Inductor current rating

I have wound an inductor for 1.5mH on ferrite core with 22SWG (0.7mm dia) = copper wire. Inductance is fine as measurement confirmed it. I sent this a= s sample to a professional inductor manufacturer to order some 50 pieces an= d he says that this wire gauge cannot handle more than 1-2A of current wher= e as my requirement is for 5A current. This came as surprise to me as the t= hickness of 22SWG wire is much thicker than the 10A fuses I have seen. What= decides the current carrying capacity of the inductor?=20

For comparision I opened pioneer and kenwood car audio head units that I ha= veat my disposal and they say 4x45W output power. They both have an LC filt= er at input of the power supply and their series inductors are also wound w= ith a 22SWG wire. To deliver 4x45watts per channel and also to cater to oth= er functions including the display the current through the inductor of the = LC filter must be more than 8-9A at car battery voltage of 12-13V. They bot= h have 10A fuse prior to the inductor.=20

Even the SWG/AWG current charts found on the net also says that 22SWG (0.7m= m dia) cannot handle more than 1-2A. Can anybody explain please?

1) How is 22SWG sufficient in these car audio systems from pioneer and kenw= ood that require close to 10A of current? 2) When they say 4X45W of RMS power, it is 180W plus say around 25W for dis= play and operation including CD motors. How are 10A fuse and 12V car batter= y supply sufficient for that wattage? 3) What else decides current carrying capacity of the inductor other than g= auge of the wire?=20

-mark

Reply to
markjsunil
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I have wound an inductor for 1.5mH on ferrite core with 22SWG (0.7mm dia) copper wire.

** What core - post a link.

Inductance is fine as measurement confirmed it.

** What core - post a link.

I sent this as sample to a professional inductor manufacturer to order some

50 pieces and he says that this wire gauge cannot handle more than 1-2A of current

** Bollocks.

What decides the current carrying capacity of the inductor?

** Two factors:

  1. Core saturation.

  2. Temp rise.

What else decides current carrying capacity of the inductor other than gauge of the wire?

** The gauge affects the resistance and hence the temp rise.

Did you calculate the air gap for the DC current level you need?

WHAT is the core - FFS post a damn link !!

... Phil

Reply to
Phil Allison

confirmed it.

pieces and

my requirement

is much thicker

the inductor?

formatting link
since you seem to not understand current capacity of wire. Since the 1 amp rating of 22 SWG is open air, you would have to de-rate it some when wound in an inductor.

--
I'm never going to grow up.
Reply to
PeterD

For a ferrite core, the functional operating current for DC will be limited by the core xsection, magnetic path length and gap size, as the core will saturate below 0.4T (4000g). Copper loss determines the temperature rise, if there isn't signifigant ripple voltage. You can check both either by running curent through your sample, or by calculation.

Fuse wire gauges have nothing to do with it.

DC inductors here will use powdered cores, not ferrite. Welcome to the world of audio equipment power ratings.

These wire current ratings are for a fixed temperature rise in free air. In a coil, the ability to dissipate loss is reduced due by reduced total surface area and possibly by the reduced permissible temperature rises of the component being created.

RL

Reply to
legg

"PeterD"

** The OP's recalcitrant winding person * seems * to be applying a rule of thumb more applicable to an AIR cored inductor of about 1.5mH.

Even then, his 1A to 2A limit for 0.7mm wire is very conservative.

.... Phil

Reply to
Phil Allison

formatting link

The one in with white color core is from Pioneer car radio and the one with black color core is the one that we winded. It has outer dia of 20mm, inner dia of 10mm and height of 8mm.

-mark

Reply to
markjsunil

copper wire. Inductance is fine as measurement confirmed it. I sent this as sample to a professional inductor manufacturer to order some 50 pieces and he says that this wire gauge cannot handle more than 1-2A of current where as my requirement is for 5A current. This came as surprise to me as the thickness of

22SWG wire is much thicker than the 10A fuses I have seen. What decides the current carrying capacity of the inductor?

my disposal and they say 4x45W output power. They both have an LC filter at input of the power supply and their series inductors are also wound with a 22SWG wire. To deliver 4x45watts per channel and also to cater to other functions including the display the current through the inductor of the LC filter must be more than 8-9A at car battery voltage of 12-13V. They both have 10A fuse prior to the inductor.

dia) cannot handle more than 1-2A. Can anybody explain please?

that require close to 10A of current?

and operation including CD motors. How are 10A fuse and 12V car battery supply sufficient for that wattage?

of the wire?

