I need a formula to prove a point

[cheapNdisgusting]

What I need to "prove" is; This is an example - Say there is a 10 amp fuse in a circuit. That fuse gets blown. I don't have another 10 amp fuse to replace it but I do have two 5a. fuses on hand. I parallel these two 5a. fuses and temporarily insert them into the circuit until I can get the proper fuse. I know that the original fuse was blown by a mistake which caused a short circuit (meaning there is nothing wrong with the circuit the fuse protects.

How can I show this? The parallel fuse part.

Thanks in advance

[haklesup]

If you can just use a single 5A fuse until you get a 10A one, thats a better solution. I wouldnt't select a 10A fuse to protect say a 8.5A circuit for example, I would use a 20A so its reasonablle to think that that 10A protected device when running normally uses less than 5A. In any case, worse you can do is blow the 5A fuse.

In theory by formula Two 5A in parallel gets you a 10A but in practice, variances in manufacture may cause one to share an unequal part of the load. at which point you should have at least 5A protection and not more than 10A but hard to predict exactly how much in between

The problem comes when it blows. One will fuse open first and then the second will blow a moment later, this will prolong the time it takes to open and probably allow additional glitches in the oversupply current which the source may not like. If its a non critical situation, you can get by but its bad practice in any book.

[cheapNdisgusting]

Hackles I do appreciate your point but this is a hypothetical situation. What I want is a way of mathematically showing the theory that it will work.

[haklesup]

A fuse is just a small value resistor engineered so that when the trip current is reached, enough heat builds up to reach the melting point of the fuse strip. The thickness of the strip and whether is it held in tension determines the fuse blow rate (fast/slow/instrument)

Mathematically, all you need to do is assume that both fuses (and their holders) are EXACTLY the same resistance and that the resistance will not change or will change in exactly the same manner between both fuses as they get hot. Start with Kirchoff's first law and from there it's just a matter of solving for current in a resistor divider where R is nonlinear (R is a function of I). It's not much different than 2 incandescent bulbs in parallel (except these bulbs will fail).

[Dean Huster]

Even if the "identical" fuses are not identical and one blows before the other because it's passing a higher current, the second one will blow quickly. Don't expect them to blow simultaneously except under a high-current short circuit condition.

[ringo47stars]

"The relationship between Voltage, Current and Resistance in any DC electrical circuit was firstly discovered by the German physicist Georg Ohm, (1787 - 1854). Georg Ohm found that, at a constant temperature, the electrical current flowing through a fixed linear resistance is directly proportional to the voltage applied across it, and also inversely proportional to the resistance. This relationship between the Voltage, Current and Resistance forms the bases of Ohms Law." A quote from http://www.electronics-tutrials.ws. It would be a good idea to consider the drain on the battery like high drain and low drain because a high drain on the battery would create excessive heat. So mathematically this works out to equate the resistance at a constant temperature as stated in the quote from electronics-tutorials.

[cheapNdisgusting]

Thanks for all the replies. I appreciate it

[haklesup]

So I was a little interested so I went about curve tracing a few fuses to open. I had a bunch of 500mA 3AG fuses lying around so I used them (littlefuse 312)



In this first graph, the tester was in curve trace mode so the point after it opens snaps to a larger voltage. Note that it starts out as a resistor of about 0.5 ohms (after subtracting 2.1 ohms for the test leads) but note how its slope decreases near the fuse point. This shows that the resistance of the fuse is getting higher just prior to blowing. See the next graph fopr a better image of that.

After peaking in current and fusing open, the curve snaps to an open characteristic (horizontal line on the X axis) but in doing so connects 2 sample points with a downward sloping line. This part of the curve is not real. Its just a discontinuity in the data with tewo points connected by a straight line



Note how my 500mA fuses are consistantly blowing at 850mA. This was a slow sweep lasting about 200ms. A much slower closer to DC sweep may have blown it a little lower but still above 500mA



This graph shows the same as the first two but the drive was in parametric mode which follows the shape of the curve a little differently


This last curve shows 2 curves in parallel. Unfortunately I can only source 1.5A and it would probably blow a little higher than that so I did not capture the fuse point but you can see the resistance of the fuses is now half what it was before.

FWIW

[MrAl]

Hello again,


Hackle:
Some very interesting curves there. It would be very nice to see the time too, which would be along the curve itself. Any way to show that too? I ask because i could compare to some formulas i have on hand, well provided we could find out what kind of material the fuse is made out of.
I guess you could test two fuses in parallel too then?

Anyone here know what kind of metal is used or better yet the melting temperature of the metal?

There are a few problems i see with trying to understand what happens with two fuses in parallel.
One is how to figure out the amount of mismatch. If both fuses in parallel are the same then they would blow at the same time and share exactly half the current each, which wouldnt tell us too much.
Another is just how perfect the two fuse holders are, or should i say imperfect.

What might help:
If there is a 10 percent mismatch where one fuse is 5 percent low and the other 5 percent high, then the higher resistance fuse carries 10 percent lower current and the lower resistance fuse carries 10 percent higher current.
With two 5 amp fuses with total current 10 amps, that would mean 4.5 amps in one and 5.5 amps in the other.
Note that this is independent of the actual fuse resistance so it makes it a little easier to understand. In other words, if we assume 0.1 ohms each (perfect) fuse it's the same as if we assume 0.01 ohm each.
If the fuses could vary by a total of 20 percent instead of 10 percent, then that means 4 amps in one fuse and 6 amps in the other. It's that simple. This means it might help to measure the resistance of several fuses and compare. Of course the variation in resistance of the fuse holder and wiring has to be taken into account also.

Perhaps we could assume some type of metal if we cant find out exactly what metal it really is.

Oh yeah one more little thing...
Since there are two fuses now the heat dissipation is twice that of a single package, meaning the blow time would be roughly 4 times the rating of one fuse. Yes, that is four times not two times. That means the fault current exists a bit longer than with only one fuse and i dont think there is any way around this except maybe to preheat the fuses.
Since the time to blow is longer, that means that for an increasing current level the current could attain a much higher level before blowing than with a single fuse. That could make it blow faster than 4 times one fuse, but it would mean a much higher fault current could exist before it blew open.