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very simplistic potentiometer question

Discussion in 'Electronic Basics' started by Barbara, Jul 1, 2005.

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  1. Barbara

    Barbara Guest

    I am just beginning to tinker with electronics. I've decided to start
    very simply and work my way up to the fun stuff. I like the thought
    of playing with LEDs for some reason so I'm starting there. I've
    figured out already that without a resistor that LEDs last about .2
    seconds before they fry. Now that I can keep one going for more than
    5 seconds I'd like to add a pot to control the brightness. I know a
    pot is a variable resistor that basically adds resistance. I had this
    idea that I would simply insert it between my resistor and the
    positive lead on my power source (a 9 volt battery in this case.) I
    bought an assortment pack of pots and was astonished to find they have
    3 leads. This doesn't fit into my notion of how they work so I am
    obviously missing the big picture here. I figure the answer is simple
    sow I've done a lot of searches on Google and yahoo and I can't find
    simple enough circuits to make sense of how they work. I need some
    basic info on how to wire a circuit using one led, a resistor, a pot
    and a battery.

  2. steamer

    steamer Guest

    --FWIW if you get into Basic Stamp ( the first thing
    they show you is how to drive an LED. After that it gets really fun, heh.
  3. Rich Webb

    Rich Webb Guest

    LEDs are (to a first approximation) constant brightness devices.
    Garden-variety LEDs (waves hands, points back at "first approximation")
    will want to drop 1.5 volts with about 15 milli-amps of current.

    The voltage drop is constant and independent of the current flowing
    through the LED (see waving hands, above), so you were "shorting" about
    7.5 volts of the battery across the very small resistance of a forward
    biased diode. That's why the LEDs had a very brief but exciting life.

    Given your 9 volt source, and the 1.5 volts dropped across the LED, you
    need to limit the current driven by the remaining 7.5 volts to about 15
    milli-amps. That works out to (7.5 / 0.015) = 500 ohms.

    There's a standard resistor value at 470 ohms that will be close enough.

    You should also check that the power dissipation capability of the
    resistor is adequate (that's the wattage rating). The resistor is
    dropping 7.5 volts and carrying 0.015 amps so it's dissipating about 120
    milli-watts. You'd want to use a 1/4 watt resistor to allow yourself
    some headroom for things like a fresh battery (that's > 9 V) and other
    device variations. If you use a smaller size 1/8 watt resistor then it
    may get too hot and, eventually, fail.

    LED brightness is typically controlled by pulsing it on and off very
    quickly. If it's only on half the time, the perceived brightness is
    about half of the 100% duty-cycle brightness.

    There are lots of ways to vary the duty cycle. A 555 timer chip is a
    good place to start. It has a wide input voltage range (will work with
    the 9 volt battery) and there are lots of sample circuits available.
    Examples at and

    Regarding the potentiometers, the two leads on the ends are connected to
    a resistive strip that's the nominal value of the pot, say 10 KOhms. The
    middle lead is connected to a wiper that moves along the length of the
    pot and so it has an output voltage that changes along the length of the
    resistive strip. If one of the outer leads is connected to the (-) end
    of the battery and the other to the (+) then you could measure a voltage
    at the middle lead that varied from the (-) to the (+) value, or 0 to 9
    volts relative to the (-) terminal.

    You can use a potentiometer as a variable resistor by connecting
    together the center and one of the outer leads. The resistance between
    the center and the remaining outer lead will vary from nothing (when the
    wiper is turned all the way towards the remaining lead) to the full
    value of the resistive strip when it's at the other extreme.
  4. Andrew Holme

    Andrew Holme Guest

    The outside pins connect to the ends of a resistive track. The middle pin
    is the wiper. You can just use the middle pin and one outside pin to get a
    variable resistor like you imagined. You'll get best results from low-value
    pots (try 1K) and, of course, you *must* retain the fixed resistor in series
    as well to avoid frying the LED when the wiper reaches the end. With this
    rather crude method, you'll find the brightness varies non-linearly along
    the track.
  5. Ban

