ECE 3144 Lecture 19

1. ECE 3144 Lecture 19 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University

2. Ideal source vo=vS vo vo vo= iSRS Ideal source Ideal source Practical source Practical source Practical source + io= vS/RS Ideal source io io - Practical source Sources Transformations Practical voltage source Practical current source If vS = iS RS If vS = iS RS vo = RS(iS – io) vo = vS - ioRS

3. Source transformations • Recall voltage source formula vo = vS – ioRSand current sourceformulavo = RS(iS – io). The voltage source circuit is equivalent to the current source circuit if they have the same i-v characteristics to the load resistor, i.e., vS = RSiS • After all, one circuit can't see another's structure, count the number of its components, notice differences in color, smell or weight. All one circuit can "see" about another is its i-v characteristic. • To another circuit, two linear circuits are equivalent (even if they are entirely different inside) provided only that the two circuits have the same relationship between voltage and current at corresponding load terminals (that is, have the same i-v characteristic).

4. Source Transformation example 1 Find I in the circuit given. The equivalent voltage source: 9 mA*5k = 45V in series with a 5k resistor. Apply KVL to the only loop: -45 +5000I+4700I+3000I+3 = 0 Which can be easily solved to find that the current I =3.3mA

5. Source Transformation example 2 Find Io in the network shown using source transformation.

6. io io + + vo vo - - (c) Thevenin and Norton theorems • Let us suppose that we need to only make a partial analysis of the a circuit. For example, perhaps we need to determine the current, voltage and power delivered to a single “load” resistor by the remainder of the circuit., which may consist of a sizable number of sources and resistors. • Thevenin theorem tells us that we can replace the entire network, exclusive of the load resistor, by an equivalent circuit that contains only an independent voltage source in series with a resistor in such a way that the current-voltage (i-v) relationship at the load resistor is unchanged. • Norton theorem tells us that we can replace the entire network, exclusive of the load resistor, by an equivalent circuit that contains only an independent current source in parallel with a resistor in such a way that the current-voltage (i-v) relationship at the load resistor is unchanged. io + vo - (a) (b) A complex network including a load resistor RL. A Thevenin equivalent network A Norton equivalent network

7. io + vo - Thevenin Theorem • This result for Thevenin theorem is not too surprising if we note that the Thevenin equivalent circuit can produce any linear (straight line) i-v characteristic for some choice of vTH(t) and RTH. • A natural question is how to determine vTH(t) and RTHfor a particular circuit. Sometimes they are measured experimentally. Sometimes they are calculated. • In either case, it is helpful to note that if we connect no load and therefore io(t) = 0, then we can determine vTH(t) from • where voc(t) is called the open circuit voltage • If we short circuit the two terminals to force vo(t) = 0, then we get • If vTH(t) ≠ 0, then ishortcircuit(t)≠ 0 and we find

8. Thevenin Theorem: cont’d • During the measurement, one difficulty is that if we short the terminals of some circuits (a wall plug, for example), the circuit can be damaged by the resulting large current. In that case, we can simply place a resistor, R*, across the terminals that causes the terminal voltage to drop below the open-circuit voltage by a measurable amount to a voltage, v*(t). By the voltage divider rule, we see, in this case that • Consider, now, the case in which vTH(t) = 0. Then, the Thevenin equivalent circuit is simply a resistor. To measure the Thevenin resistance in this case, we can apply a voltage vS(t) (usually a constant) and measure the resulting current iS(t): =>

9. io + vo - (c) Norton theorem • Based on source transformation we have learned, we can determine iN(t) and RN

10. Summary for Thevenin and Norton theorems • Thevenin’s theorem requires that, for any linear circuit consisting of resistance elements and energy sources with an identified terminal pair, the circuit can be replaced by a series combination of an ideal voltage source vTHand a resistance RTH, where vTHis the open-circuit voltage at the two terminals and RTH is the ratio of the open-circuit voltage to the short-circuit current at the terminal pair. • Norton’s theorem requires that, for any linear circuit consisting of resistance elements and energy sources with an identified terminal pair, the circuit can be replaced by a parallel combination of an ideal current source iNand a resistance RN, where iNis the short-circuit current at the two terminals and RN is the ratio of the open-circuit voltage to the short-circuit current at the terminal pair.

11. Applications of Thevenin and Norton’s Theorems • Note that this systematic transformation allows us to reduce the network to a simpler equivalent form with respect to some other circuit elements. • Although this technique is applicable to networks containing dependent sources, it may not be as useful as other techniques and care must be taken not transform the part of the circuit which contains the control variables. • How to apply these theorems depends on the structure of the circuits. • Case 1: if only independent sources are present, we can calculate the open-circuit voltage and short circuit current and then the Thevenin equivalent resistance. • Case 2: if both independent sources and dependent sources are present, we will calculate the open-circuit and short circuit current first. Then determine the Thevenin equivalent resistance. • Case 3: For circuit only contains dependent sources, since both open-circuit and short-circuit current are zero (no independent sources here to provide the controlling variables), we cannot determine RTH in this case by using voc/isc. Remember during the discussion of Thevenin theorem, we can apply an external voltage source vS(t) (usually a constant) and measure the resulting current iS(t) => RTH = vS(t)/iS(t)

12. Case 1 example 1 Find the Thevenin equivalent circuit at terminal pair a and b for the circuit shown. + This specific problem can be solved by using different approaches. We solve the problem by using source transformation technique. - Thus we have RTH = 12 and vTH = -8V

13. Case 2 example Find Vo in the circuit shown using Thevenin’s Theorem. Find Voc + Vx = 6*2k/(4k+2k) = 2V Vy = 12-4k*Vx/1000 = 12-8= 4V Voc = Vx-Vy = -2V - - Find Isc (6-Vx)/4k=Vx/2k+Isc Isc = Vx/1K+(Vx-12)/4k Vx => Isc = -0.19mA Rth = Voc/Isc=10.67k Vo = Voc*2k/(2k+10.67k) = -0.32V

14. Case 3 example Find the Thevenin equivalent of the circuit given a b Since the rightmost terminals are already open-circuit, i=0. Consequently, the dependent source is dead. For this network, since there is no independent source in the network, both voc and isc are zero. Apply a 1-A source iS externally, measure the voltage vSacross the terminal pairs. Then we have RTH = vS/iS = vS. We can see that i=-1A. Apply KVL nodal analysis at node a: a + vS => - vS = 0.6V => RTH = 0.6

15. Homework for Lecture 19 • Problems 4.19, 4.22, 4.24, 4.28