Magnetic field of a solenoid

[Klaus Kragelund]

Hi

I need to transfer some energy accross a platic barrier. I will be
using a solenoid on both sides of the barrier, one to transmit and one
to receive the energy

First thoughts is to use an air core to provide maximum stray field
(in a ferrite core solenoid the field will be concentrated in the
core, but somehow I think the air core is better over longer
distances)

The barrier is 2mm thick, and may use almost any circuit on both
sides. An HF oscillator on the primary and a simple diode
rectification on the secondary side, preceeded by a capacitor to
adjust the ressonance to get optimum perfomance.

But, I am in US right now, so I have none of my books and the internet
have not helped me in this matter.

The B field of a solenoid is:

B = u0 * uR* I * N

But how do I calculate the B field at say 10mm from the core in order
to be able to calculate the current in the recieving core.

Any ideas?

Thanks

Klaus

 

[Joerg]

Hello Klaus,

I need to transfer some energy accross a platic barrier. I will be
using a solenoid on both sides of the barrier, one to transmit and one
to receive the energy

First thoughts is to use an air core to provide maximum stray field
(in a ferrite core solenoid the field will be concentrated in the
core, but somehow I think the air core is better over longer
distances)

The barrier is 2mm thick, and may use almost any circuit on both
sides. An HF oscillator on the primary and a simple diode
rectification on the secondary side, preceeded by a capacitor to
adjust the ressonance to get optimum perfomance.

But, I am in US right now, so I have none of my books and the internet
have not helped me in this matter.

The B field of a solenoid is:

B = u0 * uR* I * N

But how do I calculate the B field at say 10mm from the core in order
to be able to calculate the current in the recieving core.

Any ideas?

I'd suggest pot core halves, assuming the available diameter in that
area is 10mm or more. 2mm isn't much of a gap, should not be a big deal
to bridge.

The classical way to transfer energy over a gap is a series resonant
converter. This basically "notches out" most of the leakage inductance
and results in a good net energy yield on the other side. How much
energy to do have to get across?

Like with all things there is a caveat: The regulatory guys. With a
series resonant converter you usually cannot stay within an ISM band
like 13.56MHz because it's only a few ten kHz wide. So it'll have to be
a rather low frequency.

Calculations are difficult. Could be done with lots of data from the
core mfg but it's better to test it in the lab and provide a really
healthy net energy margin.

 

[Uncle Al]

Hi

I need to transfer some energy accross a platic barrier. I will be
using a solenoid on both sides of the barrier, one to transmit and one
to receive the energy

How about 60 watts net power transfer at 7 meters separation with 40%
efficiency?

First thoughts is to use an air core to provide maximum stray field
(in a ferrite core solenoid the field will be concentrated in the
core, but somehow I think the air core is better over longer
distances)

The barrier is 2mm thick, and may use almost any circuit on both
sides. An HF oscillator on the primary and a simple diode
rectification on the secondary side, preceeded by a capacitor to
adjust the ressonance to get optimum perfomance.

But, I am in US right now, so I have none of my books and the internet
have not helped me in this matter.

The B field of a solenoid is:

B = u0 * uR* I * N

But how do I calculate the B field at say 10mm from the core in order
to be able to calculate the current in the recieving core.

Any ideas?

It's been done to a fare-thee-well over a couple of meters separation,

Google
Soljacic WiTricity 22,600 hits

http://en.wikipedia.org/wiki/WiTricity

 

[Klaus Kragelund]

Hello Klaus,















I'd suggest pot core halves, assuming the available diameter in that
area is 10mm or more. 2mm isn't much of a gap, should not be a big deal
to bridge.

The classical way to transfer energy over a gap is a series resonant
converter. This basically "notches out" most of the leakage inductance
and results in a good net energy yield on the other side. How much
energy to do have to get across?

Like with all things there is a caveat: The regulatory guys. With a
series resonant converter you usually cannot stay within an ISM band
like 13.56MHz because it's only a few ten kHz wide. So it'll have to be
a rather low frequency.

Calculations are difficult. Could be done with lots of data from the
core mfg but it's better to test it in the lab and provide a really
healthy net energy margin.

