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I am trying to connect, in parallel, 18 LEDs that have these specs:

LED

  • size: 10mm;
  • Voltage: 3.2V - 3.4V;
  • Current: 20mA.

Driver

From my calculation, by putting 18 LEDs in parallel I need a total current of 360mA, so by driving a (max) current of 350mA i do not need a resistor (to limit the current) and it should not burn the LEDs, however after a blink they all burned.

What am i missing?

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    \$\begingroup\$ You're using a 9V supply for a 3.4V load. That makes it not the right driver. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    May 20, 2015 at 16:12
  • \$\begingroup\$ what information from what source convinced you to try this smoke experiment? \$\endgroup\$
    – Andy aka
    May 20, 2015 at 16:36

1 Answer 1

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You have two problems.

  1. Voltage - the power supply you are using provides a constant 350mA by varying the voltage. Unfortunately, it can only go down to 9Volts. That is too much for one LED. You would need (at least) 3 LEDs in series for that to work.
  2. Parallel LEDs - this is a bad idea. LEDs don't all react the same. One will switch on at a slightly lower voltage than the others, and will get most of the current which will cause it to go BOOM in short order. Then the next LED wil get overloaded and the next and the next, and so on until you are out of LEDs.
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  • \$\begingroup\$ Just a comment on #1 JRE, the voltage doesn't matter so much as the current. You can run a single LED off of 9V (I do it all the time) but you need to limit the current to 20mA using a resistor. \$\endgroup\$
    – DerStrom8
    May 20, 2015 at 16:53
  • \$\begingroup\$ @JRE: Thank you very much for your reply. So, if I understand you well the driver supplies a minimum voltage of 9V, thus to have it working I should adjust the system this way: 54LEDs total of which: 3 series of 18 parallel LEDs -> 18 + 18 + 18), in this way the total Voltage needed is greater than 9V (9.6V - 10.2V) and the total Current needed will be 360mA (so less than what the driver can provide, 350mA, thus eliminating the need for a resistor). Right? \$\endgroup\$
    – simple mind
    May 20, 2015 at 16:55
  • \$\begingroup\$ @derstrom8 putting that resistor in would lower the voltage across the LED, I think that's what JRE is getting at. If you look at the IV curve, the LED will likely never drop 9V. \$\endgroup\$
    – Dr Coconut
    May 20, 2015 at 17:13
  • \$\begingroup\$ Sorry, I should have specified that my comment was not necessarily related to this particular question. It was just clarifying that in general the voltage is not as critical as the current when it comes to driving LEDs \$\endgroup\$
    – DerStrom8
    May 20, 2015 at 17:14
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    \$\begingroup\$ @derstrom8 Yes, that is correct. But to get the current down with a constant current source (as is used here) you have to lower the voltage - and the power supply in question can only go down to 9 Volts. \$\endgroup\$
    – JRE
    May 20, 2015 at 17:29

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