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Source Transformations Consider the Two Circuits Below Source transformations Consider the two circuits below. In particular, look at the current and voltage of RL in each circuit. Using any of the techniques we seen so far, it is easy to find iRL and vRL for each case. RS 2 k! iRL iRL + + V + R v I R R v S – L RL S P L RL 5 V 1 k! – 2.5 mA 2 k! 1 k! – 1 RL RL vRL = VS iRL = IS RL + RS 1 + 1 RL RP 1kΩ 1 = (5V)=1.67 V 1kΩ 1kΩ+ 2kΩ = 1 1 (2.5mA)=1.67mA 1kΩ + 2kΩ iRL = 1.67 mA, P = 2.78 mW vRL = 1.67 V, P = 2.78 mW Interesting: From the point of view of the resistor RL, the series combination of the voltage source and resistor RS gives the exact same result as the parallel combination of current source and resistor RP. EE 201 source transformations – 1 The series combination seems to behave identically to the parallel combination. This suggests that we may, in the right circumstances, replace one configuration for the other. Making this switch is known as a source transformation. RS IS = VS/RS RP IS VS RP = RS RS VS = ISRP RP IS VS RS = RP EE 201 source transformations – 2 These are perfectly viable substitutions. From the point of view of whatever circuitry is attached to the two terminals, the result will be exactly the same for the two source configurations. If you are trying calculate something about the two components (VS-RS or IP-RP), you cannot transform them. In transforming them, you lose the ability to calculate a specific property. This is an example of bigger, more important idea known as the Thevenin equivalent of circuit. We will introduce this later, and make extensive use of it in discussing amplifiers, etc. EE 201 source transformations – 3 Example 1 Ω R1 25 Find iR4 in the V + R2 R3 R4 S – 50 Ω 200 Ω 100 Ω circuit at right. 50 V iR4 Use a source transformation to put everything in parallel. I R1 R2 R3 R4 S Ω 50 Ω 200 Ω 100 Ω 2 A 25 IS = VS/R1 = 50 V/25 ! = 2A. iR4 Then use a current divider: 1 R4 iR4 = IS 1 + 1 + 1 + 1 R1 R2 R3 R4 1 100 Ω = 1 1 1 1 (2A)=0.267 A Easy. 25 Ω + 50 Ω + 200 Ω + 100 Ω EE 201 source transformations – 4 Example 2 R1 5 kΩ Find iR2 in the + VS R2 I circuit at right. – 7.5 kΩ S 20 V iR2 2 mA Transform the voltage source / I R R I resistor combo. ST 1 2 S 4 mA 5 kΩ 7.5 kΩ i 2 mA IST = VS/R1 = 20 V/5 Ω = 4 mA. R2 Combine the two current IP R1 R2 sources, IP = IST + IS = 6 mA… 5 kΩ 7.5 kΩ 6 mA iR2 1 R2 iR2 = IP …and use the current 1 + 1 R1 R2 divider then once again. 1 = 7.5kΩ (6 mA)=2.4 mA 1 + 1 EE 201 7.5kΩ 5kΩ source transformations – 5 Example 3 (An incorrect application) R1 1 kΩ i Find iR1 in the circuit at left. + R1 R VS – 2 IS 1 kΩ 40 V 15 mA The previous example worked nicely so use the same method. Transform VS & R1, and use current divider with the total current. i = ( + ) R1 R1 R + IST 2 IS = ()=. 40 mA 1 kΩ 1 kΩ 15 mA + It seems nice, but it is wrong because you cannot transform the component for which you are trying to find voltage or current. (To see that it is wrong, insert iR1 = 27.5 mA in the original circuit and show that there are serious inconsistencies with the currents and voltages.) EE 201 source transformations – 6 Example 3 (Redo it correctly.) R1 1 kΩ i Find iR1 in the circuit at left. + R1 R VS – 2 IS 1 kΩ 40 V 15 mA Transform IS & R2. R1 1 kΩ R2 1 kΩ Writing a KVL loop equation and solving for i gives i R1 + R1 + VS VST – – 40 V 15 V = − + = − = . This is the correct answer. + EE 201 source transformations – 7 Example 4 1.6 Ω Ω 20 – 60 V R4 R1 + + VS2 Find vR5 in the I S R R vR5 circuit at left. 3 8 Ω 5 6 Ω VS1 + 36 A – – R2 120 V 5 Ω Transform two voltage sources to current sources. (Pay attention to polarity.) 1.6 Ω 20 Ω 5 Ω R4 + I R IST2 R I R R ST1 1 2 S 3 8 Ω 5 vR5 6 A 12 A 36 A 6 Ω – EE 201 source transformations – 8 Example 4 (cont.) Add the parallel current sources into one. Combine the parallel resistors into one. 1.6 Ω + R4 Req R5 vR5 Ieq 8 Ω 30 A 2.4 Ω – Ieq = 6A – 12 A + 36 A = 30 A. Req = 20 Ω || 5 Ω || 6 Ω = 2.4 Ω. Transform Ieq & Req: Use voltage divider: 2.4 Ω 1.6 Ω = + + R R4 + eq V + R v = () eqt – 8 Ω 5 R5 . + . + 72 V – = EE 201 source transformations – 9.
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