Maximizing Power at load

Discussion in 'General Electronics Chat' started by Austin Clark, Jun 3, 2012.

  1. Austin Clark

    Thread Starter Member

    Dec 28, 2011
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    So, I had a really simple question, that resulted in a really complicated problem, that resolved into a really stupid solution. That's just the way I roll.

    I wanted to know what the load resistance should be to achieve the highest power consumption at said load given a source voltage and its ESR (Equivalent series resistance). Simple, right? Well, after some number-herding I found that the power at the load is equal to (V^2*ESR)/(ESR+Load)^2.
    To find where it's maximum lies, we want to find points in which it's slope is zero, in other words, find the point where the differential equals 0. So, I cheated and used an online calculator to find the differential, which so happened to be THIS sucker:
    -(load-ESR)*V^2/(load^3+3*ESR*load^2+3*ESR^2*load+ESR^3)
    I then set it equal to 0, and solved for the load. (That is, what does the load have to be to set this entire thing to 0, which, again, is where the maximum point lies in my power equation). Turns out it's load = ESR
    That's it. load equals ES-effing-R.
    Now that I see that, I realize it wasn't that freaking hard to come by, it just didn't occur to me earlier intuitively. I was graphing crap, flipping papers back and forth, scribbling and circling equations and garbage for like half an hour. It was still really fun, and it's awesome to know that, even though I complicated the HELL out of this, the solution still came out ok, so at least I did the complicated crap correctly... :p

    To solve a bit more intuitively, when load = ESR, 1/2 of the total power consumption is in the load (not very efficient, but regardless). Also, when load = ESR, the total power consumption is 1/2 what it could have been if the load resistance was zero. So, the load power is equal to the maximum power output divided by 4, or Maximum-Load-Power = V^2/(ESR*4). What does the load have to be to get that? Well, the voltage at the load is Vs*load/(load+ESR) (like a voltage divider), so:


    (Vs*load/(load+ESR))^2/load = Vs^2/(ESR*4) solve for load, and load = ESR :)


    I just thought this was very interesting. Some other ideas might be "where is the most efficient operating point?, that is, power-Load/power-Loss" or something similar.
    Are there any simpler solutions to this problem that anyone can find? the simpler, the more likely we'll "see" or "feel" its solution, which leads to it's mastery.
     
  2. Brownout

    Well-Known Member

    Jan 10, 2012
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    Nope. You've derived the correct solution by the correct method. It's not at all intuitive. That's the great think about math; it sometimes surprises us, but is nontheless correct. When I first learned of this, my first question was, then why isn't the load always equal to the source resistance? Turns out that maxumum power transfer is rarely the goal of amplifiers.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Jacobi's law has led some to conclude that, given a fixed load, they should match the source resistance to it in order to get maximum power transfer, which, of course, is wrong. Given a fixed load, the source resistance for maximum power transfer is zero.
     
  4. Austin Clark

    Thread Starter Member

    Dec 28, 2011
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    I'm glad you and I feel the same way :)
    I liked how I approached the problem in two COMPLETELY different ways, and still came to the exact same elegant solution. It's a wonderful feeling.

    Of course, but if the source resistance is unalterable, then you've got to make due. If it COULD be zero, you'd be able to, theoretically, draw an infinite amount of power.


    The problem with calculating the efficiency is that the MOST efficient load is an infinite one. :p So, where do you balance efficiency with power out?
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I can't think of a lot of instances where you want to deliver power to a load, but you get to pick the load resistance. In the days of vacuum tube (valve) amplifiers, a speaker was matched to the amplifier output impedance using a transformer. Nowadays, with power transistors and op amps, the best way to drive a speaker is with (near) zero impedance.
    Can you think of examples where you get to pick the load impedance?
     
  6. Austin Clark

    Thread Starter Member

    Dec 28, 2011
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    yes, heaters. Also, you can often build around specifications. A motor can be built to create the same amount of torque with the same amount of power at different voltages, right?
     
  7. Ron H

    AAC Fanatic!

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    What kind of heater is driven from a source whose impedance is high enough to warrant a maximum power transfer calculation? I'm not challenging you. I am just not very imaginative right now.:p
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Do yourself a really big favor and stop cheating like that. ;)

    Seriously, don't automatically resort to using calculators and spreadsheets and what-not to do your thinking for you.

    Given: A voltage source, V_S, with a source resistance R_S. What value of load resistance, R_L, results in maximum power delivered to the load?

    <br />
P_L = \frac{V_L^2}{R_L}<br />
V_L = V_S\left(\frac{R_L}{R_L+R_S}\right)<br />
P_L = \frac{ \left( V_S \left( \frac{R_L}{R_L+R_S} \right) \right) ^2}{R_L}<br />
P_L = V_S^2 \left( \frac{ R_L }{\left( R_L+R_S \right) ^2} \right)<br />

    P_L is maximized when its derivative with respect to the load resistance is zero.

