# Help designing an Inductor

Discussion in 'General Electronics Discussion' started by localbroadcast, Mar 3, 2015.

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Mar 3, 2015
Hello! I am new to this board, but not new to electronics! I have a pretty straight forward question, and it should be pretty easy for someone to answer it for me.. I hope that's the case anyways.

Basically, I need to build myself an inductor, because a circuit I am working on requires one with specific properties. I can't find a suitable inductor at a price point that I find reasonable, so I am going to build my own.

It needs to have an inductance of 12mH, or 0.012H. It needs to handle a peak current of 4.5 amps. It can't reach saturation of it's core before 5.5 Amps. I want it to have as low as a resistance as possible.. naturally

It will be exposed to a voltage of about 150VDC that will be pulsed on and off at 100khz at its fastest. Since the voltage is not AC, I don't think the core will need to be laminated because I don't think there will be any eddy currents caused in the core.

Basically I need help finding a formula that will help me decide how many turns, the cross sectional area of the core, the length of the coil, etc. for a DC inductor. All the formulas I've been able to find deal with the frequency of the circuit, the rate of change of current through the inductor, the time constant, etc. I need a formula that deals with a constant current, not A.C, and no time constant.. basically assuming the thing is on all the time. I will probably be using a solid steel core, and I need it to operate in the linear zone of the BH curve up to 5.5 amps.

Any help would be appreciated.!!

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3. ### davennModerator

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Sep 5, 2009
Hi
welcome to EP

everytime the current peaks and falls, a magnetic field will do the same, setting up currents in the core

at 100kHz, you don't want a solid or laminated iron core. You need to use a ferrite core

its NOT a constant current, you have already said its pulsing at 100kHz

You need to consider duty cycle ie. is it a square wave ? 50/50 cycle or something different

Also tell us some more specifics of your project and where in the circuit this inductor will be used

Inductors of that voltage and current capability are not uncommon, used in many switchmode power supplies

cheers
Dave

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Mar 3, 2015
yes, i am well aware of this. the reason i am tryimg to find the formulas for a constant current inductor is because i want to calculate an inductor that will not reach saturation of its core below 5.5 amps, regardless of the time this current is passed thru the inductor. yes my circuit will reach 100khz at times, but sometimes it wont pulsate this quickly. sometimes the inductor will reach full charge and i want to make sure its not saturated when this occures.

5. ### davennModerator

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Sep 5, 2009
so you still haven't told us what this circuit is

6. ### BobK

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Jan 5, 2010
And I predict he will not.

Bob

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7. ### davennModerator

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LOL I predict you will be right and this is where the thread will end

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
If an air core is used, the inductor will not saturate.

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Mar 3, 2015
Well that's what you get for assuming! Maybe I'll let you guess.. I'll tell you if you're right tho! Here's a couple more hints.. The purpose of the inductor is to keep the current at a constant value.. It is switched off and on with a mosfet, and has another synchronized mosfet that does the inverse of whatever the first mosfet does, acting like a freewheeling diode for the inductor, but has less power losses than a diode would..

Here's what I've found so far in my research..
β = Flux density (T), Φ = Flux (Wb), A = core cross-sectional area (m2), H = Magnetic field strength (AT/m), N = # of turns in coil, I = current (A), ℓ = axial length of coil (m), L = Inductance (H), μ = permeability (H/m)
β = Φ / A
H = [ NI ] / ℓ
L = [ N2μA ] / ℓ where μ = μoμr
μo = 4π x 10-7 (perm. of free space)
μr = relative permeability of core material
μ = β / H
β = [ μNI ] / ℓ​

If I take the formula for flux density:
β = [ μNI ] / ℓ and rearrange it to ℓ = [ μNI ] / β
And then I take the inductance formula:
L = [ N2μA ] / ℓ and rearrange it to ℓ = [ N2μA ] / L
Since both equations now solve for ℓ, we can say that:
[ μNI ] / β = [ N2μA ] / L
This simplifies down to:
NA / L = I / β
Which rearranges to:
β = IL / NA
Since we have another formula solving for flux density:
β = Φ / A
We can combine the two equations making:
Φ / A = IL / NA
Which rearranges and simplifies to:
L = ΦN / I
I am hoping that this formula for L is accurate.. because I figured it out myself and it may be wrong.. for some reason I think N should be squared, but that's not how the equation came out.. so I'm trusting the algebra... Anybody wanna confirm and double check my work?

From a βH curve I found online for solid steel, I found that the maximum value I would want for β is about 1.5 T, because higher than this, the curve starts to level out and the material becomes saturated.. So keeping it below 1.5 is the goal.
I also know that the max current through the inductor would be about 5.5 A. So at 5.5 A the flux density would be 1.5 T.

Here are the values that are known so far:
βmax = 1.5 T (From βH curve from internet source. Will be cross-referenced before actually building)
Imax = 5.5 A (max current should only ever reach 4 amps, but an error margin is built in to the design)
L = 0.012 H (this inductance was previously calculated to keep the current ripple to an acceptable range)
A = 0.0001 m2 (this area was chosen because it seems like a reasonable size that I can find or build easily)
ℓ = 0.05 m (this length of coil was chosen because it seems like a reasonable coil length)

Knowing these values, we can solve for the needed unknowns:
β = Φ / A
1.5 = Φ / 0.0001

Therefore, Φ = 0.00015 Wb

L = ΦN / I
0.012 = [ 0.00015 x N ] / 5.5

Therefore, N = 440

Assuming all the formulas I've used are correct, this means that I could build my inductor to have a core with a cross sectional area .0001m2, a coil length of 5cm, a core of steel, with 440 turns. I would use a copper conductor of a gauge thick enough to handle the 5.5A current without a large amount of resistance while still being small enough to wrap around the core. That being said, 440 turns would most likely prove difficult to actually wrap around the core that I have described. That many turns will also present a larger amount of resistance than desired.

