1
\$\begingroup\$

Planning to use a green LED (forward voltage 2V, forward current 30mA) on a 12V circuit so using as resistor calculator (led.linear1.org/1led.wiz) I bought a 390 ohms, 1W resistor.

However after connecting the resistor to the positive terminal the led refuse to light up.

Is the calculator wrong or am I making a mistake somewhere? (gather I am a complete novice)

\$\endgroup\$
5
  • \$\begingroup\$ (12V-2V)/(30mA) = 333Ohm, so your calculations are correct. The max power 1W is 3 times more than what you need, so you should be fine. You have a mistake somewhere else. \$\endgroup\$
    – yo'
    Commented Jan 25, 2015 at 19:24
  • \$\begingroup\$ Don't take it bad but... Try inverting polarity :-) \$\endgroup\$
    – Grabul
    Commented Jan 25, 2015 at 19:26
  • \$\begingroup\$ Does it need a heat sink ? what type of LED is it ? ( a clear plastic housing or for mounting on a heat sink) \$\endgroup\$
    – Spoon
    Commented Jan 25, 2015 at 20:14
  • \$\begingroup\$ @yo': Four times more. \$\endgroup\$
    – EM Fields
    Commented Jan 25, 2015 at 20:45
  • \$\begingroup\$ @EMFields I wouldn't dare myself to use a 0.25W resistor in a 0.26W setting, so I left some headroom there :) \$\endgroup\$
    – yo'
    Commented Jan 25, 2015 at 20:48

3 Answers 3

Reset to default
6
\$\begingroup\$

Ok, first let's check the resistor value. We'll do the maths so you never need trust a random website again:

$$ V_{DROP} = V_{IN}-V_{LED} = 12 - 2 = 10V $$ $$ R=\frac{V_{DROP}}{I_{LED}} = \frac{10}{0.03} = 333.\dot{3} $$ So that all looks OK.

So why isn't it working? Well, some possibilities:

  1. The LED is connected backwards. LEDs are polarized devices and have to be connected the right way around. Look for a flat edge to the rim of the LED which denotes the negative (cathode) connection. Also LEDs often have leads of different lengths. The longer is normally positive (anode).
  2. You have blown the LED. Accidentally putting 12V through the LED will pretty much instantly kill it.
\$\endgroup\$
1
  • \$\begingroup\$ I received a batch of LEDs that the molded in flat edge was where the anode connected, whether by mistake or design. That can happen too. \$\endgroup\$
    – THOMT
    Commented Jun 6, 2021 at 16:06
1
\$\begingroup\$

The ballast resistor must drop the difference between the supply voltage and the LED forward voltage with the LED current through it so, consulting Ohm's law, we have:

$$ R = \frac{Vsupply-Vled}{Iled} = \frac{12V-2V}{0.03A} \approx 333\Omega $$

So your 390 ohms should be enough to get the LED to light.

Since the LED is a diode, once past the knee the voltage it drops will stay pretty constant over a fairly wide range of currents through it, so consulting Ohm's law once again and solving for the current 390 ohms will let through, we have:

$$ Iled = \frac{Vsupply-Vled}{R} = \frac{12V-2V}{390\Omega} \approx 26 mA $$

Which should be plenty to light up the LED.

The power the resistor will dissipate will be:

$$ P = ( Vsupply-Vled) \times Iled\ =\ 10V\times 0.026A = 260mW $$

So a 1/2 watt unit would be OK. Your 1 watt resistor will be fine, though, and it'll run cooler.

I would guess that you connected the LED backwards to start with and now it won't work in either direction because you let out the magic smoke. ;)

\$\endgroup\$
1
  • \$\begingroup\$ if it looks black inside then it is dead. \$\endgroup\$
    – Spoon
    Commented Jan 25, 2015 at 20:15
-3
\$\begingroup\$

Based on the formula above, I have created a calculator that will help you calculate the resistance required - http://www.calctown.com/calculators/led-current-resistor-calculator

I did verify this value and got 333.33 Ohms.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ This looks a lot like link spam - you should answer the actual question or leave this as a comment. \$\endgroup\$
    – David
    Commented Jun 18, 2015 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.