Marcel

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  1. Error in Real Analog - Circuits 1, 4.6

    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 = Pout / Pin. 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: 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 Rth. 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 Voc and Rth represent. We then get the efficiency: η = Pout / Pin = RL * I2 / (Voc * I) = RL * I / Voc = RL * (Voc / (Rth + RL)) / Voc = RL / (Rth + RL) With Pout being the power transferred to the load, not the total power output. In the case of maximum power transfer we have Rth = RL so the efficiency becomes: η = RL / (Rth + RL) = RL / (RL + RL) = 1 / 2 = 0.5 = 50% This is consistent with what is stated in chapter 4.6: The text in bold is, however, where the textbook is wrong. If we set RL = 3 Rth we get an efficiency of: η = RL / (Rth + RL) = 3 Rth / (Rth + 3 Rth) = 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 RL 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 text in bold is only correct when Rs > RL, when Rs < RL less power is dissipated within the power supply. This is again only correct when Rs > RL, when Rs < RL an increased percentage of generated power will be delivered to the circuit. 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
  2. Error in Real Analog - Circuits 1, 4.6

    Hi JColvin Thanks 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 reasons With efficiency defined as maximum power delivered(Rload = Rth) 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: All 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
  3. 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: All of these claims only hold for Rload < Rsource, whereas Rload > Rsource 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