--
Regards, Joerg

http://www.analogconsultants.com- Skjul tekst i anførselstegn -

- Vis tekst i anførselstegn -

I actually already did a test with two pottet cores. See pictures:

http://www.microdesign.dk/tmp/misc_transformertypes.jpg
http://www.microdesign.dk/tmp/two_cores.jpg

This one is wound:

http://www.microdesign.dk/tmp/two_cores_gap.jpg
(bad mobile phone picture)

That can transfer plenty of power, but it will be costly (the turns
are not PCB traces, but a cobber wire)

I was looking for perhaps using a standard part like this:

http://www.microdesign.dk/tmp/SDR0805.jpg

At least for the primary side. The stray field would frindge around
the top, but I can get the manufacturer to make it more like a
straight rod to get an even field.
Then the secondary side could be either a coil like that or an air
core inductor (it has to be 90 degrees angled to the PCB to that
easier)

Actually I would like to be able to calculate the field, since I may
be using it in another project where the end goal is to transfer only
a signal with very low frequency. So perhaps a Hall sensor could do in
that app.

But close to the core it is the near field that decides the magnitude.
So its difficult to calculate

Regards

Klaus

 

[Klaus Kragelund]

How about 60 watts net power transfer at 7 meters separation with 40%
efficiency?

That sound almost Star-trek like.

I just need 100mW

Regards

Klaus

 

[Bert Hickman]

That sound almost Star-trek like.

I just need 100mW

Regards

Klaus

The MIT folks used about 400W from a 10 MHz vacuum tube oscillator to
perform this feat. For your application, two coils on silicon steel or
ferrite "C" cores should work as a simple transformer at line frequency
(albeit with relatively high leakage inductance from the air gap). Dual
resonant circuits are not necessary but could be used, at line or higher
frequencies, to improve overall performance.

Bert
--
***************************************************
We specialize in UNIQUE items! Coins shrunk by huge
magnetic fields, Lichtenberg Figures (our "Captured
Lightning") and out of print technical Books. Visit
Stoneridge Engineering at http://www.teslamania.com
***************************************************

 

[John Larkin]

Hi

I need to transfer some energy accross a platic barrier. I will be
using a solenoid on both sides of the barrier, one to transmit and one
to receive the energy

First thoughts is to use an air core to provide maximum stray field
(in a ferrite core solenoid the field will be concentrated in the
core, but somehow I think the air core is better over longer
distances)

The barrier is 2mm thick, and may use almost any circuit on both
sides. An HF oscillator on the primary and a simple diode
rectification on the secondary side, preceeded by a capacitor to
adjust the ressonance to get optimum perfomance.

But, I am in US right now, so I have none of my books and the internet
have not helped me in this matter.

The B field of a solenoid is:

B = u0 * uR* I * N

But how do I calculate the B field at say 10mm from the core in order
to be able to calculate the current in the recieving core.

Any ideas?

Thanks

Klaus

I've done this with parallel ferrite rods, with a tuned coil on each.
If the rods are several times longer than your 2 mm gap, and close to
the barrier, coupling should be pretty good.

A pair of C-cores would be even better, but bulkier.

How much power are you trying to couple?

John

 

[Vladimir Vassilevsky]

Hi

I need to transfer some energy accross a platic barrier. I will be
using a solenoid on both sides of the barrier, one to transmit and one
to receive the energy

First thoughts is to use an air core to provide maximum stray field
(in a ferrite core solenoid the field will be concentrated in the
core, but somehow I think the air core is better over longer
distances)
The barrier is 2mm thick, and may use almost any circuit on both
sides. An HF oscillator on the primary and a simple diode
rectification on the secondary side, preceeded by a capacitor to
adjust the ressonance to get optimum perfomance.

But, I am in US right now, so I have none of my books and the internet
have not helped me in this matter.

The B field of a solenoid is:

B = u0 * uR* I * N

But how do I calculate the B field at say 10mm from the core in order
to be able to calculate the current in the recieving core.
Any ideas?

Hello Klaus,

If the barrier is only 2mm, the best solution would be probably a magnetic
circuit made of the two half cores.

