Magnet wire for fuse

I believe that proper fuse wire has a tempco that sharpens the threshold

- when it gets past the limit, it heats and fails more suddenly than copper would. That might not matter to you, but you should know it.

About 28 awg should do.

you guys are great -- I know have 28 and 30.

wish me luck.

AWG30 is about right. Maybe jump the pads with a bit of fiberglass sleeving over it. At your own risk, blah blah.

Why don't you tack some leads on the ends of a standard 10 A glass fuse and then tack it into the circuit?

Why not do a bit of testing, (well assuming you've 10A available.) A colleague burned out about one foot of 30 AWG, with three amps, But it was in vacuum. Which raises an interesting question. How does the fuse/melt current change in air vs vacuum? I guess you'd have to specify a length too.

The length does not matter as long is it is say at least 50 times the diameter. In the center of the wire length there is pretty much no heat flowing towards the ends. ref Renee Rogers Varian trouble shooter.

Dan

ameter.

ards the ends. ref Renee Rogers Varian trouble shooter.

Thanks Dan, I assume that's in air.(?) I was thinking about the vacuum ca se too. Where I have two ends (say both at 300K, but in my case one was 77 K) and all the heat has to travel to one end or the other. (Well I guess i t would have some radiative cooling as the middle get's hot.)

George H.

Why not do a bit of testing, (well assuming you've 10A available.) A colleague burned out about one foot of 30 AWG, with three amps, But it was in vacuum. Which raises an interesting question. How does the fuse/melt current change in air vs vacuum? I guess you'd have to specify a length too.

Using Magnet wire would change the self cooling from surrounding air effect because of the enamel insulation covering the wire. I believe that NiCrome wire is usually used for fuses. You definitely need to experiment.

As for Vacuum vs. Air, Air will cool the fuse wire to a small amount due to thermal conduction.

Shaun

It's a LOT worse in vacuum, for obvious reasons, given some decent length of wire. Probably the length does't matter, once you exceed some ratio of the diameter.

I recently did a bunch of tests of connectors in a decent vacuum, they did fine at the rated current (or above), but the current was pulsed so the heat capacity comes into play too.

Best regards, Spehro Pefhany

Think about the point where it fuses. Copper melts at more than

stuck inside a ceramic kiln.

Heat transfer is proportional to the difference of the fourth power of the absolute temperatures.

You're right, things get weirder at cryogenic temperatures.. 4.2K to the fourth power is a lot different from 1300^4 (10 orders of magnitude!).

Best regards, Spehro Pefhany

Hi Shaun, Re: wire coating... It may sound a bit strange, but I was thinking that putting an insulating coating on a hot thing, might actually cool it more (in air). (I was thinking about putting heat shrink around a through hole resistor.) It's a thermal resistance vs area idea.

Yeah, I once tried to 'calculate' air conduction losses, (temperature, number of atoms, collsiion rate, mean free path... I don't remember, and I might have made some mistakes.) but I got a small number. I then tried to measure it for a resistor... which is easy 'cause it has a built in heat source. It was a lot more! Convection turns on at a few degrees or so...

George H.

case too. Where I have two ends (say both at 300K, but in my case one was 77K) and all the heat has to travel to one end or the other. (Well I gues s it would have some radiative cooling as the middle get's hot.)

it's

It's not at all clear to me if the radiative loss, or thermal transfer down the wire will be greater. Re: Radiation loss, Stefan Boltzmann law. So if you do the loss for 300K into 0K with emmisivity =1 you get ~500W/m ^2 Which is 1/2 the solar radiation, which is easy to remember. (Why is the earth ~300K?) From there you just scale things. (OK I scribbled some numbers and got ~10Watts.. (1 foot of 30AWG at 900K) b ut I'll have to check again in the morning.)

George h.

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Because the surface area is 4 pi r**2, whereas the projected area is only pi r**2. (It's only zenith sunlight that's 1 kW/m**2.)

Cheers

Phil Hobbs

For a center point source of 600K above ambient, and 6" of Cu AWG 30 wire (diameter 0.255mm) I get 80mW down each side of the wire, so

It might be interesting to calculate or simulate the temperature at which the radiative losses equal the conduction losses (assuming no convection).

Best regards, Spehro Pefhany

part.

so.

jump the fuse pads ?

x the problem -- (it's on one of the inverter boards).

Do you think the fuse is the only problem? I vote no.

Do you think the fuse is the only problem? I vote no.

Concur. An ohmmeter should find a dead short in no time.

tm

On a sunny day (Thu, 16 Jan 2014 12:31:46 -0800 (PST)) it happened mkr5000 wrote in :

Altough I have dunnit, the risk is that some piece will fly away and short something more expensive, or even start a fire, explode your home, destroy planet earth. :-)