So this may seem simple but I have wired up 6 LEDS (3 pairs of 2) with each of the pairs in parallel. Sorry if that's confusing. Anyway I am using the TSAL7400 IR LED and with 1A (which is what I am getting) it has a typical forward voltage of 2.6V and I have 3 of these in series and hence I am using 7.8V. In saying this I am supplying 9V hence I need a resistor to drain the remaining 1.2V so the LEDS don't blow. Using V = IR it should be simple and a 1.2ohm resistor should do the trick. However when testing the resistor starts smoking. Is there some practical application I am missing because theoretically it should work?

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    \$\begingroup\$ The way to not be confusing is to provide a schematic diagram. \$\endgroup\$ Commented Oct 4, 2015 at 22:40
  • \$\begingroup\$ While it would be nice to know if the desired ampere of current is per LED or the total for the two LEDs in parallel in each set, the circuit topology is entirely clear from the wording, and fully sufficient for this question. \$\endgroup\$ Commented Oct 4, 2015 at 23:13

2 Answers 2


LEDs have a negative temperature coefficient of resistance, so in order to prevent current hogging You shouldn't use a single resistor to ballast parallel strings of resistors because of the different forward voltages of the strings.

Instead, you should provide a ballast for each string and then connect each series string in parallel with the other(s).

The value of the resistor can be found by using:

$$ R = \frac{Vs - (n\times Vf)}{Iled}, $$

In your case that would be:

$$ R = \frac{9V - (3\times 2.6V)}{1A} = 1.2 \text{ ohms} $$

and the power the resistor would dissipate would be:

$$P = Iled^2 R = 1A^2 \times 1.2 \Omega = 1.2 \text { watts} $$

1.2 ohms is a standard E24 value, and Yageo makes a nice little 2 watt metal oxide film resistor which is available from Digi-Key for about USD 0.29 each.

Keep in mind though, that resistor is for one series string only, so you'll need another one for the other string of three series LEDs.


You might be using too small a wattage (power dissipation) resistor. For example a 0.5W 1/2W resistor. That will get very hot.

The resistor must be able to dissipate the power, otherwise it will increase its operating temperature, possibly until it fails.

Power = Voltage x Current.
1.2V x 1A = 1.2W.

So I would use a 2W resistor to give a bit of headroom. Is your resistor that wattage?

You could put 4 smaller power resistors in parallel (or series) to share the power dissipation, and get them into a reasonable operating power range.

  • \$\begingroup\$ I am not to sure. We have a "shop" on campus that I just bought the 1.2ohm resistor from, I will have to look at the wattage. \$\endgroup\$
    – Cameron
    Commented Oct 4, 2015 at 22:08
  • \$\begingroup\$ It is 0.6W so thats the problem. Thanks for the help \$\endgroup\$
    – Cameron
    Commented Oct 4, 2015 at 22:09
  • \$\begingroup\$ Work out the Ohm's law for three 0.6W resistors in parallel (it'll likely be easier to get values above 1 ohm), and that should work okay. \$\endgroup\$
    – gbulmer
    Commented Oct 4, 2015 at 22:12

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