How to calculate:
We assume that mu of the core material times gap width is much higher then
the length of the magnetic path. The magnetic resistance of a piece of
material is length/(mu * cross section). The coupling between the cores is
the fraction of the primary flux which gets into the secondary. Part of the
primary flux goes through the air missing the secondary, the other part
crosses the gap to the secondary. The ratio of those parts depends on
geometry, and it is approximately 2*gap width/distance between the poles (if
distance is much higher then the gap). Thus the coupling is ~ (1 -
2*gap/distance).

I can derive more accurate estimates if mu and the geometry is known.

Vladimir Vassilevsky
DSP and Mixed Signal Consultant
www.abvolt.com

 

[Klaus Kragelund]

Hello Klaus,

If the barrier is only 2mm, the best solution would be probably a magnetic
circuit made of the two half cores.

How to calculate:
We assume that mu of the core material times gap width is much higher then
the length of the magnetic path. The magnetic resistance of a piece of
material is length/(mu * cross section). The coupling between the cores is
the fraction of the primary flux which gets into the secondary. Part of the
primary flux goes through the air missing the secondary, the other part
crosses the gap to the secondary. The ratio of those parts depends on
geometry, and it is approximately 2*gap width/distance between the poles (if
distance is much higher then the gap). Thus the coupling is ~ (1 -
2*gap/distance).

I can derive more accurate estimates if mu and the geometry is known.


- Vis tekst i anførselstegn -

Hi Vladimir

That great info

Do you by any chance have a reference for where you got this info (or
was it derived from general textbook material?)

Thanks

Klaus

 

[CWatters]

That sound almost Star-trek like.

I just need 100mW

Take apart an electric toothbrush charger?

 

[Tina]

Don't overlook thinking about the plastic as part of a parallel plate
capactor if fringing fields are an issue. You'd be doing electric
field stuff rather than magnetic.

 

[Richard Tobin]

[/QUOTE]

Take apart an electric toothbrush charger?

The electric toothbrush chargers I have seen rely on one part being
*inside* the other.

-- Richard

 

[Spehro Pefhany]

Take apart an electric toothbrush charger?

The electric toothbrush chargers I have seen rely on one part being
*inside* the other.

-- Richard[/QUOTE]

See "Power with no Strings Attached", page 31 of the July 5 (#14)
issue of EDN


Best regards,
Spehro Pefhany

 

[Vladimir Vassilevsky]

If the barrier is only 2mm, the best solution would be probably a magnetic
circuit made of the two half cores.

How to calculate:
We assume that mu of the core material times gap width is much higher then
the length of the magnetic path. The magnetic resistance of a piece of
material is length/(mu * cross section). The coupling between the cores is
the fraction of the primary flux which gets into the secondary. Part of the
primary flux goes through the air missing the secondary, the other part
crosses the gap to the secondary. The ratio of those parts depends on
geometry, and it is approximately 2*gap width/distance between the poles (if
distance is much higher then the gap). Thus the coupling is ~ (1 -
2*gap/distance).

I can derive more accurate estimates if mu and the geometry is known.

Do you by any chance have a reference for where you got this info (or
was it derived from general textbook material?)


The main idea is from the general EE textbook. There is an artificial trick
to calculate the static magnetic fields: consider the magnetic path as an
electric circuit, where:

mu = conductivity
B = voltage
Flux = current

And apply Ohm and Kirchoff laws to it just like you do with electric
circuits.
Of course, you have to account for the "magnetic currents" in the air and
the complex geometry of the "magnetic conductors". However it is pretty
simple to derive the estimates which are good enough for the practical
purposes.

Vladimir Vassilevsky
DSP and Mixed Signal Consultant
www.abvolt.com

 

[Joerg]

I actually already did a test with two pottet cores. See pictures:

http://www.microdesign.dk/tmp/misc_transformertypes.jpg
http://www.microdesign.dk/tmp/two_cores.jpg

What's the cost of the milling for the circuit boards and potting or
riveting in the core halves over there? Since you only want to transfer
100mW you may be able to use this solution. I had a case with a core of
about 10mm and three turns. The coupling was quite good but only in the
tens of MHz, of course.

What is the voltage you need on the secondary side?

