- #1
nst.john
- 163
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If I am making my own inductor following the equation:
Optimizing Inductor Design: Factors to Consider
- Thread starter nst.john
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Discussion Overview
The discussion revolves around the design and optimization of inductors, specifically focusing on the effects of using an aluminum core that is hollow. Participants explore the implications of core material properties, geometry, and frequency on inductance calculations and performance.
Discussion Character
- Exploratory
- Technical explanation
- Debate/contested
- Homework-related
Main Points Raised
- One participant questions whether the magnetic permeability used in inductance calculations should be solely that of aluminum or a combination of aluminum and air due to the hollow nature of the core.
- Another participant suggests that for small currents, the aluminum core can be treated similarly to an air core, though they express uncertainty about saturation effects.
- Concerns are raised about core losses at higher frequencies, with one participant noting that aluminum will induce a current that opposes the main magnetic field, potentially reducing inductance.
- There is a discussion on the geometry of the aluminum tube affecting its magnetic properties, with one participant proposing that it may behave like a single loop inductor.
- One participant mentions the possibility of inadvertently creating an induction furnace and warns about core losses at radio frequencies.
- Another participant advises measuring the inductance after construction, suggesting a method involving a tank circuit and resonant frequency.
- There is a suggestion to use non-conductive materials for the inductor former, which have a permeability similar to air.
- One participant expresses concern about core losses at a frequency of 40kHz, while another reassures them that for a science project, minor inefficiencies may not be critical.
- A later reply challenges the initial inductance formula, stating that the effective magnetic path length includes the return path outside the cylinder and that the N^2 term may be overestimated with a low permeability core.
- There is a caution against using a conductive mandrel, as it could short-circuit the inductor, reducing its quality factor (Q).
Areas of Agreement / Disagreement
Participants express varying opinions on the appropriate treatment of the aluminum core's permeability and its impact on inductance. There is no consensus on the best approach to calculating inductance or the implications of using a hollow aluminum core.
Contextual Notes
Participants highlight limitations in the inductance formula, including the need to consider the effective magnetic path length and the potential inaccuracies introduced by the core material. The discussion also reflects uncertainty regarding the effects of frequency on core losses.
- #2
nst.john
- 163
- 1
Sorry didn't finish, following the equation L=μN2A/l where L is inductance, μ is the magnetic permeability, N is the number of turns and l is the length, and I have a piece of aluminum as my core with a μ=1.256665x10-6, if the aluminum core is hollow, meaning a conduit pipe, when calculating the inductance would μ still be just the value for aluminum or would it be a combination of aluminum and air since the pipe is hollow. Please answer quickly and thank you for your help in advance!
- #3
Jeff Rosenbury
- 746
- 145
Aren't they the same thing?
According to the chart on Wikipedia they have the same value out to 5 decimal points. I doubt a handmade inductor will be accurate that far out. I think you will be lucky to be within 5%.
You should be able to treat an aluminum core as an air core for small currents. (I don't know if it saturates like iron, but I doubt it.) You will be adding core losses though.
According to the chart on Wikipedia they have the same value out to 5 decimal points. I doubt a handmade inductor will be accurate that far out. I think you will be lucky to be within 5%.
You should be able to treat an aluminum core as an air core for small currents. (I don't know if it saturates like iron, but I doubt it.) You will be adding core losses though.
- #4
nst.john
- 163
- 1
Alrighty thanks a lot! That should be a lot of help.
- #5
tech99
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For AC, the aluminium will have a current induced into it which have its own magnetic field opposing the main one. This will reduce the inductance slightly.nst.john said:
- #6
nst.john
- 163
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How much is slightly?
- #7
tech99
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Not sure how to calculate it. Why don't you use a plastic or cardboard former? These materials all have a relative permeability of 1.nst.john said:
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- #8
Jeff Rosenbury
- 746
- 145
Me either. 
It depends on the geometry of the aluminum tube/bar. Nominally the tube will cut through the magnetic flux just like a loop of wire. It will develop a counter flux just like a loop of wire. So it should subtract an amount about equal to a one loop inductor from your multi-loop inductor (but of the length of the tube). Of course that assumes the geometry supports the tube as a full loop cutting through all the flux, which it won't. Still in a typical application it should get the lion's share, so I'd figure it for one loop.
Are you aware of induction furnaces? It is remotely possible you might be unknowingly building one. They are actually finicky, so it's unlikely, but core losses might be a problem at radio frequencies (~100,000 Hz.).
If you are concerned about the actual value, you can measure it when you are done. This is not trivial. The easiest way I know is to build a tank circuit with a known capacitance and find the resonant frequency. Then calculate back.
Almost all non-conductive materials are the same permeability as air. Wouldn't that be easier?
It depends on the geometry of the aluminum tube/bar. Nominally the tube will cut through the magnetic flux just like a loop of wire. It will develop a counter flux just like a loop of wire. So it should subtract an amount about equal to a one loop inductor from your multi-loop inductor (but of the length of the tube). Of course that assumes the geometry supports the tube as a full loop cutting through all the flux, which it won't. Still in a typical application it should get the lion's share, so I'd figure it for one loop.
Are you aware of induction furnaces? It is remotely possible you might be unknowingly building one. They are actually finicky, so it's unlikely, but core losses might be a problem at radio frequencies (~100,000 Hz.).
If you are concerned about the actual value, you can measure it when you are done. This is not trivial. The easiest way I know is to build a tank circuit with a known capacitance and find the resonant frequency. Then calculate back.
Almost all non-conductive materials are the same permeability as air. Wouldn't that be easier?
- #9
nst.john
- 163
- 1
True. I'll try that. I'm applying for the Google science fair and need it for tomorrow so this should be good. And my frequency needed is 40kHz so would that still be a problem?
- #10
Jeff Rosenbury
- 746
- 145
Yes, 40kHz will buy you some hefty core losses. How much of a problem that is will depend on your application. If it plugs into the wall and doesn't get hot you should be fine. If it is human powered... Sweat a little more.nst.john said:
If it's just a science project, don't worry. The worst that can happen is the entire thing burns and you win the pyrotechnic award.
No one will care if it uses 1/2% more power than it should.
- #11
stedwards
- 416
- 46
L=μN2A/l is not the correct equation to use, or at least not as you might think it is.
"l" is the effective magnetic path length. It includes the return path outside the cylinder. Also the permeability over the return path is the permeability of air--which is about the same as the vacuum permeability.
There's an additional problem with the formula. With a low permeability core, not all the flux passing through one loop will pass through the other N-1 turns, so the N^2 term is too large.
I would look for an online inductance calculator, or table of various inductor geometries.
Additionally, i don't think you want to use an aluminum mandrel, or any conductive mandrel. It's effectively a short circuit, transformer secondary winding of one turn. This reduces the Q.
"l" is the effective magnetic path length. It includes the return path outside the cylinder. Also the permeability over the return path is the permeability of air--which is about the same as the vacuum permeability.
There's an additional problem with the formula. With a low permeability core, not all the flux passing through one loop will pass through the other N-1 turns, so the N^2 term is too large.
I would look for an online inductance calculator, or table of various inductor geometries.
Additionally, i don't think you want to use an aluminum mandrel, or any conductive mandrel. It's effectively a short circuit, transformer secondary winding of one turn. This reduces the Q.
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