saturation Current of Toroidal Inductor

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Saturation and permiability are quite independent.
Both depend on the characteristics of the particular material, which we cannot be sure about.
All you can do is run an actual saturation test.

That is not difficult to do, and your 3 amp bench supply will be ideal to test to several tens of peak amps.

Here is a post in another thread thread in which I tell how to do it.
Scroll down to post #24
https://www.edaboard.com/threads/350084/
 
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I am using same circuit to test saturation of inductor as given below

What should be frequency of Input signal to MOS
 

Frequency will depend on the inductance and the dc voltage applied.
On time will need to be very short compared to an extremely long off time.

You really need a digital oscilloscope to monitor this because the on time will be very short, and the off time so long it will be very difficult to see with an analog scope.
 

Warpspeed said:
If that is so, Bmax would probably be 300mT and that makes it far from ideal to use as a dc choke.
Click to expand...
I have tested it and it shows saturation current for the core is showing about 16A Do you think my test results are Ok



-->Can you Please tell me how to calculate saturation current from datasheet
Lets say I have MS-184060-2
**broken link removed**
What will be saturation current for this core
 

Second curve down is the magnetising curve.
Saturation is about 1000 Orsteads
Path length is given as 10.743 cm

And there is a formula for working out N I which is number of turns and amps.
 
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M.Rehan said:
I have tested it and it shows saturation current for the core is showing about 16A Do you think my test results are Ok
Click to expand...

It does not depend on the current; it depends on the N (no of turns in the solenoid) and the magnetising current (I) and more accurately, their product. They produce a magnetic field that induces the magnetization in the magnetic material.

You will find B-H curves in the datasheet but not the current.
 
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I have read somewhere that saturation occurs when Permeability µi is about to 90% of of initial Permeability?
)
 

I have read somewhere that saturation occurs when Permeability µi is about to 90% of of initial Permeability?)
Click to expand...

Learning also involves unlearning. Good knowledge must give space to better. Rational thinking is a great asset. Consider the initial permeability is 100. 90% of initial permeability will be 90 - the core has already saturated!!

Perhaps you meant something else. If what is written is not what is meant or intended, we shall only have confusion. Permeability is often given as a derivative.
 
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M.Rehan said:
I have read somewhere that saturation occurs when Permeability µi is about to 90% of of initial Permeability?
)
Click to expand...
Its like politics and religion............
There are just rapidly diminishing returns.

90% or 95%, or 99.9999999999999999999999999%
When you get there.
There just ain't no more.

But probably anything over about 90% means the inductor is virtually saturated.
And you can stick a fork in it, cos its done.
 

Need to contradict some previous statements about core saturation which sound somehow ignorant. Previous posts referred to an iron powder core which has a rather soft saturation characteristic. Small powder core inductors are often operated at 90 or 80 % of initial permeability. The other point with powder cores is that the useable flux is often limited by core losses rather than saturation.
 

FvM said:
Small powder core inductors are often operated at 90 or 80 % of initial permeability.
Click to expand...

Warspeed said that saturation value of magnetic field H will be taken as 1000 Oersted is he right?
lets say I have 20 turns
then I saturation = H*le/0.4*Pi*N
le = 10.743cm
N = 20 turns
Hsat = 1000 as said by warspeed
I sat = 427 A
Am I right? this core will saturate at 427 A current for 20 turns





Or

According to FvM
I saturation = H*le/0.4*Pi*N
le = 10.743cm
N = 20 turns
Hsat = 30 as said by FvM (10% decrease of initially Permeability)
I sat = 12 A

Which one is correct?
 

It is getting near saturated at 200 Oe, H = I t / lg (SI units) where lg is the mean mag path of the toroid, I = amps, t = turns,

1 Oe = 79.58 At/m (H), so 200 Oe = 15916 At/m (H) plug into above formula to get I.

- - - Updated - - -

[answer I = 85A for near saturation]
 

M.Rehan said:
Which one is correct?
Click to expand...
Ninety percent of initial permiability remaining is nowhere near saturation.
Its only a ten percent loss of inductance.
And it is sometimes a typical design figure for many things.
Fifty percent of initial permiability is also a common design figure for swinging chokes.

Saturation is TOTAL saturation, or hitting a brick wall at 100 % total loss of all inductance.
Because the saturation characteristic is quite soft, it probably never actually gets to total loss of all inductance.
So only ten percent of initial permiability remaining is probably regarded as being pretty much saturated.

Yes you could probably run 400 amps through it, and the wire would get quite hot, and it would still have one or two percent of the initial inductance remaining.
But is that total saturation ?

Its why you need to look at that curve and decide for yourself how far you want to run it up, and how much loss of inductance you are prepared to tolerate.

Here we are only talking about inductance, and not core loss, which is an entirely different subject, which is also very important.
 
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Warpspeed said:
Its why you need to look at that curve and decide for yourself how far you want to run it up, and how much loss of inductance you are prepared to tolerate.
Click to expand...
I am taking about inductor for SMPS and looking for best engineering practices for saturation used mostly

Easy peasy said:
1 Oe = 79.58 At/m (H), so 200 Oe = 15916 At/m (H) plug into above formula to get I.
[answer I = 85A for near saturation]
Click to expand...

You are right that
1 Oe = 79.58 At/m (H),

but in datasheet it is clearly mentioned that
that
H is in Oe
Idc in Amperes
 

Best engineering practice depends on the purpose, and economics.
There are no hard and fast rules about inductor design, its whatever you feel is appropriate.

Efficiency, temperature rise, total loss, and saturation limit are all limits we need to consider, but there are no iron clad rules.

Some circuits are designed to run pretty hard up into saturation in order to work properly.
Some simple oscillators actually work on that principle.
Ferroresonant voltage regulators work that way, and so do some types of saturable reactors and magnetic amplifiers.
 
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Easy peasy said:
1 Oe = 79.58 At/m (H), so 200 Oe = 15916 At/m (H) plug into above formula to get I.
[answer I = 85A for near saturation]
Click to expand...


In datasheet it is clearly mentioned that
that
H is in Oe
Idc in Amperes


Please someone answer it
Question is posted in Thread #32
 

M.Rehan said:
View attachment 130416
Click to expand...
Clearly the inductance keeps falling up to and possibly beyond 1,000 Orsteads.
COMPLETE TOTAL saturation is the point where inductance effectively reaches zero.
That is zero on the vertical scale of that graph.

When a jug is full, it is full right up to the brim.
You might say anything over 90% is totally full, and that is the industry accepted norm.
But others might wish argue about that.
 

Warpspeed said:
Clearly the inductance keeps falling up to and possibly beyond 1,000 Orsteads.
COMPLETE TOTAL saturation is the point where inductance effectively reaches zero.
Click to expand...

Inductance will never become zero because the once the core is saturated, there will be no more additional magnetization and the core will effectively act like a air-core inductor. However, compared to a 100-1000 fold change due to magnetic induction, this (reduced inductance) is effectively zero.
 
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Industry standard for Imax on chokes is 10% drop point so 30 Oe for this example.
 
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