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So I'm making a parallel LED circuit using THREE CREE x-lamp LEDs which typically run at ~ 3.1v @ 350 mA, but they can max out at 1000 mA. I'm putting a resistor for each LED.

So my logic may be flawed here but this is how I'm thinking (sorry if I seem stupid):

My power source is 12v and 1 amp, which means if I use three equal resistors, each LED will draw .333 mA, close to the typical value.

Calculation of resistor: R = V/I ------ R = (12-3.1) / .333 (3) = 8.9 ohms (I believe I'm supposed to multiple the denominator by amount of lights I have, correct? So I would use a 10 ohm resistor.

Wattage of resistor: P = VI ------- P = 8.9 ohms * 1 amp = 8.9 watts. So I would use a 10 ohm, 10 watt resistor.

Is this correct?

Also, say I had only one LED. would it draw that maximum amount of amps from the power source, so 1 amp?

Sorry, I am really new to this stuff. I don't know if I'm doing the correct calculations or not.

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  • \$\begingroup\$ Is there a specific reason why you want to run them in parallel? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 29 '15 at 8:05
  • \$\begingroup\$ I dont know if this is a good reason but I'm planning on controlling their light levels with arduino so I need one light per pin. So if I ran it in series, wouldnt controlling 1 pin control all the lights at once? \$\endgroup\$ – txmb95ads Aug 29 '15 at 8:09
  • \$\begingroup\$ Do you have a LED power supply that limits the current to 1 A or a standard power supply that may malfunction above 1 A? If it's the latter don't assume that it cannot output more than 1 A. It probably can, it will just get damaged in the process. \$\endgroup\$ – Damien Aug 29 '15 at 9:29
  • \$\begingroup\$ Your LEDs don't "max out" at 1000 mA. They may burn out if your circuit drives them with more current than the manufacturer recommends, but they won't draw any more or any less current than your circuit allows them to draw. How much current to you want to feed them? That's the question. \$\endgroup\$ – Solomon Slow Aug 29 '15 at 20:32
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Your calculations for both the resistor value and wattage are off.

For the resistor, you show the formula R = V / I, but then calculate using the entire current for all three LEDs (1 A) instead of the current for just one LED.

So it should be

$$R = \frac{V}{I} = \frac{(12 - 3.1)}{.333} = 26.7\space Ω$$

For the wattage, you are trying to use the formula P = VI, but then you substitute R for the V and use the current for all three resistors (1 A) instead of just one.

So the correct calculation would be

$$(12 - 3.1) * .333 = 2.96\space W$$

or you could use:

$$P = I^{2} R = (.333)^{2} * 26.7 = 2.96\space W$$

So you could use a 27 ohm, 5 W resistor.

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If you had only one LED which dropped 3.1 volts with 333 milliamperes through it and you were driving it with a 12 volt supply, you'd need a series ballast resistor to drop the remaining 8.9 volts.

Then, since in a series circuit the current is everywhere the same, that 333 milliamperes would pass through both the LED and the ballast and the value of the resistor would be:

$$ R = \frac{Vs-Vled}{Iled} = \frac{12V - 3.1V}{333mA}\approx 27\Omega $$

Since your supply is a voltage source it'll maintain its output voltage at 12 volts as long as you don't exceed its 1 ampere rating, the effect being that you can connect any number of LEDs and their associated ballasts across the one supply and each pair will think it's connected to its own independent supply.

Consequently, each of your LEDs would be ballasted with a 27 ohm resistor which would limit the current through each LED to 333 milliamperes, for a total of one ampere for the array.

Each resistor would dissipate about:

$$ P = IE = 333mA \times 8.9V \approx 3\text{ watts,} $$

so it would be prudent to use 27 ohm, 5 watt resistors for your ballasts.

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  • \$\begingroup\$ Thanks for the help. I was under the impression from watching a few resistance calculation videos in parallel that you divide by the sum of the currents, which, in my case, would be 1000 mA. But I guess that is wrong. I just noticed your calculation was assuming I'm in series, but I'm in parallel. Wouldnt it be 12V - (3.1 + 3.1 +3.1)/333 if it were in series? \$\endgroup\$ – txmb95ads Aug 29 '15 at 16:47
  • \$\begingroup\$ @txmb95ads: Each of the LEDs is connecter in series with a 27 ohm resistor, so each series string draws 333 milliamperes when connected to a 12 volt supply. Then, because each string draws 333 milliamperes, when they're all connected across the supply in parallel, the sum of the currents drawn by the three strings will be one ampere. \$\endgroup\$ – EM Fields Aug 29 '15 at 20:02

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