size fuse for load rather than overcurrent protection

[lots o tots]

Hello,

i dont know how (and concerned it may be too time consuming) to build an overprotection circuit for a load controlled by an arduino pro mini with other peripherals attached to the arduino.
I figured, the next easiest thing to do was to use a fuse in the circuit, but i dont know how to size the fuse.

The power supply for the load will be an old computer laptop power supply.  this one is rated 19.5V at 3.5V max.  i am expecting the load drawing to be about 0.5-2 amps at that voltage.  I am not sure at this point.  the load is a 60 watt household incandescant light bulb, but as i understand it, the resistance changes with voltage, so i don't know what hte resistance would be and cannot calculate this.

as a side note, should i have something in place  (another fuse) to protect the arduino and other peripherals?
is all this even necessary?
if you think there is a much easier solution, please let me know.
-LOT

[rstofer]

How do you run a 120V incandescent lamp on 19.5V?  Certainly not anywhere near full brightness...

The hot resistance is easy to calculate:  P = E² / R.  For a 120V lamp, E=120V and P=100W (say) so R = 144 Ohms

The cold resistance you can measure with a DMM.

You might just use linear interpolation to get a rough guess of the resistance at 19.5V - or just measure the current when the lamp is getting 19.5V.  Current will give you a better estimate.

There's no reason not to add fuses.  Either way, one versus two, is better than nothing.

[lots o tots]

How do you run a 120V incandescent lamp on 19.5V?  Certainly not anywhere near full brightness...

The hot resistance is easy to calculate:  P = E2 / R.  For a 120V lamp, E=120V and P=100W (say) so R = 144 Ohms

The cold resistance you can measure with a DMM.

You might just use linear interpolation to get a rough guess of the resistance at 19.5V - or just measure the current when the lamp is getting 19.5V.  Current will give you a better estimate.

There's no reason not to add fuses.  Either way, one versus two, is better than nothing.

the purpose is to draw a load, not to light anything.  from what i understand, the resistance significantly increaes when the metal is hotter.  i figured at 19.5 volts, the resistance value would be someone in between the hot and cold values, but probably much closer to the cold.

i will definatley measure it.  but when adding a fuse, i have to consider inrush current, maybe other factors, so i am not sure which curren tvalue to use.

[tszaboo]

Fuse is not used to protect electronics.
Fuse is there to protect cabling to not catch on fire.

[lots o tots]

Fuse is not used to protect electronics.
Fuse is there to protect cabling to not catch on fire.

part of the concern is that the load is going to be placed inside a box.  the box will be maintained at a temperature above room temperature.  the load will shut off and on.
I am just trying to determine if any safety needs to be built into the circuit.  I was going to size the cables properly.  I thought the fuse could provide an overcurrent protection.  is this not the case?

[ovnr]

I am just trying to determine if any safety needs to be built into the circuit.  I was going to size the cables properly.  I thought the fuse could provide an overcurrent protection.  is this not the case?

You size the fuse to protect the cabling, and make sure the cabling is generous for the load. Fuses are not meant to be precision devices, and a 1A fuse can be perfectly fine with 2A for a surprisingly long time, and conversely, blow at 0.9A if you keep it running for long enough. Thus, if your load is 1A, you'd stick in something like a 2A fuse, and cabling rated to more than that.

If you want to protect your device from a fault condition that isn't massively excessive current draw (5A for a nominally 1A load, for instance), you need something else. If your worry is overheating: Get a thermal fuse, or monitor the temperature somehow. If you're specifically worried about the current: Get an electronic fuse; it'll have a sharp cutoff point.


This is a classic case of "I am doing something and X sounds like a good solution, and people say nice things about X so I'm going to do that. You cannot seriously be saying that X isn't going to do the thing I want it to do?". Tell us what you're doing instead of (or in addition to) how you want to solve it. In this case: What's the load, and what's the safety concerns with it?

[rstofer]

You can easily get the inrush current for the lamp from the cold resistance I = V/R
You probably can't get the inrush current for the PS unless you have some decent instrumentation.
I would probably let the PS take care of itself and just fuse the circuit connected to the lamp.  Maybe 125% of the inrush current.
If the fuse doesn't last, increase it a bit.

[Wimberleytech]

Whatever you do, at least buy 100% tested fuses.

[Zero999]

Hello,

i dont know how (and concerned it may be too time consuming) to build an overprotection circuit for a load controlled by an arduino pro mini with other peripherals attached to the arduino.
I figured, the next easiest thing to do was to use a fuse in the circuit, but i dont know how to size the fuse.

The power supply for the load will be an old computer laptop power supply.  this one is rated 19.5V at 3.5V max.  i am expecting the load drawing to be about 0.5-2 amps at that voltage.  I am not sure at this point.  the load is a 60 watt household incandescant light bulb, but as i understand it, the resistance changes with voltage, so i don't know what hte resistance would be and cannot calculate this.

as a side note, should i have something in place  (another fuse) to protect the arduino and other peripherals?
is all this even necessary?
if you think there is a much easier solution, please let me know.
-LOT

I presume you mean the power supply gives 19.5V and is rated to 3.5A?

If the load draws a maximum of 2A, then the internal protection circuitry inside the power supply should provide enough protection. Add a 3A or 4A fuse if you like, but there's no guarantee it will blow, even if there's a short circuit.

[csar]

Is the fuse of the multimeter used to protect electronic components?

[dmills]

Only from gross overloads, and even that is not its primary purpose.

The fuse is mainly there to protect the user and to a lesser extent the test leads from massive overcurrent conditions once the electronics has failed catastrophically or if the user is trying to measure the current of a voltage source (Been there, done that!).

The things are not in any real sense precision devices and are often under sized by people who do not understand them, use them to protect cables and limit I^2T let thru that is what they are good for.

If overheating is a concern, use a thermal fuse (quite a different thing) rated for well above your intended maximum temperature.

Regards, Dan.