Help with simple circuit....

Discussion in 'The Projects Forum' started by rochler, Jun 3, 2008.

  1. rochler

    Thread Starter New Member

    Jun 3, 2008
    Howdoo folks,

    I'm a total noob with electronic circuits but I want to build a very simple circuit for a test lamp (to be used for a foucault tester).

    Basically I want wire up a 'super bright' 5mm LED to a battery, switch & pot so I can control the LED intensity somewhat. Yes, I hear you laughing, but I'm stuck on a few points:

    I want to run the LED from a 9V battery for size & convenience, but the LED only requires forward voltage of between 3.2 & 3.6 (min/max). Am I correct in assuming that this can simply be fixed by using the appropriate amount of resistance? The forward current of the LED is 20mA, so I calculated that I would need around 270 Ohms resistance - is that correct or am I way off?

    If correct, what Pot. will be suitable - and can the Pot. provide the necessary resistance without requiring any additional resistors in the circuit?

    Hope someone can stop laughing long enough to help me hehe.... :D

    Cheers, Fred....
  2. SgtWookie


    Jul 17, 2007
    Hi Fred,

    No, it's not a silly question at all. It does come up rather frequently, though.
    LEDs usually have a Vf(typical) and Vf(max) specification for a given maximum current.
    The Vf(max) means simply that when 20mA is being passed through the LED, the Vf will not exceed the specified value.
    Vf(typ) is basically the average voltage you might get at the specified current if you tested a whole slew of the same batch of LEDs. They can vary a fair bit.

    But your basic problem now is to find out what value resistor to use to keep the current from exceeding the maximum current of 20mA. We'll use Ohm's Law.
    R = E/I (or, Resistance = Voltage/Current)
    We'll customize the equation a tad bit:
    Rlimit = (Vsupply - Vf(LED)) / I(LED)
    Rlimit = (9v - 3.2v) / 20mA
    Rlimit = 5.8 / 0.02
    Rlimit = 290 Ohms
    So, the minimum resistor size you can use is 290 Ohms.
    But, it's a bit more complicated than that, because not all 9v batteries are equal. Some may put out around 8.4v even when new. Others might put out nearly 10v. Additionally, you can't typically buy a 290 Ohm resistor, as it's not a standard value.

    Anyway, here's a link to a table of standard resistor values:
    If we look up under E24 values, we see that 300 Ohms is a standard value. So, we'll use that for our minimum resistance.

    Let's check to see what wattage the resistor will have to handle:
    P = EI (Power in Watts = Voltage x Resistance)
    P = 5.8 x .02
    P = 0.116 Watts
    We usually double that to make sure our resistors will stay cool.
    P = 0.232 Watts
    You'll be fine if you use a 1/4 Watt resistor.

    But how dim do you want the LED to go? Let's say you want to be able to reduce the current down to around 3mA.
    Rlimit = (9v - 3.2v) / 3mA
    Rlimit = 5.8 / 0.003
    Rlimit = 1933.333... Ohms
    So, if you subtract the original 300 Ohm resistor from that, you'd need a pot that was around 1633 Ohms. Unfortunately, they don't make them in oddball sizes like that. You could get a 1K Ohm pot or a 5k Ohm pot at your local Radio Shack.
    Let's see what a 1k pot in series with the 300 Ohm resistor would give you.
    Ohm's Law again:
    I = E/R
    I(LED) = 5.8/1300
    I(LED) = 4.46mA - that's not bad.

    Or, you could order something closer from someplace like Mouser, Jameco, Electronic Goldmine, or many other places online.

    So, basically - your circuit is the battery, a switch, the LED, a 300 Ohm resistor that prevents you from turning up the current too much, and a 1k or 5k pot all in series.
    Last edited: Jun 3, 2008
  3. Wendy


    Mar 24, 2008
    Don't be afraid to ask questions here. We all started somewhere, and I don't have space to list the boners I've pulled. I think you'll find it is a pretty friendly group.
  4. rochler

    Thread Starter New Member

    Jun 3, 2008
    O.k. I've got a plan now gracias... :)

    I can't find 300 Ohm resistors locally - we have Tandy & Dick Smith here in Aust. but they don't stock all sizes. I think the smallest they have is 1k, so I might just get 4 x 1.2k and wire them in parallel?

    I should be able to get a 1k pot. no problem, but does it matter if LIN or LOG? What should be rating be in mW? In the catalogs they show as '200mW' and '200VDC' etc. - what does that mean or doesn't it come into play in this application?

    Also, in what order should I wire them up - I presume Battery->Switch->Res->Pot->LED->Battery.

    Thanks for your patience & advice - much appreciated!

    Cheerz, Fred.... :D
  5. Gadget

    Distinguished Member

    Jan 10, 2006
    Dick Smith, Tandy, Jaycar etc etc all stock e24 and e12 resistors....
    try either a 270 ohm or 330 ohm....
    (Dick smith cat numbers... R 0560 and R 0562 for the e24 metal film type)
    (R 1060 and R 1062 for the E12 Carbon film)
    Jaycar have a 1/2 watt 300 ohm metal film type.. cat RR-0559
  6. SgtWookie


    Jul 17, 2007
    4 x 1.2k in parallel would work!

    Actually, a log pot would work better in this case. You'll have to do a bit of fiddling to figure out which "end" of the pot to use. There are three terminals on the pot; one for each end of the resistor and one for the wiper. From one side, it'll seem like the LED goes from ON to OFF very suddenly. The other way it will dim very gradually.
    Take another look at my first reply; I'd already calculated above that the maximum power dissipation (with just the 300 Ohm resistor) is 116mW.

    That will work fine.
  7. rochler

    Thread Starter New Member

    Jun 3, 2008
    You've all been very helpful in clarifying what I need to do - thanks again for the very detailed & informative answers. It's great to have a site like this.... :):):)