    Ban Guest

    So you have already calculated the resistor: If it is a white LED, it might
    have a forward voltage of 3.3V at 20mA current. The 9V -3.3V= 5.7V must be
    dissipated. R=Voltage/Current = 5.7V/20mA = 285 ohms. So 270 will also do.
    Now you can take a pot of low value (1k) in put it in series with that
    resistor and you can vary the current from 4.5mA to 21mA. But 0mA can not be
    If we connect the pot in a different way, we can vary the LED-current from
    18mA to 0. (3.3mA run through the pot)
    The best is to use a 1W wire-wound pot for this purpose to not burn the
    wiper or the carbon trace. Also a linear not log characteristic should be
    chosen, they have the best power rating.
    | |
    | .-.
    --- | |
    - | |270
    --- '-'
    - |
    --- 9V |
    - .-.
    --- | |<------.
    - | |1k |
    --- '-' |
    - | V white LED
    --- | -
    - | |
    | | |
    | | |
    (created by AACircuit v1.28 beta 10/06/04
    view\fixed font
  6. Guest

    You need some bqsic info on both LEDs and potentiometers. LEDs are
    junction diodes that have all the normal properties of any other
    diodes. When you apply reverse voltage to them, they conduct only a
    small leakage current, till that voltage gets to some breakdown level,
    and then the current rises very fast. In the forward direction, the
    current rises with an exponential relation to the voltage. Equal
    increments of voltage increase raises the curent by equal multiples.
    every 60 millivolts or so increases the current by a factor of 10.
    That means that once the forward voltage is in the normal working
    range, very small additional voltage will drive way too much current
    through them, and you see a flash and it is all over, as you have

    The series resistor acts as a current stabilizer, using up all the
    extra voltage as the LED current passes through it, also. Resistors
    have a linear relation between current and voltage, so that doubling
    the voltage across them simply doubles the current through them. This
    linear relationship is called Ohm's law.

    Potentiometers are resistors with a connection to each end of a strip
    of resistance material and also a sliding contact that can be moved to
    any spot along that strip. The sliding contact is almost always the
    center of the three terminals.

    If you want to use one as a variable resistor (make connections to a
    variable length of resistance material) you use one of the end
    connections and the sliding contact.

    Most LEDs have a maximum current rating of 20 milliamps (.02 amperes)
    and pass this current when there is a volt or three across them. The
    voltage depends on the color. Red photons have less energy each, so it
    takes electrons falling through about 1.5 volts to produce them. Blue
    photons are about twice as energetic, so electrons have to fall through
    about 3 volts to produce them.

    So, based on the color of your led and the total voltage available, you
    can calculate the minimum resistance needed to keep from overloading
    the LED. Leds say you have a red LED and a 9 volt battery. This means
    that you have about 9-1.5=7.5 volts to burn up across the resistor.
    7.5volts/.02amperes=350ohms. Ohms is just a label that means volts per
    ampere. So if you put something close to that in series with the LED,
    you can expect it to shine with about rated light output and still last
    a long time.

    Then you can connect a pot in series with that pair (using the center
    and one end connection) to allow you to turn the current down lower
    than the rated current and see how the light varies. I recommend you
    get a $20 digital multi-meter at Walmart and make a table of LED
    current versus LED voltage to get familiar with the exponential diode
    current voltage relationship I spoke of earlier. You can keep the
    meter set on voltage and use the voltage drop across the fixed resistor
    (dividing the voltage by the resistance to calculate the curent) to
    make this table for various pot settings.
  7. Byron A Jeff

    Byron A Jeff Guest

    Define fun stuff?
    Not a bad start.
    But have you figured out why? I'll take a minute to explain. LEDs required
    a minimum voltage to light. Also there is a maximum amount of current that
    can flow through is before it burns up. Without the current limiting resistor
    the LED sees too high a voltage and allows too much current through and it
    blows like a fuse. POOF!

