R + R L S b 7 Practical Voltage Source • As RL (open-circuit), V0 VS RS a I0 VS RL V0 VS V0 = RS 1+ RL b 8 Practical Voltage Source • The current through the load resistor is RS a I0 VS RL V0 VS I0 = RL + RS b 9 Practical Voltage Source • As RL (open-circuit), I0 0 RS a I0 VS RL V0 VS I0 = RL + RS b 10 Practical Voltage Source • As RL 0 (short-circuit), I0 VS / RS, I0 ISC RS a I0 VS RL V0 VS I0 = RL + RS b 11 Practical Current Source • Consider the practical model of a current source consisting of an ideal current source in parallel with a resistance. a IS RS b 12 Open-circuit voltage • The open-circuit voltage is found by applying Ohms Law . a IS IS RS VOC VOC = ISRS b 13 Short-circuit Current • The short-circuit current is found by installing a short-circuit link. The current takes “the path of least resistance. a IS RS ISC ISC = IS b 14 Short-circuit Current • From another point of view, the short-circuit link forces the voltage across the resistance to be zero, consequently no current can flow through the resistance and IS must flow through the shortcircuit link. a IS RS ISC ISC = IS b 15 Practical Current Source • Consider a practical current source driving a variable load resistance. a I0 IS RS RL V0 b 16 Practical Current Source • The current through the load resistance is found by applying the current divider formula. a I0 IS RS RL V0 ISRS I0 = R + R S L b 17 Practical Current Source • As RL (open-circuit), I0 0 a I0 IS RS RL V0 ISRS I0 = R + R S L b 18 Practical Current Source • As RL 0 (short-circuit), I0 IS a I0 IS RS RL V0 ISRS I0 = RS + RL b 19 Practical Current Source • The voltage across the load resistance is: a I0 IS RS RL V0 ISRSRL V0 = R + R S L b 20 Practical Current Source • As RL 0 (short-circuit), V0 0 a I0 IS RS RL V0 ISRSRL V0 = R + R S L b 21 Practical Current Source • As RL (open-circuit), V0 ISRS • V0 VOC a I0 IS RS RL V0 ISRS V0 = R S + 1 RL b 22 Source Transformation • Given a practical voltage source, can we find an equivalent practical current source? RV a I0 VS RL V0 VSRL V0 = RL + RV VS I0 = RL + RV b 23 Source Transformation • Given a practical voltage source, can we find an equivalent practical current source? a I0 IS RC RL V0 ISRCRL V0 = R + R L C ISRC I0 = R + R L C b 24 Source Transformation • Given a practical voltage source, can we find an equivalent practical current source? VSRL VS Voltage Source: V0 = R + R , I0 = R + R L V L V ISRCRL ISRC Current Source: V0 = R + R , I0 = R + R L C L C VS = ISRC , RV = RC 25 Source Transformation RS a a VS VS RS RS b b 26 Source Transformation RS a a IS RS ISRS b b 27
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