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Principles of Computer Engineering: Lecture 5: Source Transformation

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Introduction Source transformations are useful method of circuit analysis It is theoretically possible to replace any given arbitrary linear circuit containing any number of sources and resistances with either a Thévenin equivalent or Norton equivalent circuit A simple source transformation allows a voltage source with series resistance to be replaced by a current source with a parallel resistance and vice versa. Both circuits behave in the same way to external loads. a b

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Thévenin Equivalent Circuit All linear circuits can be modelled by an independent voltage source and a series resistor The voltage of the source is the open circuit voltage of the network across ‘a’ and ‘b’ The resistance is determined from the short circuit current

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Norton Equivalent Circuit All linear circuits can be modelled by an independent current source and a parallel resistor The current of the source is the short circuit current through ‘a’ and ‘b’ The resistance is determined from the open circuit voltage (same as Thévenin resistance)

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Example 1 Find both Thévenin and Norton equivalent circuits for the circuit below

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Example 2 (TQ5) Find both Thevenin and Norton equivalent circuits for the circuit below

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Summary Thévenin equivalent circuit Norton equivalent circuit Examples Questions

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Tutorial Question 6 Solve for i o using source transformations

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Principles of Computer Engineering : Lab Exp 5: Source Transformation Dr. Steve Alty

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Network Simplification 1. Construct circuit network 2. Measure open-circuit voltage 3. Measure short-circuit current 4. Determine Thévenin equivalent circuit 5. Determine Norton equivalent Measur ed V oc = v s I sc = i s R=V oc /I sc

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Network Simplification 1. Use equivalents to work out effect of 1k resistor as a load (across ‘a’ and ‘b’) on Thévenin and 4.7 k resistor as a load on Norton. 2. Measure actual effect of resistors using DMM 3. Analyse circuit and calculate mathematically the equivalent circuits

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Results R ab (k Ω ) Predicted V ab (v) Measured V ab (v) Predicted I ab (mA) Measured I ab (mA) 1 (v s =??, R=??) 4.7 (i s =??, R=??)

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Proof Theoretical calculate the V oc and I sc Open circuit R eq =1 + (5.7||5.7) = 1+2.85 = 3.85 kΩ I s = 15/3.85 = 3.9 mA V 1k = 3.9×1 = 3.9v V 5.7k = 15-3.9 = 11.1v V b0 = (1/5.7) ×11.1 = 1.95v V a0 = (4.7/5.7) ×11.1 =9.15v V ab = V a0 - V b0 = 9.15-1.95 = 7.2v = V oc

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IsIs I1I1 I3I3 I2I2 I4I4 I ab Short circuit R eq = 1+(4.7||1)+(1||4.7) = 1+0.825+0.825 = 2.65 kΩ I s = 15/2.65k = 5.66 mA V 0.82k = 5.66×0.825 = 4.67 v I 1 = 4.67/1k = 4.67 mA I 2 = 4.67/4.7k = 0.99 mA I 3 = 4.67/4.7k = 0.99 mA I 4 = 4.67/1k = 4.67 mA I 1 =I ab + I 3 I ab = I 1 - I 3 = 4.67 – 0.99 = 3.68 mA = I sc

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These are some of the most powerful analysis results to be discussed. They permit to hide information that is not relevant and concentrate in what is important.

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