# Physical Wire Size Confusion

I'm going to get this wire for a simple project, and I need to get some heat shrink for it. My problem is that I can't figure out the diameter of the individual insulated wires. There are going to points in the project where the cable is going to be split, and soldered onto LED leads.

The problem is that I can't figure out the diameter of the individual insulated wires. I think it's .145", according to the datasheet.

I think this heatshrink will work, with the wire above, but I want to triple check before I order.

• Those links don't work for me. Could you re-list them? – helloworld922 Jun 16 '13 at 21:31

According to this selection guide, you should pick a tubing size which is ~20-30% larger than the thickest diameter object which the heat shrink will cover. Additionally, you want the minimum reduction ID to be smaller than the minimum diameter object so you'll get a snug/tight fit.

I'm assuming this is the tubing datasheet you're interested in: FIT-221B-3/16

I was not able to find the wire in question, but let's assume that your statement about a 0.145" diameter for the wire + insulation was accurate.

The tubing has a minimum expanded diameter of 0.187", which is ~29% larger than the wire (including insulation).

The maximum recovery ID is 0.093", which is more than enough to securely cover the insulated portions of the wire (0.93" < 0.145").

There are still one outstanding questions to answer:

Presumably you actually want to solder the wire to something (in your case, an LED). This means there will be portions of the wire without insulation, and there will be some lead thickness which needs to be taken into account.

Let's take for example this red LED. It has square leads with a maximum side length of 0.525mm, or 0.0207".

Suppose the wire thickness was 8 AWG, 0.145" total thickness (quite thick insulation). Then the cross-sectional profile looks something like this:

Ignoring the solder, the thickness of the wire and the LED lead is:

$$\sqrt{(\frac{0.1285 in.}{2} + 0.0207 in.)^2 + (0.0207 in.)^2} + \frac{0.1285 in.}{2} = 0.1517 in.$$

This is still ~23% smaller than the tube expanded ID, so we're good here. Likewise, the minimum recovery ID checks out too because the wire itself is larger. Adding on solder might make it a bit tight of a fit, but it should still fit (just don't glob on too much solder).

Now on the other end of the spectrum, suppose we have a 22 AWG, 0.145" total thickness (quite thin insulation). This time the cross-sectional area looks like:

And the minimum diameter is now 0.046". This time, we will have problems because the reduced heat shrink won't tightly wrap around the section containing the wire and the LED lead. In this case, we would need shrink tubing with a shrinkage ratio of 4:1.

So in conclusion (tl;dr): It might work, but you'll need to know more about the wire dimensions, as well as what you're attaching the wire to. For the 22 AWG case you'll need a much higher shrinkage ratio to tightly wrap the actual joint. For a thicker wire, say 8 AWG the proposed tubing should be fine. There could be other factors as well such as heat shrink tubing material properties, but I doubt for the majority of projects these are a major concern.