I need to wire 5 led's in series with a switch and variable resistor to control brightness.

These I think

I think the circuit will need to look something like this (Ignoring the values the editor added!) but it's been a very long time since I've done anything like this.


simulate this circuit – Schematic created using CircuitLab

I don't know how to calculate how much battery power will be required, or if it will need additional resistence.

How do you go about solving these kinds of equations?

  • \$\begingroup\$ I'll put the resistor before the first LED, after the switch \$\endgroup\$ – KingsInnerSoul Sep 18 '13 at 13:53
  • \$\begingroup\$ @KingsInnerSoul How exactly does the position of the resistor matter? \$\endgroup\$ – Anindo Ghosh Sep 18 '13 at 13:53
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    \$\begingroup\$ I'm assuming you will attach the wiper to either end of the potentiometer. You should also add an additional series resistor to set the maximum current allowed when the potentiometer is set to zero ohms. This is the "smallest value of resistance" referred to in Brian's answer. This will protect your LEDs. \$\endgroup\$ – Tut Sep 18 '13 at 13:56
  • \$\begingroup\$ a convention which i am used to, and thats how I usually see them in many other schematics. \$\endgroup\$ – KingsInnerSoul Sep 18 '13 at 13:57
  • \$\begingroup\$ @KingsInnerSoul Current flow convention has nothing to do with resistor position convention. Neither is there any recognized convention for which side a resistor goes in, nor does it make the least bit of difference to circuit operation. Your comment merely confuses EE-inexperienced readers, while adding no value whatsoever. \$\endgroup\$ – Anindo Ghosh Sep 18 '13 at 14:05

The product summary tells you the typical and max. forward voltage and maximum current.

Multiply typical and max. voltages by no. of LEDs, then choose a power source (battery) with higher voltage than max * no. of LEDs.

Subtract typical voltage * no. of LEDs from battery voltage; this is the voltage you can expect across the resistor.

Divide that voltage by the max. operating current to give you the smallest value of resistance you need. You can increase it to reduce brightness.

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  • \$\begingroup\$ That sounds great! I'll run these numbers later today. Really appreciated. \$\endgroup\$ – asawyer Sep 18 '13 at 13:58
  • \$\begingroup\$ Divide that voltage by the max. operating current How do you determine max operating current? \$\endgroup\$ – asawyer Sep 18 '13 at 16:20
  • \$\begingroup\$ See first line of answer. The product summary is on the page you linked to. \$\endgroup\$ – Brian Drummond Sep 18 '13 at 16:37
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    \$\begingroup\$ The math is OK but 0.5V/0.02A = 25 ohms is the resistor you would ask for. HOWEVER if the LEDs you get have Vmax=4V ... 18V < 20V so no light. Which is why Vmax determines your supply ... in your case, 27V and take it from there. \$\endgroup\$ – Brian Drummond Sep 18 '13 at 17:11
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    \$\begingroup\$ You got it right. But to explain: when designing, you have to take variability into account. 100 ohms could over-stress "typical" LEDs (with a fresh battery!) though you might get away with it. 200 ohms would be safe under all combinations of LED at a (surprisingly small) cost in brightness (and more battery life!) The answer is a shorthand process to accomplish that : the comments explain why. \$\endgroup\$ – Brian Drummond Sep 19 '13 at 8:01

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