Some Questions about Potentiometers and Resistors

Discussion in 'General Electronics Chat' started by TehYoyo, Apr 20, 2013.

  1. TehYoyo

    Thread Starter New Member

    Apr 20, 2013
    11
    0
    I have a few questions for you guys.

    1. What does the watt rating mean on a resistor?

    2. If I (according to a calculator) need a 330 ohm, 1/4 watt resistor, is that the same as a 165 ohm (330/2), 1/2 watt resistor?

    3. How do I know how what potentiometer I need? For instance, I'm going to use potentiometers to control an RGB LED strip (link). It's 2 amps for the whole strip at 12VDC. However, I'm planning to cut it down to 1/30 of its original size. What resistor do I get then and how do I calculate that?


    Thanks,
    TehYoyo
     
  2. bob800

    Member

    Dec 10, 2011
    48
    3
    1. Watt ratings refer to the amount of watts (volts * amps) the potentiometer can handle before overheating.

    2. No. You may substitue a pot with a higher wattage rating, but the resistance should match. The watt rating has no effect on resistance.

    3. You could stick a resistor or pot in parralel, but you would end up wasting energy as heat and you would need a large wattage rating. The best solution would be to use a transistor, and then divide the transistor´s base voltage with the potentiometer (outer ends go to GND and VCC, center goes to transistor). Like this:

    [​IMG]
     
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  3. TehYoyo

    Thread Starter New Member

    Apr 20, 2013
    11
    0
    1 and 2 I understand. Thanks.

    I understand the schematic that you posted, but where would I hook up the lights? On the bottom side of the transistor (referring to the bottom of the schematic)?

    Also, I was looking at Mouser's collection of potentiometers. The max they have are 8-watt potentiometers. Not for this project, but for another, I want to control 55-watt lights... can I not do that? Is there another source of high-watt potentiometers?

    Thanks,
    TehYoyo
     
  4. bob800

    Member

    Dec 10, 2011
    48
    3
    The positive "end" of the lights would attach to Vout, with the negative attached to ground. For this, the resistance of the potentiometer doesn't really matter, as it simply divides the voltage... A 1K or 10K should do the trick at 1/4 W or even less.

    They do not carry potentiometers above 8W because potentiometers are not meant to be configured in the manner you have in mind. They are supposed to control a transistor, which in turn varies the voltage or current to the lights. For a 55W light you might want to consider using a MOSFET instead (they can generally handle higher loads).
     
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  5. TehYoyo

    Thread Starter New Member

    Apr 20, 2013
    11
    0
    Thanks bob800, and to all the people coming here from my "Which way is a switch oriented" thread. I looked at this article which explains in more depth what's going on in bob800's schematic. It uses pretty much the same schematic:

    [​IMG]

    A few questions:
    1.) Could I put a switch in series with the potentiometer, b/w the pot and the +voltage?

    2.) How do I know whether or not the pot is safe (the transistor/MOSFET is changing the voltage/load/canyouexplainwhatwordtousehere enough so that the switch/pot doesn't get overloaded)?


    Thanks,
    TehYoyo
     
  6. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    1 yes

    2 make sure no rating if pot is exceeded (current, voltage, power)
     
  7. TehYoyo

    Thread Starter New Member

    Apr 20, 2013
    11
    0
    How do I do those calculations (a link to an article would be great)?
     
  8. Mike33

    AAC Fanatic!

    Feb 4, 2005
    349
    25
    You use Ohm's Law, TehYoyo...I = E/R.
    I=current in amperes, E = voltage in volts, R is resistance in Ohms.

    You may rearrange this equation to suit any unknown quantity. If you know two values, you can find the third. You can do I*R=E, E/R=I, and so on. This is perhaps the MOST important equation needed for all electronics work!! The power equation to find out your component ratings comes after this.

    At a most rudimentary analysis, the maximum amount of current that can flow between your +6v and thru the base of the transistor is:

    E/R = 6V/1000 ohms = .006 amperes (or 6 milliamps).

    You next determine the power rating needed here. That is I^2*R, or I*E (voltage * current = Power, in Watts).

    6V * .006 amperes = .036W.

    That is VERY little power! As to be expected at a transistor base. The small 'standard' resistors we mostly use are rated at 1/4W. As you can see, the power flowing here is far below even that.

    Most pots are rated for 1/4W, so you are safe to use about any one in that position. Most pots are used in low-power areas like this, rather than getting the big high-watt rated ones, because they are EXPENSIVE when you go above this power rating.

    Now, the ACTUAL current drawn by the transistor base may be much lower, mind you, and will be determined by the rest of the circuit....working the problem this way means you know the MAXIMUM that could EVER flow through the base, and so you are safe.

    Play around with this way of finding answers to how much current there is in a circuit, what the power being dissipated is - it is VERY important!! :)
     
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