#22 has a fusing current of 41 amps. Open air, it will be fine at 5 amps.

What matters in an inductor is how heat from the inner windings manages to diffuse out to the surface, and how the surface is cooled. The easy thing to do is test your prototype inductor and see how hot it gets at 5 amps, mounted as it will be in use.

You can measure the resistance for fun. It will increase about 0.4 per cent per degree C rise.

The specs on audio gear are pretty much all lies.

--
John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   
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Reply to
John Larkin

Can't see the link, but you're not going to be able to build a 1mH, 5A inductor on a core of that size, no matter what core material is being used (not even gapped amorphous $$).

Typical sizing of commercial builds, using powdered cores:

formatting link

If all you're doing is duplicating the Pioneer application, then perhaps examining the original part and it's intended function more closely will make the task easier.

RL

Reply to
legg

>
formatting link

The inductor in pioneer is part of LC filter at power supply input to filter out automotive transients on 12V line.

Is there a way to calculate current saturation point from the core dimensions , wire dia and no of turns or I need to have measurement setup to do that?

-mark

Reply to
markjsunil

Wire diameter is immaterial to core saturation. Wire diameter is what makes the wire melt. Core saturation is what makes the inductance go down with high current (which, depending on your circuit, may lead to even higher currents and the wire melting, but...)

Core saturation (which is the only place I can think of "saturation" as making sense here) is a strong function of the number of turns, the core dimensions, and the core material (and a weak function of how the turns are arranged). If the application is critical, you do _not_ want to use mystery cores from junk radios -- you want to buy cores with known characteristics.

If you can find Amidon on the web go looking through their site -- they used to have some nice equations that would let you figure out that sort of thing -- IF you know the core material.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
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Reply to
Tim Wescott

ia) copper wire. =A0Inductance is fine as measurement confirmed it. I sent = this as sample to a professional inductor manufacturer to order some 50 pie= ces and he says that this wire gauge cannot handle more than 1-2A of curren= t where as my requirement is for 5A current. This came as surprise to me as= the thickness of 22SWG wire is much thicker than the 10A fuses I have seen= . What decides the current carrying capacity of the inductor?

haveat my disposal and they say 4x45W output power. They both have an LC fi= lter at input of the power supply and their series inductors are also wound= with a 22SWG wire. To deliver 4x45watts per channel and also to cater to o= ther functions including the display the current through the inductor of th= e LC filter must be more than 8-9A at car battery voltage of 12-13V. They b= oth have 10A fuse prior to the inductor.

7mm dia) cannot handle more than 1-2A. Can anybody explain please?

nwood that require close to 10A of current?

isplay and operation including CD motors. How are 10A fuse and 12V car batt= ery supply sufficient for that wattage?

gauge of the wire?

per ampacity chart 22 Awg used as chassis wiring, maximum is 7 amps used as power transmission, maximum is 0.92 amps to get above 5 A, for power transmission supposed to use 14 Awg.

Calculate the heat you generate [Q of coil] and be certain you can get that heat out.

If a wire is properly heat sinked, you can run a LOT through it. In IC's current density in those bonding wires is equivalent to running

100A through a 36 Awg wire!
Reply to
Robert Macy

Your u0 is off by a factor of 10 (10E-7 means 10 * 10^-7).

A good approximation is 1.256 microhenries / meter.

--

Tauno Voipio
Reply to
Tauno Voipio

out automotive transients on 12V line.

To calculate the DC bias on an inductor you can use the simplified formula:

B = u0 x ur x N x I / le

B = flux density in Teslas u0 = permeability of free space (4 x pi x 10E-7) ur = relative permeability of the material (from mfr of core matl) N = number of turns I = current in Amps le = magnetic path length in meters

For a high permeability core material with a gap, this becomes

B = u0 x N x I / lg

where lg is the gap length alone, again in meters.

As saturation approaches, ur will approach unity and lg will begin to look like le. Because inductance is also dependent on permeability, ALL forms have an inductance x current limitation. These limits are illustrated in a core mfr's Hanna curves. Worth googling or fetching from a mfr's catalog.

