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Help with finding the right resistor

  1. Jul 27, 2015 #1
    so i need to reduce a dc 12v to dc 5v. do you know what type of resistor i should use? also could you please suggest an online calculator or an equation so that i can find the resistance i need in the future.
     
  2. jcsd
  3. Jul 27, 2015 #2

    billy_joule

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    https://en.wikipedia.org/wiki/Voltage_divider

    Note that:

    A voltage regulator like the LM7805 might be a better solution depending on your application.
     
  4. Jul 27, 2015 #3

    davenn

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    using a resistor is far from a good way to do it. This is because a resistor or resistive voltage divider as billy_joule suggested will also limit the current and that may be a bad thing for what you are trying to achieve

    billy_joule's second suggestion is the best .... use a voltage regulator and ensure it can handle the current your project needs
    And yes there are ways to boost current requirements via a regulator

    for current requirements up to ~ 0.75A ( 750mA) a 7805 reg is ideal
    for higher currents and much more efficient operation use a switching DC-DC buck converter
    eg ....
    http://www.ebay.com.au/itm/DC-DC-Bu...13?pt=LH_DefaultDomain_15&hash=item27f68c89dd

    it can take a wide input voltage range and has an adjustable output and is capable of 3 Amps

    cheers
    Dave
     
  5. Jul 28, 2015 #4

    meBigGuy

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    I like both previous posts and third the suggestion that you drop the voltage with a voltage regulator.

    You need to compute the power dropped in the regulator and make sure it can dissipate the heat (that would be true for a resistor also).

    The power is 12-5 = 7V times load current. If 100ma. then that is 0.7 watts. Thermal resistance junction to air (to-220 package) is 65C per watt. so the junction will be 65 * 0.7 = 45C above ambient without a heat sink. Max junction temp is 125C. How hot will the environment be?

    Another way to look at it. Say a 50C ambient, that leaves 125C - 50C = 75C. 75C / 65C/W = 1.2 watts 1.2 watts/7V = ~160ma. Above that you need a heat sink.
     
  6. Jul 28, 2015 #5
    thanks for the tip i think i will use the lm7805
     
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