On Thu, Jun 26, 2008 at 08:42:51AM -0400, Tom?s ? h?ilidhe wrote: > > (This post is kind of therapeutic for me, I'm trying to understand > things as I go along; it helps to feel like I've an audience when I'm > writing something) Have at it. I'll add my comments. > > Having studied basic Physics, one of the first things I learned about > was the conservation of energy: > "Energy can neither be created nor destroyed, but rather only > converted from one form to another" In addition you'll up the entropy during the conversion. Some of that energy will get converted to unusable heat. > The main gem of information I take from this quotation is basically that > you can't just pull energy out of thin air, it has to come from > somewhere. "There's no free lunches" as they say. That's exactly right. As Issac Asimov stated in his famous essay on energy and thermodynamics: "Thermodynamics: you can't win, you can't even break even, and you can't get out of the game." > So let's say you have an AA battery, which is 1.5 V. You say to me that > you're going to step the voltage up to 5 V. Well my first reaction to > this is that a 5 V battery is more "powerful", so where the hell is that > extra energy going to come from? Oops! Power has two components: voltage and current. You figured this out below. > Anyway here's what went on in my head when I was trying to get a grip on it: > * Power = Energy per second > * Power is measured in watts, which is joules per second > * Power = Voltage x Current = volts x amperes > * 1 ampere = 1 coulomb per second > * Therefore Power = volts * coulombs / seconds > * Therefore Energy = volts * coulombs All correct. > Now if I'm right, the energy contained in a battery is equal to the > amount of charge in it (i.e. coloumbs) multiplied by the voltage (i.e. > volts). Right. But usually the total energy is measured in Watt-hours, right? > So let's say I have an AA battery rated as follows: > Voltage = 1.5 V > Capacity = 500 mA hr = 1800 coulombs > > The energy contained in this battery would then be 2700 joules. Or 750 mW hr. > So... if I follow the whole "energy conservation principle", then that > means that if this battery were to somehow become a 5 V battery, then > it's capacity would have to drop to 540 coulombs, which is 150 mA hr. Yes and no. Remember that you're going to lose energy in the conversion because no conversion is 100% efficient. Most likely you'll be able to recover 90-95% of the energy losing 5-10% of it to heat. So instead of 750 mW hrs you'll get back between 675-712 mWhr. > Let's say you give me a 200 ohm resistor and a 1.5 volt battery. You say > to me "Put 25 mA through that resistor". My first reaction would be "I > can't, the battery isn't powerful enough". Better would be that the battery does not have sufficient voltage to generate that much current across that resistance. It's not the fact that it doesn't have the power. It's the fact that the voltage part of that power equation is insufficient. The most current you can get across that resistance at 1.5V is 1.5V/200ohms = 0.0075A or 7.5 mA. > How much truth is there in saying "The battery isn't powerful enough"? > Is it possible to take that 1.5 volts, step it up to 5 volts, and make > it put a steady 25 mA through the resistor? Yes. But at a cost of course. You're using the energy in the battery at a rate 3.3 times faster than if you pulled it out at 1.5V. So the battery will be exhausted faster. In addition you'll lose out on some energy to do the actual conversion too. > If so, then I imagine it would have to work something like as follows: > * The current through the resistor would be 25 mA > * This means the output current from the voltage regulator is 25 mA. > * If we pretend that the regulator is 100% efficient, then "power > in" has to be equal to "power out". Though we know that it's not. > * So if "power out" = 25 mA x 5 V = 125 mW, then the "power in" has > to be 125 mW also. > * "power in" = 125 mW = current x 1.5 V > * Therefore, current coming from the battery is 83 mA Correct. Which means that this works so long as the internal resistance of the battery is less than 1.5V/83mA = 18 ohms. Now if the internal resistance of the battery is more than that (like 9V batteries have high internal resistances) then you won't be able to get the 25mA out @ 5V like you want. > * I haven't a clue how efficient step-up regulators are, but I'll > take a wild guess of 95 %, so that would bring the battery current to > about 87 mA. That's in the ballpark. > So is that how it all works? If you want to use a single AA battery to > put 25 mA through a 200 ohm resistor then: > * 87 mA has to come from the AA battery > * The regulator steps up to 5 V and gives out 25 mA > > That right? Bingo. And since the capacity of the battery is 500 mAHr, this system will run for 500mAHr/87 mA = 5.7 hours. > I just have one other unrelated question, but it's to do with batteries. > Let's say I have two AA batteries, each of which is 500 mA hr. If I put > them in series to form a 3 V battery, then what's the capacity of this 3 > V battery? Is it 500 mA hr at 3 V? (The energy conservation principle is > telling me that yes it is 500 mA hr at 3 V, but I just want to be sure) Yes. Now the one thing you may miss in all of this is an effect called the Peukert effect. For batteries it simply states the faster that you draw energy out of the battery, the less total energy you can extract. So that 500 mA hr rating would be for a specific current draw for a specific number of hours, like a 20 hour rating for example. At 20 hours you can extract energy at a 25 mA rate. However if you extract it faster, then the amount of energy available goes down. For example at the 87 mA rate the capacity may only be 304 mA Hrs giving only 3.5 hours of operation at your example. Hope this helps, BAJ -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist