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Voltage Question — Parallax Forums

Voltage Question

kenwtnkenwtn Posts: 250
edited 2008-02-08 15:58 in BASIC Stamp
I have a homework board which has a 5 volt regulator build onto it. My question is I m going to be using a 7.5 battery pack from RC car as power supply. I understand that these packs are designed to unload very quickly. So my quest is should I do something to protect the board before I connect the battery pack?

Comments

  • Mike GreenMike Green Posts: 23,101
    edited 2008-02-08 15:47
    These RC battery packs are indeed designed to provide a lot of current when it's demanded, like from a motor working hard. The homework board itself draws relatively little current. The 5V regulator is designed to limit the amount of current it will draw. It will turn itself off if it becomes too hot or if the amount of current drawn through it is too high.

    The usual thing that is done to protect devices from excess current is to use a fuse. You can get "in-line" fuse holders at Radio Shack and auto parts stores designed for use in an automobile. You could put one of these in the positive (usually red) wire from the battery pack and use a 1/2 Ampere fuse.
  • kenwtnkenwtn Posts: 250
    edited 2008-02-08 15:47
    I guess my concern heat problem on the regulator. Maybe I should build a off board voltage regulator and heat sink the regulator?
  • Chris SavageChris Savage Parallax Engineering Posts: 14,406
    edited 2008-02-08 15:48
    Ken,

    There’s really no need to do anything specific to connect your battery. It cannot expend power any faster than dictated by current draw, which, on that board will be negligible unless you have a short somewhere.

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    Chris Savage
    Parallax Tech Support
  • kenwtnkenwtn Posts: 250
    edited 2008-02-08 15:50
    Thanks for all the information, I appreciate the fast replies
  • Mike GreenMike Green Posts: 23,101
    edited 2008-02-08 15:56
    Parallax builds their products to work as described. The Homework Board has a regulator with a 1/2 Amp current limit and they specify the board for use with power input up to 9V. The regulator has to dissipate (9V - 5V) x 1/2A or 2 Watts in the worst case. I think you'll find that the regulator will get warm in those conditions, but be well within the temperature limits of the part. In your case, you'll be using a 7.5V supply, so the power dissipated under worst case conditions will be about 1/2 that [noparse][[/noparse](7.5V - 5V) x 1/2A = 1.25W].

    If you plan to control something that requires more than 1/2A at 5V, you will need some kind of add-on regulator. Usually things that need that kind of power (like motors) work off an unregulated supply anyway.
  • kenwtnkenwtn Posts: 250
    edited 2008-02-08 15:58
    Mike and Chris you have put my mind at ease and thanks for all your help.
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