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Power Supply

  1. Nov 5, 2008 #1
    I have a 24 V lead acid battery that I am going to use for a power supply. I need to get 5 volts so that I can power an AVR some other stuff. I was originally intending to use a 7805 to get 5 volts, but I am concerned that this may be too much power dissipated.

    I think I need about 200-300 mA and I am I not mistaken that is 19V * 300mA = 5.7 W which is awfully high. However, we do essentially have a very large heatsink that we can use because this will be attached to a large metal frame, but I think that may still be too much...

    It was suggested to me to use a switch mode power supply, but unfortunately, I think be both too costly as well as taking up too much space.
    Would a power resistor and a 1 W zener diode be a better choice than the 7805?

    Does anyone have any suggestions for what I can use to get a steady 5 volts? ( I cannot compromise on the 24 V source)
     
    Last edited: Nov 5, 2008
  2. jcsd
  3. Nov 5, 2008 #2

    MATLABdude

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    These guys have some "all-in-one" 7805 drop-in replacements:
    http://www.hvwtech.com/products_list.asp?CatID=163&SubCatID=276

    They're pricey, but they would probably work if you didn't want to design your own switcher. You might also want to find the company that actually makes those instead of ordering from Canada. Most of the switch-mode ICs are quite well-documented with reference circuits, pertinent calculations and even part recommendations (take a look at Digikey for some buck voltage conversion ICs by Intersil or TI).

    You *could* use the power resistor (properly spec'd and heat sinked) but you won't have great load regulation. But then again, that might be enough.
     
  4. Nov 5, 2008 #3

    mgb_phys

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  5. Nov 6, 2008 #4

    Redbelly98

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    MATLABdude and mgb_phys made some good suggestions. I'll just mention another option. Use a 7805 with a 47Ω, 10W resistor between the 24V supply and 7805 input. At 300 mA this will drop the voltage to about 10V at the regulator. Power dissipated will be 5W in the resistor and 1.5W in the regulator.
     
  6. Nov 9, 2008 #5
    Buck converter?
     
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