Amt of current to saturate ferrite rod

Yes, to the degree that you can "saturate" any material. Isn't saturation an asymptotic property that can't be fully achieved this side of infinity? So, depends on your definition of saturation.

I don't know if 'asymptotic' can really apply to non pulsing DC current influenced things. There are rumours of it appearing in cyclical stuff...you know...tangentially speaking.

Mind you, if you take numerous periods of infinity and stick them end to end, a definite pattern would certainly show up. Neglect the half life of the rod (and any friction.). Me? I don't have time for such foolishness, so I'm graphing only *one* infinity.

I do know for certain that unapplied dc is a completely different animal than the 'applied' dc being assumed by the original poster.

Kinda like calculus vs shop math.

mike

At least for moderate length rods (roughly 6 mm dia. x 30 mm long, often seen in computer SMPSs), the change in inductivity is about half, so you'll hardly notice it. The change is just as sharp as any ferrite inductor though.

At an aspect ratio like that, you'll probably see more gradual saturation, as the center becomes saturated first and the whole thing progressively becomes an air-cored inductor.

Depending on what you're doing, rods often aren't wound all the way to the end, because inductivity drops -- SWAG, fringing sucks down maybe half the field out past 80% from the center. You occasionally also see multilayer windings, which are great if the wire is thin, but only up to a winding height about equal to the diameter. Think of the fields of a bar magnet, they loop around pretty well from the ends, but any space within those arcs is prime real estate for coil winding purposes.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

"Paul Nelson"  wrote in message 
news:4ed80cd2.925531@news.tpg.com.au...
> Is it technically possible to saturate a garden variety ferrite rod
> (10mm dia. x 120mm long) with a single layer winding?
>
> This assumes applied DC.
>
> Paul Nelson

snipped-for-privacy@lightrax.com (Paul Nelson) wrote in news:4ed80cd2.925531 @news.tpg.com.au:

Yes.

But I didn't know that even those can be hit with AMT :-)

--
SCNR, Joerg

http://www.analogconsultants.com/

If you just want static saturation then use neodymium magnet. The electromagnetic saturation will consume some power to the point that the winding will start decapitating some heat so to avoid that multi layering with thicker wire must be used.

Mathew Orman

--
No sense losing your head over a little heat, eh?

As others have pointed out, yes.

Also, as others have pointed out, how will you define saturation?

Addressing your subject line:

1 amp. (assumes 100 oersteds) (955 turns of wire over the 120mm) or 10 amps. (assumes 100 oersteds) (95 turns of wire over the 120mm) or etc.

Of course, the 100 oersteds is probably overkill for what you're asking, but who can tell at this point?

And if you need more current, use an edgewound ribbon conductor. I agree, the question is vague.

You might apply the formula:

B = 4 x pi x 10^-7 x N x I / le

B = saturation flux density in Teslas ( ferrite ~ 0.3) N = number of turns I = DC current in Amps Le = magnatic path length in Meters - ( length of air path between rod ends)

As the ferrite saturates, le doubles, before B can continue to climb.

In practical situations, you'd be more concerned with temperature rise in the copper. Because of the long gap, inductance and Q values are pretty low, until you hit RF frequencies, but are better than air-cored structures of comparable size.

RL

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