Correct Fuse Rating for 30W CFL Lamp With Power Factor Consideration?

Hello I am trying to work out what fuse i need to individually fuse a 30w cfl lamp

I thought that it would be power divided by voltage equals current

(P) 30 / (V) 240 = (I) 0.125

so i thought 0.125 amp delay fuse would be the right one allowing the extra voltage for the light ignition thingy.

Then i read somewhere that i should include the power factor in this calculation.

The power factor can be as low as 0.5 on cfls i have read. So does that mean i double the current of my fuse or half it?? or was i right originally?

Any help would be much appreciated

You may be interested in "this thread(multiple question marks)":http://diablo.incgamers.com/forums/archive/index.php/t-547742.html

I would do two things, one - you need to measure the true power through the circuit - which is the fusing current. you might want to go 2x on the normal current and figure for inrush so a slo-blow fuse is in order. Now... I would also call littlefuse and buss and ask them about this because the apps engineers likely have a better rating system.
measure the load with a power analyzer say a voltech PM1000 - www.voltech.com
hope that helps

Perhaps these will shed some light!!!

* *http://www2.electronicproducts.com/Selecting_the_right_fuse-article-farrschurter-jun2007-html.aspx*
* *http://www.bdfuses.com/derating.php*
* *http://www.digikey.com/PTM/IndividualPTM.page?site=us&lang=en&ptm=25015*

Found by doing a Google search on "how to select the proper fuse"

[You, too, can Google search with impunity!!!!!!!!!!!!!!]

Thanks kevin.

I dont have access to one of those analyzers im afraid. But i can take some current readings with something a bit simpler and see what my results are and take it from there.

Thanks again

I think the intent of multiple questions marks is to indicate severe perplexity--not to be confused with intense inquisitiveness.

A fuse is rated not only by a current rating but by I squared T. The fuse needs to be robust enough to not open during normal operation and circuit startup yet it needs to be such that it will open during catastrophic circuit failures.

The way to find the I squared T and current is to measure the current using an oscilloscope and a current probe. This should be performed for normal operation, during startup, and during single component failures of the circuit. The single component failures include any one component opening or shorting.

The first, essential thing you must know is that in general, fuses are there to save you from a fire, not to save the circuit. A fuse is just a thin wire that must heat up and melt or vaporize to open. A failure must have already occurred, so you won't save whatever failed, and generally a semiconductor under very high current will be damaged long before a fuse will blow.

The second thing for you to know and understand is that power factor affects the -power-, but current is still the same.

A circuit drawing 10A at 10V (both rms) with a power factor of only 0.1 may be using only 10W of power, but it is still drawing 10A of current. That is only 100% true if you stick with clean sine waves for voltage and current.

However, with CFLs the situation is more complex. Most CFLs have high harmonic content. The internals use a diode and capacitor to convert to DC, causing the only current drawn to be at the peak of the incoming voltage. So CFLs don't have a power factor as much as they have nonharmonic content. Since a fuse takes a lot longer than a half-cycle to blow, if it draws 1A during the peak cycle but averages out to 125mA it probably won't blow the fuse.

The fuse should be rated at 2x the load, or it is going to fail after a relatively short time. There is also the matter of inrush current when it initially charges the internal capacitors.