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I am using 75leds having forward voltage 3.3-3.4V and 60mA forward current. I connected 3 in series with a 39ohm 1/4watt resistor. My power source is a 12-0-12v_3A transformer with a rectifier circuit. The problem is when i connect the leds the resistor begins to smoke. Is it because of my power source?

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  • \$\begingroup\$ Tell us about your rectifier circuit. Two diodes with center tap grounded? Filter capacitor value? \$\endgroup\$
    – AlmostDone
    May 16, 2018 at 12:31
  • \$\begingroup\$ 25V_4700µF Filter capacitor \$\endgroup\$ May 16, 2018 at 12:39
  • \$\begingroup\$ Measure the voltage across the resistor. \$\endgroup\$ May 16, 2018 at 13:26
  • \$\begingroup\$ Schematic or it didn’t happen. \$\endgroup\$
    – winny
    May 16, 2018 at 16:14

3 Answers 3

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Most likely, the voltage is too high, and the 39 ohm resistor isn't high enough to limit the current. Have you managed to wire the transformer to give 24V, when you thought you were going to get 12V?

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  • \$\begingroup\$ I am geting 12v i have checked \$\endgroup\$ May 16, 2018 at 12:14
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  • The voltage through the circuit is 12 - 3 * 3.3 = 2.1 V
  • The current is V = I * R <=> I = V / R = 2.1 / 39 = 54 mA
  • P = V * I = 2.1 * 0.054 = 0.11 W

You might accidentally give 24 V (12 V - - 12 V) which would results in 0.22 W. And due to some inaccuracy (from the LEDS and/resistors) it might be above 0.25W which is too much.

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    \$\begingroup\$ Assuming full wave center tap rectifier arrangement, wouldn't the voltage be closer to 16.8? With such a low value current limiting resistor, any small increase in voltage has a crowbar effect: V is now 6.9V; your 54mA becomes 177mA; power is then 1.22W, enough to let the smoke out of the resistor! Good demonstration of the value of an active current source though. \$\endgroup\$
    – AlmostDone
    May 16, 2018 at 12:40
  • \$\begingroup\$ I have tested by connecting 4 leds in series with the same resistor....it is now working fine, but the resistor gets very hot although it is not smoking anymore. To be safe i have connected 5 Leds, but the brightest ness seems to be a little less. \$\endgroup\$ May 16, 2018 at 12:46
  • \$\begingroup\$ @AlmostDone Thanks for this clarification, I assumed a 12V power supply would give 12V. (also I see now the comment in Simon B's answer where Kangkan Kakaty verified it is 12V, so it seems contradictory). \$\endgroup\$ May 16, 2018 at 12:54
  • \$\begingroup\$ @KangkanKakaty Notice however, that you are playing with 'limits', either on the too less power (so one LED does not lid fully) or too much power (resulting in hot/melting reistors). It's best to find the real cause. Check for example in your circuit that it is 12 V and around 0.06 A (preferably with a scope if you have one). \$\endgroup\$ May 16, 2018 at 12:56
  • \$\begingroup\$ @KangkanKakaty That's why I asked him in a previous comment first what his rectifier and filter arrangement was. If he indeed has a FWCT then his measurement of 12V is incorrect. He could have measured while loaded, but I seriously doubt it would have been 12.00, so what was it? \$\endgroup\$
    – AlmostDone
    May 16, 2018 at 13:00
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Since your power supply is UNregulated it will drop 50% from no load 18V to full load 12V . If done with 2 diodes correctly from a tapped 24 V 30VA transformer. If your XFMR is 3A and you draw only 1.2A , it should yield about 5% higher but if the cap is too small, it will be much lower.

In any case a correctly chosen series R should be 4W not 1/4W unless 1 per string, which may be best.

Therefore you need a low side constant current limiter which can be done with a .7 V current shunt then an NPN Vbe to drive with collector to base of a 2nd power transistor with no series R but has base pullup to conduct ~75/4s x 0.06A = 1.2A Power dissipation can be expected around 2+ W so a heat sink is needed. All 20 or less x 4Series strings are necessary to operate.

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