    A good discussion of this is here.
    You can do that, but you're not going to get the expected result.
    They all do. The two outer ends are like a regular fixed resistor. The
    middle lead is known as the wiper and it's resistance varies vs. the other
    two leads. So if you use one outer lead and the wiper lead, you'll get
    the variable resistance you're looking for.
    See above. That should be good enough to get started.
    Well that's a different kettle. LEDs are highly non-linear with respect
    to current and brightness. Also as you've already learned they burn up
    if too much current goes through them. And they go out if there is too
    little voltage.

    What you can do is hook up a low resistance pot (50-100 ohms) as a voltage
    divider. You do this by tying the outside two leads to the two leads of
    your 9V source and pull the variable voltage from the wiper lead. Be aware
    that if you do this with a 9V batter that it'll go out pretty quickly because
    the pot will be burning 10-20ma of current the entire time you're

    BTW dimming LEDs usually isn't done this way. It's usually done by turning
    them fully on/full off very fast. It's a technique known as Pulse Width
    Modulation (PWM).

    Hope this helps,

  8. tlbs

    tlbs Guest

    One other thing to add to all the good information, above.

    This table is just a "rule of thumb". You need to read the datasheet
    for each LED for the exact current ratings and voltage drops.

    RED -- 1.5 V 10 to 20 mA (some as low as 1 mA)
    AMBER -- 2.0 V 10 to 20 mA
    GREEN -- 2.2 V 10 to 20 mA and even higher
    For the following colors it is best to read the datasheet
    BLUE -- 3.5 V to 5V(on some) 20 to 50 mA and higher
    WHITE -- same as blue

    Note that some of the BLUE LEDs have very low reverse voltages, and can
    be damaged by connecting them backwards to the 9V battery.
  9. Bob Myers

    Bob Myers Guest

    To a better approximation, they are devices whose brightness is
    current-controlled rather than voltage-controlled (as you noted,
    the drop across such things is fairly constant - since it is, after
    all, a diode). Control the current, and you control the
    brightness, with a *reasonably* linear relationship between the
    two. (Well, actually, it's luminous intensity that has a nice
    relationship with current, and not "brightness," but now I'm
    definitely getting in WAYYYYY too deep....) You DO, of course,
    as noted always want to limit the current so as not to exceed the
    device's maximum power- handling capabilties (the limit is generally
    power, not current, which is why higher peak currents are permissible
    for most devices in pulsed applications). Pulsed operation as a means
    of varying brightnessis often desirable, because it lets the device
    operate in a more efficient manner (relative efficiencies for
    LEDs tend to go up with current, at least for a while), but might
    be more difficult for the beginner hobbyist to implement than a
    simple current-controlled, constant-on mode.

    So a variable resistor CAN be used to vary LED brightness, as
    long as you ensure that you never can get to the point where there
    is so little resistance in the circuit that the maximum permissible
    steady-state current (as determined by the LED's power-dissipation
    limit) is exceeded. A potentiometer configured as a variable
    resistor, plus a fixed resistance in series, will generally do nicely.

    So let's figure on about a 2V drop on a red LED operating within
    the usual range of currents, and a 9V battery; if we want to be
    able to vary the current up to, say, a safe peak of 30 mA for the
    particular device in question, that's

    (9V - 2V)/ 30 mA = 7/0.030 = 233.3 ohms.

    With the nearest standard-value fixed resistor at 240 ohms (for
    5% parts), we'll err on the side of caution and choose that. Now,
    adding a 470 ohm potentiometer configured as a variable resistor
    (center terminal tied to either of the outside two), we can vary the
    current from 30 mA down to:

    7V/(240 + 470 ohms) = 7/710 = 9.9 mA,

    for roughly a 3:1 range in current and therefore intensity. If you
    want more range, you can always drop the bottom end by using
    a larger potentiometer (at the cost of being able to set the mid-range
    and higher brightness levels to as fine a degree of control).

    Bob M.
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