-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 > I have a vested interest in getting lights to be brighter for less > - Unfortunately I > don't have the money to go to HID lights. > > So - is this true or is it a crock !! > > Basically what Illustro does is this. Using a standard bulb, be it > Tungsten, Halogen or Xenon (but not HID), a 24 volt pulse is fed in > rather than a constant12 volts. By switching from 0 to 24 volts up > to 2,500 times a second you get a light that is 2.5 times brighter. > For example a 60-watt standard Halogen bulb will give out the > equivalent light of a 180-watt bulb. And all this for the same > amount or amperage! Well, I can explain the principle on which it rests, but as to whether it is true or not, I cannot tell you for sure. It seems too obvious to be in violation of thermodynamics, but it may be. The principle is not a difficult one. It is simply a situation where the RMS power exceeds the average power. In your example, the average power is: (24*.5)^2/R The RMS power for a 50% duty cycle like you propose is Prms = (RMS voltage)^2/R RMS voltage = sqrt(V^2*Duty) So, V^2*.5/R = 24^2*.5/R Now, the comparison between average power and RMS power for this case is then: (24*.5)^2 : 24^2*.5 .5^2 : .5 .25 : .5 1 : 2 So you double your effective power. By the same principle, for any rectangular wave that goes from 0V to some other voltage, the average power can be compared to the RMS power by a ratio of duty% : 1 or 1 : 1/(duty%) Now, with that principle set, it seems to me that it could not work, because a system could be set up where a resistor generated heat using extremely high voltage spikes, with extremely low duty cycle, ex: 20KV, with a 0.1% duty cycle through a 1 ohm resistor, generating 20000^2/1 * .001 = 400kW effective for (20000*.001)^2 = 400W Even losses due to ineffective thermal conversion units would pale in comparison to a factor of 1000. Thus, while this should work fromt he equations, I don't see how it can, considering that it would violate the second law of thermodynamics. Could someone explain why this doesn't work to me? Because from the equations, I don't see why it wouldn't. My best guess is that the power draw on what ever your power source happens to be is the RMS power. Thus you don't gain anything. Basically, to answer your question, if you put any stock in the second law of thermodynamics, the principle has to be bull, but, darn it, it *should* work! Hope that helps a little... Now I'll be told to re-take freshman physics, I'm sure. Dazed and a little confused, - --Brendan -----BEGIN PGP SIGNATURE----- Version: PGPfreeware 6.5.8 for non-commercial use iQA/AwUBPVFNvgVk8xtQuK+BEQJPkQCcCX7CpuU7/JFR5s7c5Xsh9GK1Sl8An1H1 +yHVnykFMF9DQjmcTF2mUGz4 =wTJM -----END PGP SIGNATURE----- -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.