Marcel 0 Posted September 14, 2015 Hi,I'm following an EE class where we are using your Real Analog - Circuits 1 textbook and lectures. I've been very pleased so far, but I happened to notice a giant bummer in chapter 4, section 6 - Maximum Power Transfer. The whole chapter is based on the misunderstanding that maximum power transfer also implies maximum efficiency, which is obviously wrong.A few qoutes:The power delivered to the load is maximized if the load resistance is equal to the Thevenin resistance of the circuit supplying the power. When this condition is met, the circuit and the load are said to be matched. When the load and the circuit are matched, 50% of the power generated in the circuit can be delivered to the load – under any other circumstances, a smaller percentage of the generated power will be provided to the load.(page 40)The same amount of power is converted to heat within the power supply; this is the reason many power supplies contain a fan to actively disperse this heat to the atmosphere. If the circuit’s input resistance is not equal to the source resistance, less power is transmitted to the circuit and a correspondingly greater amount is dissipated within the power supply(page 40)Again, a reduced percentage of the power generated by the source will be delivered to the circuit when the circuit and source are not well matched.(page 41)All of these claims only hold for R_{load} < R_{source}, whereas R_{load} > R_{source} will lead to higher efficiency and less power dissipated in the supply.I believe there are more cases of this misunderstanding appearing in the text, those are just a few examples. Regards,Marcel Share this post Link to post Share on other sites

JColvin 327 Posted September 17, 2015 Hi Marcel,Thank you for your feedback! Let me see if I can help clarify some things.Within this particular section, the goal is to determine the largest amount of power that can be delivered to the load so that it can do some form of work for us. In my mind then, the maximum efficiency of the circuit (where I calculate efficiency shown here) would be when the maximum amount of power is delivered to the component (and correspondingly the maximum amount of power that the supply could provide) so it can do as much work as it can.So, if we wanted to calculate the power used/going through a component, we would want to use the general power equation. Presuming we had circuit such as the one in Figure 4.15 on page 38, we can then show the Thevenin equivalent of this circuit in Figure 4.16, where V_{OC} is the Overall circuit voltage (and is an ideal voltage source because R_{S} is accounted for), R_{TH} is the Thevenin resistance (the resistance of the voltage source), and V_{L} and R_{L} are the voltage across the load and the resistance of the load, respectively.We can calculate the voltage across the load via a voltage divider (eq. 4.13)Using our power equation, we can then calculate the amount of power delivered to the load (equation 4.14)We can then graph this equation by picking a V_{OC} and a R_{TH} (R_{S}) and varying the load resistance both above and below the value of R_{S}.We can then determine where at what value of R_{L} this peak occurs by differentiating the equation we had with respect to R_{L} and setting it equal to zero (so that we find the peak).The equation is satisfied if either V_{OC}^{2} is zero (which would imply no voltage, and thus no power by P=V*i) or by having the numerator be equal to 0. Following along with the text on page 39, we see that we will achieve this peak when R_{L} = R_{TH}. The same result is achieved if we instead solved for the maximum power delivered by our source. We can confirm this by choosing a V_{OC} and plugging in various combinations of R_{L} and R_{S}.Thus in my mind, when we do not get the maximum power we cannot achieve the maximum efficiency of the circuit, but again this is all based on the assumption that efficiency is referring to how much work we can get out of the component receiving energy from the source.If instead efficiency is determined by the minimum amount of power dissipated by the source, then yes, I definitely agree that a R_{L} greater than the R_{S} would be more efficient. However, it follows that with larger and larger values of R_{L}, less current will be flowing through the circuit, leading to the conclusion that the power supply that dissipates the least amount of heat is also the one that is hooked up to an infinite load (an open circuit), which I find to be of limited use.Please let me know if you have any questions.Thanks,JColvin Share this post Link to post Share on other sites

Marcel 0 Posted September 17, 2015 Hi JColvinThanks for your thorough reply. I follow your chain of thought, however I'm not so sure that the way you define efficiency is the commonly accepted one. To be less diplomatic I would probably say that it is plain wrong.From the Wikipedia article on electrical efficiency we have:Efficiency = (Useful power output) / (Power input)In the case of the non-ideal voltage source, the power input will be the power delivered by the voltage source and the useful output will be the power consumed by the load. As the load resistance approaches infinity, the efficiency approaches 100%. However, in a real world scenario that won't be that case, as any mains connected power supply will consume some amount quiescent current and any battery that I know of will have some amount of self-discharge. Thus the practical source impedance should be somewhat lower than close-to-infinity to achieve the maximum theoretical efficiency. And of course in the real world, efficiency will not necessarily be first priority since you probably want the source to deliver however much energy you need for your load to perform some task. But using a power supply / source with low internal impedance will always improve efficiency.I believe the current text of chapter 4.6 is problematic for a number of reasonsWith efficiency defined as maximum power delivered(R_{load} = R_{th)} the maximum proportion of generated energy delivered to the load should be 50%, with the remaining 50% wasted as heat dissipated in the source. Did you ever hear of anyone advertising a power supply with 50% efficiency? That will lead to confusion.The text encourages impedance matching in cases where it is definitely inappropriate. Suppose you are designing a circuit with a built-in power supply. If the power consuming part of the circuit has a fixed impedance, you could get the idea from the text that you should then match that impedance in the supply part of the circuit. However that will actually lead to both less efficiency and less maximum power delivered (higher total resistance). The only case where impedance matching to maximize power makes sense is if you have a source with fixed impedance and you absolutely need to draw the maximum amount of power. Frankly, I don't think that's very often. In our world of highly power efficient mobile gadgets and the issue of global warming, how often will you be satisfied with 50% efficiency? To quote a few other sources dealing with maximum power transfer:The theorem results in maximum power transfer, and not maximum efficiency. If the resistance of the load is made larger than the resistance of the source, then efficiency is higher, since a higher percentage of the source power is transferred to the load, but the magnitude of the load power is lower since the total circuit resistance goes up.en.wikipedia.org/wiki/Maximum_power_transfer_theorem 2015-09-17 07:01 UTCThe Maximum Power Transfer Theorem is not: Maximum power transfer does not coincide with maximum efficiency. Application of The Maximum Power Transfer theorem to AC power distribution will not result in maximum or even high efficiency. The goal of high efficiency is more important for AC power distribution, which dictates a relatively low generator impedance compared to load impedance.www.allaboutcircuits.com/textbook/direct-current/chpt-10/maximum-power-transfer-theorem/ 2015-09-17 07:04 UTCAll About Circuits is an open electronics textbook that is continually peer reviewed by its readers and contributors. I sincerely hope that this will make you consider revising the text in chapter 4.6.Thanks for an otherwise great and very understandable textbook!Regards,Marcel Share this post Link to post Share on other sites

Commanderfranz 30 Posted September 17, 2015 Hi Marcel,The example in the text book is a closed system, therefore to apply efficiency is misguided as efficiency is always the same in a closed system, because Pout = Pin in a closed system by definition.Even if I applied efficiency to this circuit it would go as follows. The conclusion in the textbook states that R_{L} = R_{TH} for maximum power transfer.Since R_{L }and R_{TH} are in series their currents will be the same, that is I_{L} = I_{TH.}Efficiency is defined as P_{out}/P_{in.}P_{out} = R_{L}(I_{L})^{2} since the load is our output.P_{in} = R_{TH}(I_{TH})^{2} since the source is the input.Thus efficiency = (R_{L}(I_{L})^{2})/(R_{TH}(I_{TH})^{2} ) which is equal to 1, or 100% since the voltages and currents are equal.This is not 50% as you stated, so is not equal to the power transfer. So it is not implying that maximum efficiency is equal to maximum power transfer.You are correct in stating that maximum power transfer does not imply maximum efficiency, however, you are incorrect in stating that the textbook implies that it does. Above I proved that the derivation does not imply maximum power transfer equals maximum efficiency. I think what is bothering you is this statement in the beginning of the chapter, which is the only part that brings up efficiency. For example, in an audio system it is important that the amplifier transfer as much power as possible to the loudspeakers. Otherwise, the amplifier generates power which is not used for any productive purpose 1 and the efficiency of the overall system suffers.This statement is true, it is saying the maximum power transfer will increase efficiency, however, it does not state that they are equal in any way. Power transfer is a factor in efficiency of this system, but there could be many other factors as well. I hope this clears up your confusion. Kaitlyn Share this post Link to post Share on other sites

Marcel 0 Posted September 18, 2015 Hi Kaitlyn,Thanks for your reply. In my previous posts I didn't write out the math explicitly, because I thought it wouldn't be necessary. I will now explain it more thoroughly and also comment on all the sentences in chapter 4.6 that I find to be problematic or just plain incorrect.I disagree with you in that it shouldn't make sense to talk about the efficiency of a closed system. I'm not trying to violate the laws of physics here, rather it's the definition of efficiency that's the issue. You say that efficiency = P_{out} / P_{in}. Although some other sources claim the same, it's not entirely correct. If it were, all closed systems would be 100% efficient, and the only time interconnected systems would have less / more efficiency would be if energy was stored in the system or retrieved from the system. Electrical energy converted to heat in one system wouldn't change its efficiency(even if the system had an entirely different purpose), since the heat energy would simply be part of the power output. That leads us to the correct definition of efficiency, here from two different sources:Efficiencytechnicalthe ratio of the useful work performed by a machine or in a process to the total energy expended or heat taken