If the ripple voltage being filtered is high, the current through the choke will vary, and so will the flux. Large flux excursions produce core loss that will contribute to temperature rise of the part.

- To calculate the first approximation dc copper loss, you calculate the length of the conductor, then the conductor resistance (tables will give ohms/meter if you're lazy). Losses are straight I^2 x R.

- To calculate the temperature rise, divide the total dissipation (milliwatts) by the approximate shape surface area (cm^2) to get rise above ambient in degreesC (+/- 10%), in free/unforced air.

You can ignore the presence of a hole in a toroid structure or unfilled winding area space, in determining dissipator shape.

RL

Reply to
legg

"Robert Macy"

per ampacity chart 22 Awg used as chassis wiring, maximum is 7 amps used as power transmission, maximum is 0.92 amps

** Wot utter bollocks.

.... Phil

Reply to
Phil Allison

formatting link

** Nothing shows up.

** I asked for a link to the core.

Did you BUY one or find one in junk pile?

What was your " winder" going to use ?

... Phil

Reply to
Phil Allison

Not disagree with that comment about the values listed, but my comment that it came from an ampacity chart is true the chart is shown here

Copper Wire Gauges and their Electrical Current Load Limits Page 1. Diameter mm. Ohms per km. OOOO. 0.460. 11.684. 0.049. 0.161.

380.000. 302.000. 107.219. OOO. 0.410. 10.404. 0.062. 0.203. 328.000. 239.000 ...
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-

shown as page 1 with some table and NO explanation.

Reply to
Robert Macy

copper wire. Inductance is fine as measurement confirmed it. I sent this as sample to a professional inductor manufacturer to order some 50 pieces and he says that this wire gauge cannot handle more than 1-2A of current where as my requirement is for 5A current. This came as surprise to me as the thickness of

22SWG wire is much thicker than the 10A fuses I have seen. What decides the current carrying capacity of the inductor?

haveat my disposal and they say 4x45W output power. They both have an LC filter at input of the power supply and their series inductors are also wound with a

22SWG wire. To deliver 4x45watts per channel and also to cater to other functions including the display the current through the inductor of the LC filter must be more than 8-9A at car battery voltage of 12-13V. They both have 10A fuse prior to the inductor.

dia) cannot handle more than 1-2A. Can anybody explain please?

that require close to 10A of current?

display and operation including CD motors. How are 10A fuse and 12V car battery supply sufficient for that wattage?

gauge of the wire?

Audio requirements are nothing like the max power quoted. Even if the amp could deliver 4x45 watts true RMS ( which is very doubtful, knowing how audio power is inflated ), the actual power load even on loud typical audio would likely be no more than 1/10 or less of that, else there would be severe clipping on peaks. So they can get away with much lower rated inductors, on average. Even your estimate of 25W for display and other ancilliaries sounds much too high. My guess would be a couple of watts at most.

Yes there is derating of the wire in a coil, since current carrying capacity of a wire is usually taken in 'free space'. But the audio limitations far outweigh the winding wire derating.

--
Regards,

Adrian Jansen           adrianjansen at internode dot on dot net
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Reply to
Adrian Jansen

"Adrian Jansen"

** The current capacity of copper wire depends on the coating material - ie what temperature can the coating stand.

For example, PVC coatings can stand from 75C up to 105C, while silicone, Kapton and Teflon may go to 250C or more.

Enamel coated winding wire varies from 90C to 220C.

For the same copper cross section, the current rating may vary over a range of 10:1 when generous safety margins are included as is the case with AC power cables.

For a ferrite cored inductor, continuous ( DC or low frequency) current handling will depend on core saturation mainly, then the grade of enamel used and finally the Curie temp of the particular ferrite.

If cool air is flowing over the core, the safe current rating can be double or more compared to still air.

... Phil

Reply to
Phil Allison

I stand corrected. Not good enough that I alone knew what I mean. Should have been

u0 = permeability of free space (4 x pi x 1E-7)

I prefer this to any four digit approximation. I already have pi on call and like to stick to MKS. Units are henries per meter, here.

RL

Reply to
legg

That's right. The approximation is usually good enough for practical calculations, as the relative mu of magnetic materials is hardly ever good for more than two digits.

--

Tauno Voipio
Reply to
Tauno Voipio

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