# Clarification of Onderdonk's Fuse Equation Assumptions

1. Oct 6, 2014

### tempneff

Hi all,

I am trying to get an understanding of Onderdonk's Fuse Equation:

$I_{fuse}=Area*\frac{\sqrt{log\left(\frac{T_{melt-T_{ambient}}}{234-T_{ambient}}+1\right)}}{33*Time}$

Empirically I do not see that this function accurately depicts the behavior of the wires I am using. For example, I hooked up a 37 gauge copper wire to a current source with 3A. I verified the values in this chart of Onderdonk's values for copper against the equation and found that they agree. I should have seen the wire break within a second, however it glowed but did not break. It took nearly 5A to melt it.

I'd like to know what assumptions were made for this equation to work? Such as length of wire, ambient pressure maybe...

Thanks in advance, I love this site.

2. Oct 6, 2014

### Baluncore

What units did you use for the area and for the temperature parameters?

The wire needs to be sufficiently long so that the middle section is not cooled by the mounting connections.

Forced or convective airflow will cool the wire and so require higher currents before fusing. I do not know the original airflow assumption.

A ceramic or glass tube will reduce thermal radiation and lower the current needed to fuse the wire. You would need to correct the Tambient if in a tube to the temperature of the tube inside wall.

3. Oct 7, 2014

### tempneff

Temperature is in Celsius, area is in circular mils. I didn't think to try longer wires, I have been using wires I thought would be comparable to standard electronic fuse ~3/4 inch.

4. Oct 8, 2014

### sophiecentaur

It could be worth while,doing the experiment with the wire under various conditions - enclosed in a tube, wrapped in fibreglass etc. to find when you could actually achieve fusing. You may have been 'borderline' with your test, even though the dissipated power at 5A is 25/9 times the power with 3A (approx) - which doesn't look to be very borderline, I admit.

If you are doing these experiments with a view to actually applying the results for a circuit design, it would be wise just to work on recommended practice, which is much more conservative. Alternatively, if you want to make a fuse, then I believe Tin is the preferred metal.

5. Oct 8, 2014

### zoki85

Looks like this is an aproximate formula and 37 gauge is very thin. Maybe, the formula better aproximates melting of thicker wires?

6. Mar 23, 2015

First, the formula is incorrect. The " 33*Time " should be under the square root radical. Second the denominator 234-Ta should be 234+Ta. The two primary assumptions are (a) no cooling, not from convection, conduction, nor radiation (Thus rapid heating assumption), and (b) time is the time TO the melting temperature, not that plus the time to actually melt the copper. Thus, the relevant time is short, from one to 10 seconds, depending on who you read.

7. Mar 23, 2015

### sophiecentaur

As a bit of theory, this is interesting but, if a fast fuse action is needed in practice, there are many better ways of doing it.
I'd bet that all that work was done before the luxury of alternatives was available for protecting sensitive stuff.

8. Mar 23, 2015

Actually, Onderdonk was considering the copper wires that supported the poles and insulators of high-voltage transmission lines. He wanted to determine the minimum wire size necessary to carry a short-circuit for a time period long enough for the circuit protection devices to kick in. His equation dates back no later that 1928. We have found no references earlier than that.

9. Mar 23, 2015

### zoki85

Hmm, but I would think that melting point of wires is much above allowed termal stress of the insulators.

10. Mar 23, 2015

The problem was arcing across the insulators because of dust or moisture. Here is where this is discussed: E. R. Stauffacher, “Short-time Current Carrying Capacity of Copper Wire,” General Electric Review, Vol 31, No 6, June 1928