ECE 125, April 26, 2010 I. Thevenin's Theorem This theorem ...

ECE 125, April 26, 2010I. Thevenin’s TheoremThis theorem provides a circuit simplification technique that focuses on terminal behavior. Wewant to replace a circuit network at terminals a-b with a voltage source in series with aresistance. The voltage source will have value V th and the resistance will have value R th .To findTo find the Thevenin equivalent circuit,1) Open the circuit at terminals a-b and find the open circuit voltage (V oc ). This voltage willbecome V th.2) Find Rth by deactivating independent sources (Short any voltage sources and open anycurrent sources).Example. Find V o using Thevenin’s Theorem.This analysis will be done by starting “opening” the circuit near the 1 kΩ load resistor. We willfind the Thevenin equivalent circuit by following the steps above. The load resistor will bereattached to the equivalent circuit (V th in series with R th ).Step 1. Find V oc . In this case, the 2 kΩ resistor does not contribute to V oc because the circuit isopen at the cut x-x. An open circuit draws no current so there is no voltage drop across the 2 kΩresistor. Because of this, V oc is the voltage across the 6 kΩ resistor.By voltage division, Voc = 12 (6/12) = 6 V.

Step 2. Find R th. Shorting the voltage source yields the circuit below:Shorting the 12 V source places the two 6 kΩ resistors in parallel (3 kΩ) and that combination isin series with 2 kΩ. R th is equal to 3 kΩ + 2 kΩ = 5 kΩ.The circuit network to the left of the cut x-x can be replaced by the Thevenin equivalent (6 V inseries with 5 kΩ). The “load” resistance (1 kΩ) can now be attached to the equivalent circuit.By voltage division,V o = 6* ( 1 kΩ)(5 kΩ + 1 kΩ)= 6(1/6) = 1 V.II.Maximum Power TransferThevenin equivalent circuits are useful in determining the maximum power that can be deliveredto a load. In a simple series Thevenin equivalent circuit, the power delivered to the load is givenby:P L = i 2 * R LIt is useful to determine the value of R L that willmaximize the quantity of power delivered to aload resistance. This maximum value can bedetermined by differentiating P L with respect to R L and setting the derivative equal to zero. Thederivation will be shown to you in ECE 225. Setting dP L /dR L = 0 results in R s = R L . R s issimply the resistance we have been calling R th .

Another example.For the following network, find RL for maximum power transfer as well as the maximum powertransferred to this load.Step 1. Find V ocRemove the load resistance by opening the circuit just to the left of the load and determining V oc .V oc is the voltage across the 4 kΩ resistor. Doing a source transformation of the 4 mA sourceand the series resistance (2+1 kΩ = 3 kΩ) yields a 12 V source in series with the 3 kΩ:By voltage division, V oc = 12 * (4k) = 12* (4/9) = 16/3 V.(4k + 3k + 2k)Step 2. Find R thOpen the 4 mA source and determine R th at the place that you “opened” the circuit (this is notthe same place that you are opening the 4 mA source).In this circuit, 5 kΩ (2 + 1 + 2) is in parallel with 4 kΩ leading to R th = 20/9 = 2.22 kΩ.

Putting this all together…We set the load resistance = 2.22 kΩ for maximum power transfer and now can calculate P L.P L = I 2 R L and I = 5.33(2.22+2.22) kΩSo= 1.2 mAP L = (1.2 mA) 2 (2.22 kΩ) = 3.2 mW.You could also calculate P L by using the equation V L 2 /R LAnother example.Problem E5.7 from the text. Find R L for maximum power transfer and find P maxTo find V oc , open the circuit at a-b and note that V oc is the voltage across the 12 kΩ resistorbecause the 2 kΩ resistor is connected to the open circuit. In this case, V oc = 6 (12/18) = 4 V.Shorting the 6 V source puts the 6 kΩ and 12 kΩ in parallel and that combination is in serieswith the 2 kΩ.

So R th = 4 kΩ + 2 kΩ = 6 kΩ yielding the circuit below:P max = 0.667 mW