Inefficent?
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Inefficent?
I was thinking, wouldnt it make more sense to just take 4 double a batteries, run em in series, use a 4805 regulator and 2 capacitors, and make it more efficent, because its easier to take 6 volts down to 5 instead of taking 3 and turning it into 5? Sure, itd take up the extra room of the extra two double a batteries, but the pcb would be way smaller (all you need space for is a 4805, and maybe 1 or 2 capacitors). Plus, wouldnt it cost less? j/w
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Isn't it a 7805? Anyway, I made a homemade MintyBoost for <$5 including etching the PCB. The only expensive piece can be sampled, as for efficiency, I'm pretty sure this is more efficient, I could be wrong, but I'm not really the person to tell you. There's a whole section with math, but I don't understand it...
You can see for yourself here:
http://www.ladyada.net/make/mintyboost/process.html
Good luck making sense of those numbers they just make my head hurt
You can see for yourself here:
http://www.ladyada.net/make/mintyboost/process.html
Good luck making sense of those numbers they just make my head hurt
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- Joined: Tue Jan 09, 2007 9:58 pm
Actually, by sampled, he means going to (in this case) Maxim Semiconductor's website, and actually sampling the chip. The process is similar to purchasing the chip, but it is free. The catch of course is you can only do it once every few months, and only a handful of parts at the time.
As for efficiency, the Mintyboost should be more efficient. The mintyboost is effectively converting amperage into voltage, where the 7805 converts the excess voltage to heat, and keeps current the same. Correct me someone if I'm wrong, but if 1A at 3V goes into the Mintyboost, around 500mA and 5V comes out, going from 3W of power to 2.5W, only a half a watt lost (using Ladyada's 85% efficiency calculations). Now, if you input 500mA at 7.5V (the 7805 has a minimum input voltage of 7V, the extra batteries idea would require a 5th AA, for 7.5V), the output would be about 500mA at 5V. This represents an input of 3.75W, and an output of 2.5W, a loss of 1.25W. The second equation does not calculate in the efficiency of the 7805 either.
So, 0.5W lost with the Mintyboost vs. 1.25W lost using the 7805, over twice as efficient. Granted, you will get more charge time out of the 5 battery solution, because you have more power stored in the 5 batteries than in just 2, but it is not more efficient.
As for efficiency, the Mintyboost should be more efficient. The mintyboost is effectively converting amperage into voltage, where the 7805 converts the excess voltage to heat, and keeps current the same. Correct me someone if I'm wrong, but if 1A at 3V goes into the Mintyboost, around 500mA and 5V comes out, going from 3W of power to 2.5W, only a half a watt lost (using Ladyada's 85% efficiency calculations). Now, if you input 500mA at 7.5V (the 7805 has a minimum input voltage of 7V, the extra batteries idea would require a 5th AA, for 7.5V), the output would be about 500mA at 5V. This represents an input of 3.75W, and an output of 2.5W, a loss of 1.25W. The second equation does not calculate in the efficiency of the 7805 either.
So, 0.5W lost with the Mintyboost vs. 1.25W lost using the 7805, over twice as efficient. Granted, you will get more charge time out of the 5 battery solution, because you have more power stored in the 5 batteries than in just 2, but it is not more efficient.
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Please be positive and constructive with your questions and comments.