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# ECA - Source Transformation

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### ECA - Source Transformation

1. 1. INTRODUCTION A large complex circuits Simplify circuit analysis Circuit Theorems ‧Thevenin’s theorem ‧Circuit linearity ‧source transformation ‧ ‧ ‧ Norton theorem Superposition max. power transfer
2. 2. Source Transformation A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa
3. 3. Source Transformation vs vs = is R or is = R
4. 4. Source Transformation Vs = Rs I s Vs Is = Rs
5. 5. Source Transformation  Equivalent sources can be used to simplify the analysis of some circuits.  A voltage source in series with a resistor is transformed into a current source in parallel with a resistor.  A current source in parallel with a resistor is transformed into a voltage source in series with a resistor.
6. 6. Example  Use source transformation to find vo in the circuit.
7. 7. Example
8. 8. Example we use current division to get and 2 i= (2) = 0.4A 2+8 vo = 8i = 8(0.4) = 3.2V
9. 9. Superposition  The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.
10. 10. Steps to apply superposition principle 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.   1. 2.  Turn off voltages sources = short voltage sources; make it equal to zero voltage Turn off current sources = open current sources; make it equal to zero current Repeat step 1 for each of the other independent sources. Find the total contribution by adding algebraically all the contributions due to the independent sources. Dependent sources are left intact.
11. 11. Superposition - Problem 2kΩ 4mA 12V – + 2mA 1kΩ I0 2kΩ
12. 12. 2mA Source Contribution 2kΩ 2mA 1kΩ I’0 I’0 = -4/3 mA 2kΩ
13. 13. 4mA Source Contribution 2kΩ 4mA 1kΩ I’’0 I’’0 = 0 2kΩ
14. 14. 12V Source Contribution 12V 2kΩ – + 1kΩ I’’’0 I’’’0 = -4 mA 2kΩ
15. 15. Final Result I’0 = -4/3 mA I’’0 = 0 I’’’0 = -4 mA I0 = I’0+ I’’0+ I’’’0 = -16/3 mA