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Resistor Trouble

Discussion in 'General Electronics Discussion' started by Christian, May 30, 2016.

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  1. Christian


    May 30, 2016
    Hello, I'm very new to circuitry and the forum and I am wondering if I could get a few answers about how a resitor works because my current build seems to be working poorly...

    I have a battery pack with an output of 10.5V and 1400ma, the goal is to add a resistor to drop the Voltage for 10.5V to 6V.

    Ive done the math and I should need a resistors of about 4.3Ohms. However when I try and hook up the resistor (10 Ohms) I see little voltage change, if any at all.

    What am I dong wrong?
  2. Gryd3


    Jun 25, 2014
    You are using a resistor to drop voltage for an unknown and/or dynamic load.

    You cannot use a resistor this way.
    You can use a resistor to alter the voltage level for 'signals'.
    You can use a zener diode and resistor to alter voltage level for very low power devices.
    You cannot use either method above to alter voltage level for higher load devices *unless* they are simple... perhaps an LED or light-bulb.

    You should be using a 'voltage regulator' .
    There is 'linear' and 'switch-mode' types.
    Linear is cheap, easy to use, but inefficient.
    Switch-Mode is a little more expensive, easy to use (but not always build), and much more efficient.

    So. Look online and buy a pre-made switch mode regulator, or get a linear regulator like an LM7806
    Last edited by a moderator: May 30, 2016
    (*steve*) and davenn like this.
  3. davenn

    davenn Moderator

    Sep 5, 2009
    I have hilited the VERY important parts in Gryd3's post..... the bits you really need to note well :)

    Gryd3 likes this.
  4. BobK


    Jan 5, 2010
    You are confusing battery capacity mAH with current mA. They are not the same thing.

    The battery capacity (mAH) tells you how many mA you can draw from the battery for 1 Hour before it is exhausted.

    Ohm's law tells you how much voltage is dropped when a given amount of current flows through the resistor. In your example, if you draw 1400 mA through a 4.3Ω resistor the voltage drop will be 1.4 * 4.3 or about 6V.

    But when you just place the resistor in series with the battery, and you are powering nothing (drawing no current) the current is zero, so the voltage drop is zero.

  5. davenn

    davenn Moderator

    Sep 5, 2009

    all true, but it doesn't deal with the issue of producing a stable 6V supply from the 10.x V battery
    that's where Gryd3's post was spot on ;)

  6. BobK


    Jan 5, 2010
    Which is why I didn't bother to answer that.

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