in.define:efficiency on google.com (2015-09-18 07:14 UTC) based on Oxford American College Dictionaryhttps://en.wikipedia.org/wiki/Electrical_efficiency (2015-09-18 07:17 UTC)The keyword here is useful, which means that the power output used for the calculation is really just a fraction of the actual power output. For most electrical systems, except electrical resistance heaters, the not useful part of the power output is heat dissipation. But really the term is subjective. If you suddenly decide that your stereo is a heater, it's efficiency will change. Now let's go back to the thevenin equivalent circuit connected to a load:The power input to the circuit is the power produced by the voltage source. The useful power output, is the power consumed by the load. The not useful power output is the power consumed by R_{th}. Of course that is my definition(and the one that appears to me to be the commonly accepted one when talking maximum power transfer), but I think it makes sense that we want power to be consumed in the load, not in "that other circuit" that V_{oc} and R_{th} represent.We then get the efficiency:η = P_{out} / P_{in} = R_{L} * I^{2} / (V_{oc} * I) = R_{L} * I / V_{oc} = R_{L} * (V_{oc} / (R_{th} + R_{L})) / V_{oc} = R_{L} / (R_{th} + R_{L})With P_{out} being the power transferred to the load, not the total power output.In the case of maximum power transfer we have R_{th} = R_{L} so the efficiency becomes:η = R_{L} / (R_{th} + R_{L}) = R_{L} / (R_{L} + R_{L}) = 1 / 2 = 0.5 = 50%This is consistent with what is stated in chapter 4.6:The power delivered to the load is maximized if the load resistance is equal to the Thevenin resistance of the circuit supplying the power. When this condition is met, the circuit and the load are said to be matched. When the load and the circuit are matched, 50% of the power generated in the circuit can be delivered to the load – under any other circumstances, a smaller percentage of the generated power will be provided to the load.http://www.digilentinc.com/Classroom/RealAnalog/circ1/text/RealAnalog-Circuits1-Chapter4.pdf (page 40, mid) (2015-09-18 09:40 UTC)The text in bold is, however, where the textbook is wrong. If we set R_{L} = 3 R_{th} we get an efficiency of:η = R_{L} / (R_{th} + R_{L}) = 3 R_{th} / (R_{th} + 3 R_{th}) = 3 / 4 = 0.75 = 75%Which is the same as saying that 75% of the power generated in the circuit will be delivered to the load. So the claim that under any other circumstances, a smaller percentage of the generated power will be provided to the load is wrong. That is what I meant, when I wrote that the text says that maximum power transfer implies maximum efficiency, which is wrong. In fact, as R_{L} approaches infinity, the percentage of power delivered the load approaches 100%. My conclusions are consistent with every other article I have read on the web that deal with maximum power transfer and efficiency. The other errors I found in chapter 4.6 are as follows:The same amount of power is converted to heat within the power supply; this is the reason many power supplies contain a fan to actively disperse this heat to the atmosphere. If the circuit’s input resistance is not equal to the source resistance, less power is transmitted to the circuit and a correspondingly greater amount is dissipated within the power supply.http://www.digilentinc.com/Classroom/RealAnalog/circ1/text/RealAnalog-Circuits1-Chapter4.pdf (page 40, bottom) (2015-09-18 09:40 UTC)The text in bold is only correct when R_{s} > R_{L}, when R_{s} < R_{L} less power is dissipated within the power supply. Again, a reduced percentage of the power generated by the source will be delivered to the circuit when the circuit and source are not well matched.http://www.digilentinc.com/Classroom/RealAnalog/circ1/text/RealAnalog-Circuits1-Chapter4.pdf (page 41, mid) (2015-09-18 09:40 UTC)This is again only correct when R_{s} > R_{L}, when R_{s} < R_{L} an increased percentage of generated power will be delivered to the circuit. Often, it may not be feasible to match the load with the power supply. For example, when we are testing circuits in our lab assignments we do not generally attempt to maximize the power delivered to the circuit – this is typical when prototype circuits are being tested. One simply recognizes that excessive power is being dissipated within the power supply and that the overall system is not functioning efficiently. If, however, the power supply and associated circuit are being designed as part of an integrated overall system one will generally attempt to match the power supply to the rest of the system.http://www.digilentinc.com/Classroom/RealAnalog/circ1/text/RealAnalog-Circuits1-Chapter4.pdf (page 41, bottom) (2015-09-18 09:40 UTC)The first text in bold is wrong most of the time for the same reason as the previous quotes. A lab supply will generally have a very low impedance compared to the load.The second bold text encourages bad, power inefficient design. One should generally strive for a power supply with as low impedance as possible, in order to achieve high energy efficiency.What the author doesn't seem to understand, is that the maximum power transfer theorem really isn't that useful in most modern circuit design. How often do you want to draw the maximum amount of power from a source, while wasting half the energy? Solar cells is the only example I can think of where this might make sense. But I believe they produce less power when hot, so it's probably more complex than just applying the theorem. I'm sure there are other applications, but it's definitely not the universal good design practice that it is presented as in the text. I'm looking forward to seeing you correct these mistakes in the book.Regards,Marcel Share this post Link to post Share on other sites