This one is wound:

http://www.microdesign.dk/tmp/two_cores_gap.jpg
(bad mobile phone picture)

That can transfer plenty of power, but it will be costly (the turns
are not PCB traces, but a cobber wire)

The winding and potting into the core halves would have to be contracted
out. If that is not an option I think it would have to be 13.56MHz or
27.12MHz ISM in this case, with just trace inductors. At 27.12MHz you
may even be able to get away without a core but with more (free...)
turns on the board. However, in that case the regulatory folks would
have to check whether such ISM band usage is legal in all the countries
where the product is going to be marketed.

ISM requires a narrow tolerance clock source, either crystal driven or
resonator driven. Personally I prefer resonators for anything that might
get banged around a bit.

I was looking for perhaps using a standard part like this:

http://www.microdesign.dk/tmp/SDR0805.jpg

At least for the primary side. The stray field would frindge around
the top, but I can get the manufacturer to make it more like a
straight rod to get an even field.
Then the secondary side could be either a coil like that or an air
core inductor (it has to be 90 degrees angled to the PCB to that
easier)

Should be in line and same orientation. But this would be a custom part.

Actually I would like to be able to calculate the field, since I may
be using it in another project where the end goal is to transfer only
a signal with very low frequency. So perhaps a Hall sensor could do in
that app.

But close to the core it is the near field that decides the magnitude.
So its difficult to calculate

Precise calculations would require a software like EESOF (Agilent) but
my experience is that simulations are of limited value with air-coupled
inductors. Too many variables. It may be better to measure and slap on a
huge margin. For calculating a single coil scripts like this help:

http://smirc.stanford.edu/spiralCalc.html

If all this has to be small, cheap and transfer 100mW or more I'd look
at higher frequencies. In the range below 100kHz things become large.
For example, the cores for charging electric toothbrushes and stuff like
that are almost a cubic inch in volume. Many of those operate around
50-70kHz.

 

[Joerg]

Take apart an electric toothbrush charger?

The electric toothbrush chargers I have seen rely on one part being
*inside* the other.
[/QUOTE]

Not necessarily. On the Philips Sonicare they a butting up against each
other, with plastic in between. The two ridges in the base are only
there to prevent the toothbrush from falling over.

 

[John Larkin]

Not necessarily. On the Philips Sonicare they a butting up against each
other, with plastic in between. The two ridges in the base are only
there to prevent the toothbrush from falling over.

Does that run at line frequency, or is there electronics?

John

 

[Joerg]

Does that run at line frequency, or is there electronics?

I held the analyzer to it and AFAIR it was running around 60kHz. At line
frequency you wouldn't be able to achieve enough coupling.

Those toothbrushes work great, BTW.

 

[Kevin G. Rhoads]

I need to transfer some energy accross a platic barrier. I will be

using a solenoid on both sides of the barrier, one to transmit and one
to receive the energy

1) Curve both solenoids into half-circles, then you have an
air-core toroid with the plastic sheet separating the two halves.
Better yet, take a toroid and a glass-cutter and score it and
break it into two halves and use those as the cores. (ferrites
will break like glass, use glass "cutting" techniques)

2) Don't cross-post unless absolutely necessary. It was not
necessary for this question.

3) I hope you are using a full wave bridge on the output side
and not some !@#$%^&* half-wave rectification scheme. Resonating
the secondary is not necessary with a ferrite core toroid arrangement.
I'm not sure it would be necessary with highly sub-optimal windings,
but it is likely to involve you in lily-painting. Optimize the
winding geometry and core, then paint your lilies if you absolutely
feel you must.

4) You could also do this with capacitive transfer. (I should
probably keep my mouth shut and not mention this, but ...)

5) Two right-cylindrical solenoids is an absolutely atrocious
way to transfer energy. The coupling coefficient is negligible.

6) Calculate NOTHING -- build, measure, adjust. The calculations
needed to get decent accuracy are just NOT owrth the time and effort.

YMMV - MHOO - HTH
Kevin

 

[CWatters]

Take apart an electric toothbrush charger?

The electric toothbrush chargers I have seen rely on one part being
*inside* the other.[/QUOTE]

Well yes. But I'm a great believer in changing the problem to make the
solution easier

They so work across a plastic barrier - just not a flat one. So deform the
barrier!