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EE 215 Fundamentals of Electrical Engineering Lecture Notes Thévenin Equivalents 8/3/01 Rich Christie Overview: Along with "How many h's and f's in Kirchhoff?" (2 each), "Where does the accent go in Thevenin?" is a leading EE trivia question. For the record it's Thévenin. His equivalent can help simplify complicated problems, and is a powerful circuit concept. Thevenin Equivalent: Thevenin was a French engineer who had an idea: Take any linear circuit. Take any two points (terminals). Pretend that the rest of the circuit is in a black box. Regardless of what is actually in the black box, we can pretend that what is in there is a voltage source with a resistance in series. This is called the Thevenin equivalent. 18 V + 5A Rt + 3! 3! - vt 3! The equivalent is mostly useful when we are concerned about one circuit element, often representing the output of the circuit. For example, suppose the right-most 3 Ω resistor is the circuit load. We want to know the power delivered to the load. We can compute the Thevenin equivalent seen by the load - that is, the Thevenin equivalent of the circuit consisting of everything except the load. 18 V + 5A Rt 3! 3! + - vt 3! 3! 1 The process of finding a Thevenin equivalent depends on what's in the circuit. Simple Thevenin Equivalents: Easiest case: Independent sources and resistors: 1. Find the open circuit voltage - that's the voltage with nothing attached to the terminals. Pretty clearly, this is the Thevenin equivalent voltage vt. In fact, people often write voc instead of vt to help them remember. We want the terminals of the black box (the dotted line) to be open, which means removing whatever is connected to them that is not in the box: 18 V + + voc 3! 5A - Now we can find voc from superposition. 3! - voc = !18 + 3 " 5 = !3V 2. Find the Thevenin equivalent resistance Rt by zeroing all the independent sources and combining resistances. Rt = 3 + 3 = 6! 6! + - -3V By adding the load back in, it's easy to see that the load voltage is 1V (OK, -1V) and the power consumed is 1/3 W. 3! Note that the terminal voltage changes when something is connected. 2 Try finding the Thevenin equivalent seen by the center 3 Ω resistor: Rt = 6! 18 V + Voc = 18 + 5 ! 6 = 48V 3! 5A 3! More Systematic Thevenin Equivalent: A little harder: Independent and dependent sources, or resistors that do not combine easily. 1. Find open circuit voltage voc the usual way. 2. Find short circuit current isc. That's current with the terminals short circuited. (This may be dangerous in real life!) 3. Then Rt = voc (and, yes, this will work for easy circuits too) isc Example: Here, short circuit current is easy, 10 ! 2 A 10 ! 2A 10 ! 4A 10 ! ! 2 A 10 2A isc = 4 A For open circuit voltage, we could use nodal analysis, but in this special case we can appeal to symmetry. The current must evenly divide between the two upper resistors - and between the two lower ones, too. Then voc = 2(10 + 10) = 40V v 40 Rt = oc = = 10! isc 4 if we put a 3 ohm resistor on this circuit, the power consumption would be a bit different! 3 10 ! + 40V - Let's try the dependent source thing. Find the Thevenin equivalent seen by RL. 2! 5V + 3! (Remember to remove RL from the circuit before taking the equivalent!) a RL vab/4 b 2! 5V 3! a isc + vab/4 Hmm, isc is probably easiest, because vab is zero. Then 5V isc = = 1A 2+3 b 2! 5V + 3! a vab/4 b For voc, note that vab is also the voltage across the dependent current source - and the voltage across the 2 Ω resistor will be determined by the current source. Then KVL around the loop is 4 vab + vab = 0 4 v ab = 10V = voc v 10 Rt = oc = = 10! isc 1 !5!2 10 ! + 10V - Thevenin Equivalent with Dependent Sources: Finally, what to do about dependent sources only. Basically, voc is zero and it's just a matter of finding Rt. But the dependent source won't do anything unless another source is connected to the terminals. Normally a 1A current source is used, and the terminal voltage is found. Then Rt = vab 1A Example: - 2+vab/2 2! 1+vab/4 + 3V - 3! + a 1A vab/4 1A b Now we could solve this circuit using nodal or mesh analysis, but I'll be heuristic, and try to write everything in terms of vab, which we are trying to find. (This doesn't always work!) Note that the current through the 3 Ω resistor is 1 A, so the voltage across it is 3V. Then the current through the 2 Ω resistor is 1+vab/4, so the voltage across it is 2+vab/2. The voltage across the 1A source is vab, so KVL gives 5 " vab + 3 + 2 + vab =0 2 vab =5 2 vab = 10V Rt = 10 = 10! 1 It's just the same equivalent resistance as the previous problem with the source zeroed, which is sort of what we expect. But voc is zero: 10 ! Note that circuits with dependent sources can give rise to negative Thevenin equivalent resistances. Norton Equivalent: There's another kind of equivalent that for some reason takes second place to the Thevenin equivalent. That's the Norton equivalent. (Heyyyyyy Norton! - any Jackie Gleason fans in the crowd?) Norton equivalents are current sources in parallel with a resistor: (Hint: one of these circuits is NOT a Norton equivalent!) Rt in Req voc + - Let's try to find Norton equivalent parameters in terms of Thevenin equivalent parameters, i.e. find the Norton equivalent of the generic Thevenin equivalent. Open circuit voltages should be equal: 6 voc = Req in Short circuit currents should be equal in = voc which is how to find the Norton current. Rt Subbing in voc = Req voc Rt Req = Rt So actually the Norton is the dual of the Thevenin, it just has short circuit current instead of open circuit voltage, and the same equivalent resistance! So to find the Norton equivalent, you find short circuit current instead of open circuit voltage, and then do the same things (including 1A current sources if needed) to find the equivalent resistance. Or, find Thevenin equivalent and convert (which is longer to do but easier to remember, so it's often what I find myself doing). Find Norton equivalent of: 10 ! + - 10V 1A 10! Maximum Power Transfer: Suppose we are designing a stereo. We made this great amplifier, and now we want to pick speakers. It turns out that we can model a speaker (badly!) as a resistor, RL. In this case, picking the speaker just means choosing RL. We want speakers that dissipate the maximum power in RL, because in this case, dissipated power translates to sound output, and we want to rock the neighborhood - like, from Seattle to Portland. So what value of RL dissipates the maximum power? We need to know something about the amplifier we just built. Specifically, we need to know the Thevenin equivalent (!). So suppose we know voc and Req. Then the complete circuit looks like: 7 OK, so Req 2 & voc # ! R pL = i RL = $ $R +R ! L L " % eq 2 voc + RL - Is the formula for the output power. How do we find the maximum with respect to RL? Take derivative wrt RL and set to zero, solve for RL. Works for any function. Any differentiable function, anyway. (Oh, man, I thought I was done with calculus!) dp L RL d = voc2 dRL dRL (Req + RL )2 = voc2 d !2 RL (Req + RL ) dRL [ ] = voc2 (1)(Req + RL ) + RL (! 2 )(Req + RL ) (1) !2 !3 And set the derivative equal to zero ' $ 2 RL 1 voc2 % ! =0 2 3" ( ) ( ) R + R R + R L eq L &% eq #" 1! 2 RL =0 Req + RL 2 RL = Req + RL RL = Req ! All this math leads to a simple yet important result: Maximum power transfer occurs when the load resistance equals the equivalent resistance. Turn that thing down, dude! And it works for Norton as well as Thevenin, since the equivalent resistances are the same. 8