[SOLVED] Some inductor Calculations

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Freiheitpt

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Hi there,

I wanted to ask, how can I find out what is the max operating current of a homemade inductor.
I have already a small program wich tells me how many windings it needs to have for a certain core material and size.
Is there any formula you can use wich tells you the max operating current?
(my core material is yellow/white in the color code, i think 43.5nH)

Thanks a lot,
Marc Wagner
 
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There is:

(1) safe ampacity, based on how hot your wire gauge gets with a given metal mass / layers of windings / duty cycle, etc...

and

(2) saturation current, based on how intense a flux field your core metal can support (if you use a core). There are complicated formulae for calculating this.

Link to charts of wire gauges showing ampacity:

http://amasci.com/tesla/wire1.html
 
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Thanks you 2 for answering.

Well, I would like to know the safe ampacity. With this I mean, the max operating current with wich the inductor still works as a inductor.
I know that my wire (24 AWG) is rated for 0.5/0.8 amps. but i have seen some inductors with the same or yet thinner wire and could drive something like 5 or 10 amps.
Yes I use a core with the following specs:

µi = 75
AL = 43.5 nH/N^2

Od = 0.7 inches
Id = 0.35 inches
Height/Width = 0.24 inches

for 100µH I have 48 turns (calculated)

For a frequency of 50Hz (European mains AC frequency), I get a reactance (impedance) of 31.4 mOhms

Is there anyway I can calculate at least the safe ampacity (1) with this data?


ark5230: with that link, it looks a bit strange that value.
with those calculations I get almost 9 amps for a 20 AWG wire (0.8 cm), and in the AWG table it says that this wire has a max ampacity of 2amps.
Please note that I dont want the max current the wire can handle, but how much current the inductor can handle.


Again, thanks a lot!
Marc Wagner
 

how much current the inductor can handle
Click to expand...

... to the point its inductance changes appreciably by the self heating. The later depends on many factors but in any case its current density (A/mm2) is likely much lower than of the wires transmitting power for example (as in transformers).

Kerim
 

Freiheitpt said:
I mean, the max operating current with wich the inductor still works as a inductor.
Click to expand...
To work that out, you need to know the saturation flux density (Bs) of the core.

The specifications for the core may not give Bs, but it should give the core material e.g. N87 or T35. If you know the material, you can look up Bs for that material.
 
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    KerimF

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Yes, godfreyl pointed out an important effect (saturation) when current is increased to relatively high values in case the core is not air. And this may change the initial inductance to a much greater extent than the physical deformation does due to self heating.
 

Here's a link to another wire gauge table. Its ampacity values are much different than the other tables.

**broken link removed**

Possible reason for the disparity: One source may base ampacity on a higher temperature which makes the wire get too hot to hold, while the other source may use a lower temperature where copper's electrical characteristics start to change even though it is only warm.
 

BradtheRad said:
Here's a link to another wire gauge table. Its ampacity values are much different than the other tables.
Click to expand...
I expect it depends on the application. For example with a mains power cable, the thick rubber/plastic sheathing provides good thermal insulation. At the same time the wire must remain cool enough not to melt or burn the rubber/plastic.

OTOH, in an inductor made with enameled copper wire, the cooling is better and the temperature can be much higher.
 
well, my core is aproximatly T-68 and the material is 26 (color code: yellow & white)
 

There's some useful information here: **broken link removed**, especially this PDF: **broken link removed** (see pic below).

:???: I'm not an expert at this, but as best as I can figure it out....
For your inductor, N = 48 turns and path length is about 4cm, so from the equation below, H = approximately 15*I, so I = 0.067*H.

From the graph, permeability reduces to 80% of it's initial value when H = 20, i.e. Current = 1.3A, and permeability reduces to 50% when H = 50 i.e. Current = 3.3A.

 
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    KerimF

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Well, I have read around, and, I think I got the answer.
The maximum operating current has something to do with the AWG wire table.
I noticed that all inductors are rated something between 1 to 2 amperes less than the maximum amperes rated for that wire for chassis wiring.

Could this be right?
 

Yes, that's half right. There's two limits; the thickness (AWG) of the wire, and saturation of the core.
 
godfreyl said:
Yes, that's half right. There's two limits; the thickness (AWG) of the wire, and saturation of the core.
Click to expand...

so, I have to see which is lower, the saturation of the core, or the AWG current rating. right?
the lowest value will be the rating of my core, i suppose.
 

Freiheitpt said:
so, I have to see which is lower, the saturation of the core, or the AWG current rating. right?
Click to expand...
Yes. If you're using 24AWG then that's probably the limiting factor, but you can always use thicker wire.

edit: For high power, high frequency work e.g. SMPS transformers, you also have to worry about core heating due to core losses.

If it's just for high DC current with a small AC ripple, you only have to worry about wire heating and core saturation.
 
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its only for the mains. 50hz 230V AC

(interference surpressor for dimmer - triac firing)
 

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