    <br />
P_L = V_S^2 \left( R_L \right) \left( R_L+R_S \right)^{-2}<br />
\frac{dP_L}{d_RL} = V_S^2 \left[ \left( R_L+R_S \right)^{-2}  -2R_L\left( R_L+R_S \right)^{-3} \right] = 0<br />
\left[ \left( R_L+R_S \right)^{-2}  -2R_L\left( R_L+R_S \right)^{-3} \right] = 0<br />
\left( R_L+R_S \right)^{-2} = 2R_L\left( R_L+R_S \right)^{-3}<br />
\left( R_L+R_S \right) = 2R_L<br />
\left( R_L+R_S \right) = 2R_L<br />
R_S = R_L<br />

    Take every opportunity you can to practice your manual math skills. Notice that, in the above, I didn't invoke the derivative chain rule for quotients. Why? Because it is too complex for my small brain to remember! But the chain rule for products is easy enough that I can always recall it even after years of not using it. So if I have a quotient, I simply bring the denominator upstairs by negating the exponent and keep going.

    Also, while not apparent in the above, I screwed up the actual taking of the derivative but caught it two lines later because the units didn't work out. Tracking backward, the units took me straight to the scene of the crime, making it easy to spot where I had screwed up.

    If you are going to graph crap, be sure to graph crap that is meaningful to the problem. In this case, graph the power in the load resistor as a function of the load resistance. Then spend some time making its behavior make intuitive sense. In this case, you should conclude that for very small load resistances the power is low because, while there is a large current, there is not enough resistance for that current to deliver much power to (since P = I^2*R) while, for very large resistances there isn't enough current to deliver much power to the resistor. An intermediate value of load resistance achieves a balance: small enough resistance to get some current but still keep enough resistance for that current to have something to dump power into.

    This is circular logic. Basically, you are saying, "When the load is equal to Fred, then Sue is true. So, what does the load have to be for Sue to be true? Not surprisingly, it turns out that it has to be Fred."
     
    Austin Clark likes this.
  9. crutschow

    Expert

    Mar 14, 2008
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    Good point. That's often a trick question on an exam after the students have studied the Maximum Power Transfer Theorem.
     
  10. Austin Clark

    Thread Starter Member

    Dec 28, 2011
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    Thanks a lot for your response, you brought up a few very good points.
    I recently graduated high-school and honestly I just haven't taken the time to officially learn calculus, in fact everything I know about it I learned myself by experimenting and playing around with interesting problems, I only understand the concept and I can solve very simple problems. I do, however, almost always at least ATTEMPT to simplify and solve equations algebraically beings that I have almost mastered it. Given my limited instruction in mathematics, I'd say I'm relatively capable :)

    Yes, graphing "crap" was really just a poor choice of words on my part, I do graph relevant functions, and I like how graphically you can find drastically different perspectives on the same problem, and can use vastly different solving techniques to reach the same conclusion. Sometimes I do accidentally graph pretty useless functions, but usually only on a failed hunch, and it's not like it hurts anything :)

    Lastly, I realize now that that really was circular logic at the end, I'm surprised I didn't catch that. I wrote that last solution on-the-fly beings that I only thought of it while at the computer at that time, and just sorta went with it because I didn't see any issues at the time. I'll look at it some more after awhile and see if I can learn anything interesting from it.
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    I don't know if I recall you pointing this out before or not. Given the lack of a calculus background, it is a lot more reasonable to quickly resort to a graphing calculator or symbolic math tool.

    I think you will do great in college (you BETTER be going to college!). Your approach to problems is quite solid with relatively few rought edges given your current level of education. I think you will really enjoy calculus, but be on the lookout for getting saddled with an instructor that teaches the what of the mechanics and not the underlying principles or the why of how the mechanics are derived and how they work. If you can, take a class in analytical geometry before you take calculus.
     
  12. Austin Clark

    Thread Starter Member

    Dec 28, 2011
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    Unfortunately, MOST instructors (as it seems to me) do like you said, teach the methods but not the concepts. It's been both frustrating and enjoyable actually, because I've been forced to learn everything, effectively, from scratch, which is a welcomed challenge. I love the feeling I get when I REALLY get something, and I figured it out on my own. Sometimes I wish I lived in an earlier age, so I could get some credit for discovering some of these things... heh.
    Anyways, as I was saying, I absolutely plan on going to college, but I don't feel as though it will be as effective as self-studying. I'm mainly going for the crummy receipt (otherwise known as a diploma), job connections, and relations with others interested in the same things as I. That's also the reason I joined this forum, people and principles.
     
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