# Heating Nitinol while driving LED

Edited in my LED module current requirements, fixed needed resistance and added a schematic to better illustrate what I'm trying to do. Thanks Tom and asmydolf.

Okay, it might seem like a dumb thing to do, but I'd like to light up a LED at the end of a conductive wire, which in turn I would like to heat up. The conductive wire is Nitinol, which is used to move the LED to a new position, hence the reason for driving them together. The LED is right now a module, rated 12V and 0.36W, so it should require 0.03A, and the Nitinol wire is 0.1mm Dia., and 30cm long. I made some calculations, and to heat to Nitinol wire in 5 seconds I need 0.35A, 36 Ohm, which will create 4.4W - meaning a reasonably priced resistor can be found for that. I thought of running both the Nitinol wire and the LED in parallel, as I hope my sketch will show. Alas, just running the LED module at the end means no heating, and when I place a potentiometer across the LED terminals to divide the current I either short out the LED while heating the Nitinol wire, short out the wire while driving the LED, or getting the potentiometer burning while trying to do both...

simulate this circuit – Schematic created using CircuitLab

Anyway, any and all help on the matter will be greatly appreciated, I'm trying to help my wife's project off the ground in between my exams, and I was never really good with electricity.

Thanks, Uri

• What is the current rating of the LED module? – Tom Carpenter Jul 20 '15 at 1:12

You say you need 4.4W, at 0.35A inside the wire, so that makes 12.6V (rounded).

But, if your wire is only 2.9Ohm, that would mean (P = V^2 / R):

V = sqrt(P * R) = sqrt( 4.4W * 2.9 Ohm ) =~ 3.57V

I = sqrt(P / R) = sqrt( 4.4W / 2.9 Ohm ) =~ 1.23A

I'll use my values for voltage and current, but leave you with the thoughts and formulae so you can redo it if you are 100% sure you were right about 0.35A (impying 12.6V for the wire, making about 24V for the total).

That means, if the LED module needs 12V, means the wire+LED should be powered with 12V + 3.57V =~ 15.6V, but 15V would probably do well enough.

You then need to adapt the LED end to handle the 1.23A, of course it normally wants only 20mA maximum, the module has an internal resistor that makes that happen for 12V, but if you power it through the wire with 15V, the wire will not add much resistance (2.9Ohm you say ?), which will cause the internal resistor to not limit enough any more, and the LED will in the best case be brighter with a slightly shorter life span (or die in the worst case, depends on how many chips and what resistor).

So you need to put an extra resistor in parallel to the LED that steals away 1.21A at 12V.

From R = V / I =~ 12V / 1.21A =~ 9.9 Ohm.

But that resistor needs to then be: P = I * V =~ 1.21A * 12V =~ 15W

Which means it will also dissipate a lot of heat, and not be very small at all.

If your wire really is 2.9 Ohm and needs 4.4W to get hot enough for your aims, you should seriously consider replacing the LED module with one that has no internal resistor, or is set for 3.5V or such. That would at least allow you to make the resistor:

R =~ 3.5V / 1.21A =~ 2.9 Ohm

P =~ 3.5V * 1.21A =~ 4.3W (still a lot!)

As you can see, in either case you are much better off adding a thin extra wire (possibly loosely wrapped around the Nitinol wire??) to power the LED separately.