As noted, the area and coil length were estimated without much knowledge to base it on.. The cross sectional area for instance was made 0.0001m2 because this seems like a size that is small enough to get the materials at a reasonable price, and large enough to be able to put it together without a microscope. Since these values aren't based on any specific requirement of the circuit, I can change them as I see fit. This will allow me a way to get the number of turns from 440 down to something reasonable like 25 or 50. I will also search for some cores that are already built and have a known BH curve, possibly of ferrite like was mentioned earlier by someone else. If I find one that is a reasonable price and size, I can redo my calculations off of those dimensions.

As long as my formulas check out... I have the hard part done. Now all I need to do is tweak the dimensions to get the numbers to work out how I like

Any thoughts??
Find attached the bh curve I used. If anyone has better BH curves that include more materials, I'd love to see them. Thanks for the support team!

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10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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So its a buck or boost constant current driver with synchronous rectification.

The current through the inductor will not be constant. It will vary between two values which straddle your desired current.

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Mar 3, 2015
Yup, you guessed it. It's a DC - DC buck converter, constant current driver. Yes, it will straddle 2 current values which will appear as a ripple to the load. Depending on the value of the inductor, this ripple can be minimized, thus the importance of the inductor I am planning to make. I didn't want to get into the details of the circuit because I wanted to stick to the question I've asked by starting this thread... Too many times I post a simple question, and instead of helping me solve the specific question, people want to delve into the other aspects of the circuit which I have already solved or plan to tackle at a different time.. I know it's hard to answer a simple question when you don't know the bigger picture... Just I know how people tend to stray away from the actual problem I'm working on at the moment and try to find issues with parts of the project that really don't relate to the issue at hand.. It just makes for a thread that really never answers the original problem, and just goes off in a direction that really doesn't help anyone..

That being said.. using an air core would give the same magnetic properties as when a ferrous core gets to the saturated point, except that it behaves this way at all stages of induction.. Not just at high currents. So it would be like using a core that is saturated 100% of the time. Not really what I'm after.. But thanks for thinking outside the box.

Any insight as to weather or not the formulas I used in my calculations are accurate?

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Your understanding of an air core is incorrect. It will maintain the same inductance at any current up to the point the wires melt.

I made a couple of switchmode buck regulators using air cored inductors precisely because the peak current was not well defined and I didn't want the core to saturate under high transient loads.

However, since we now know that your design is constant current, as long as the rest if your design is capable of cycle by cycle current limiting you should not have to worry. Simply design the inductor for some suitable factor greater than your regulated current.

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Mar 3, 2015
No.. I'm pretty sure I've got it understood pretty well.. I don't think I explained my thoughts correctly. The permeability 'μ' for air is constant no matter what the current, whereas the permeability of a material like steel will reach a point where it cannot hold any more flux per unit area. It's like the magnetic field can't be crammed in there any tighter. You can increase the current, but you aren't getting any more magnetic field generated for that extra power exerted. With air, you don't reach this limit.. you can increase the current endlessly and you will still get an increased flux. The benefit of using a core like steel is that with a relatively small increase in current, you get a large increase in magnetic flux created. With an air core, the increase in magnetic flux is a more gradual increase with current. So for the same value of power input, you will be getting more magnetic flux generated with a steel core, and thus more inductance... thus, a more efficient current regulating device, given that you are operating within the range of the core prior to saturation.

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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No, that's false. It all comes down to inductance. You need more turns with an air core, but the degree of regulation is the same. You will have more losses due to the lack of magnetic shielding and the potentially higher resistance, but aside from this the major difference is the core doesn't saturate.

In any case, you know the max current to a reasonable level of certainty so it is not an unknown. Just use that to determine the requirements of the core.

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15. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Oh, I think your error ( or one of them) is where you end up with inductance being determined by current.

It is proportional to the area and the square of the number of turns, but not current.

16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Mar 3, 2015
That is true if you are using an air core.. But if you are using a ferrous core.. the inductance most definitely is influenced by the current. Check out a BH curve.. the H stands for ampere turns per meter. That means the number of amps x the number of turns / the length of the coil. More current, higher H value. As H changes, so does the permeability. "μ". Inductance is influenced by the value of μ. In an air core, μ is constant. In a core of any number of other materials with a BH curve, current will influence the inductance. Why else would people want to avoid the saturation point of the core? It's because after the saturation point, more current does not equate to more magnetism.. thus, it is wasted current, and efficiency takes a nose dive.

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Mar 3, 2015
Here's a quote from another forum just to further prove my point.
"
Of course if you increase H further on (by exciting the core with more current), the curve flattens out, showing the nonlinear relationship with B, hence mu will also change, it will reduce to a lower value, it is said: the core starts saturating.

rgds
unkar"

19. ### BobK

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Jan 5, 2010
If you are needing 12mH for your inductor, you are probably using too low a switching frequency. IC switch mode controllers use frequencies in the 100s of KHz and require only 10s of microhenries for the inductor. It is much easer to find a 47uH inductor than can handle 5.5A than it is to find (or make) a 12mH one.

Bob

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20. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Sure, but the intent is typically to use a core material with low hysteresis, and in the portion of the operating region where the curve is most linear.

Hence my suggestion that since you know the current, this is no longer a concern for you.

(unposted from yesterday)