Which potentiometer to use?

Discussion in 'The Projects Forum' started by dmc0162, Dec 18, 2007.

  1. dmc0162

    Thread Starter Active Member

    Dec 13, 2007
    51
    1
    I am new to this stuff so bare with me...

    I built a voltage regulator with a 7805 regulator and 2 10uF (35v) capacitors.

    The regulator will have an input of 12.6 volts from a car battery. I tested the regulator with my multi-meter, and with an input of 12.6v, the output was 5.2 so that was good.

    Now, I need to know what potentiometer I can use to be able to vary the voltage from 1.0-3.0 volts. The highest I will need it to go is 3 volts, so is there a resistor I can use to first get down to 3.0 volts, and then what potentiometer can I use to vary it down to 1.0 at lowest? This is all through a car battery as the 12v power source, and a 7805 voltage regulator, so I am not sure if you know the resistance or not. I was going to include the resistance in this post, but I am so new to this stuff that I don't actually know how to find the resistance of it.:(


    If anyone could please help that would be great. If you need the resistance to answer, I will try to find it with my multi meter if someone explains how to check it. Thanks

    -Joe
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    We need to know how much current you need to supply at 1-3V. What are you actually trying to do?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The LM78xx series are intended primarily as fixed-voltage regulators.

    While you CAN use external resistors to INCREASE the output voltage, DECREASING the output voltage becomes complicated; in this case you would need to connect the reference pin to a negative voltage.

    For a more simple solution, look at the LM317 (up to 1.5A) or the LM350 (up to 3A). These regulators were designed to have adjustable outputs from around 1.2V to in excess of what your 12v battery can supply.

    A problem you're going to run into is heat. If you're drawing very much current at those low voltages, the remaining power will be spent in the voltage regulator by generating heat. Without proper heat-sinking, the regulator will rapidly overheat and turn itself off. In the meantime, you'll have a room heater.

    There are schematics in the datasheets for these devices for building switching-type power supplies, which will be far more efficient (and generate much less waste heat) than using a simple linear regulator IC. Yes, it will be more complicated, but your battery will last a great deal longer on a charge, and you will spend less on air conditioning. ;)

    For Applications Notes and datasheets, one source is National Semiconductors' website:
    http://www.national.com
    Look up the LM117 (that's a higher-temp spec than the LM317) - in the application notes (AN-181), there's a 3A switching power supply schematic shown on page 4 that's adjustable down to 1.8V.
     
  4. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    As SgtWookie suggested, you should use a LM317 regulator. Probably you have to redesign the whole thing, but it is better than using a pot.

    With a pot you have two not so good choices when driving significant loads. Or you use a high resistance pot and the load will drive your pot/voltage divider, or you use a low resistance divider and waste a lot of energy. The voltage divider should be biased by a current 20 times greater than the one biasing the load.

    Just to remember that the LM317 cannot give voltages lower than 1.25V, but I think you can solve this with a diode in series with the load (make the discount of 0.7V). I don't agree with diodes in series to cause drop a voltage drop, but I can't see other solution for now.
     
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