I am trying to power up the following 22 uH inductance coil (secondary
coil) with 115nF capacitor in parallel at the resonant frequency of
100KHz using Helmholtz coil. The load connected to the secondary
requires 100 mA at 3.3 V.
The resonant frequency is 100 KHz. The problem is that the secondary
coil gets really hot after like fifteen to twenty minutes.
How can I minimize this heating? Should I lower the Q of the secondary
coil system ?
I do not have the means to measure the magnetic field generated by the
Helmholtz coil.
Helmholtz coil is getting around 4A of current measured with an
ammeter.
I have also tried coils with values less than 22uH, smaller radius
and different resonant capacitor. I am driving the Helmholtz coil
using Full
H bridge.
> I am trying to power up the following 22 uH inductance coil (secondary
> coil) with 115nF capacitor in parallel at the resonant frequency of
> 100KHz using Helmholtz coil. The load connected to the secondary
> requires 100 mA at 3.3 V.
> The resonant frequency is 100 KHz. The problem is that the secondary
> coil gets really hot after like fifteen to twenty minutes.
> How can I minimize this heating? Should I lower the Q of the secondary
> coil system ?
> Thanks
> jess
You need to reduce the losses in your coil.
You have losses in both the core material and the wire.
Look for core material that has low loss at 100khz.
Wind the coil with larger wire or find the proper Litz
wire for 100khz and use it.
This site has data on sizes of Litz to use for different frequencies.
http://www.litz-wire.com/applications.html For 100khz it says to use either 38 of 40 gauge wire. Then you need to
see how many strands of wire to put in the bundle. That will be determined by the area available and how much current you're pushing.
You might try removing the wire from the coil you have and rewinding it with larger wire or two strands of smaller wire that have a larger
circular diameter.
Mikek
My guess is that you have core saturation or, the bridge rectifier you are using isn't fast enough? is the bridge a schottky type or some UF
type? Indication would show this if the bridge, too, is getting hot!
For a basic test, resistor load the secondary with nothing else to put
it to maximum load. If the coil is still getting hot, then you need to select a coil with different core material designed to operate at those speeds.
The core most likely is experiencing eddy currents and thus, is the same thing as induction heating.
Are you serious? That is a *signal* diode. Ifrm is .35A (350 mA). You're pushing 2A average thru them? Jeeze! It's hard to believe you haven't smelled the odor of defeat.
Amdx is more likely to be right in suggesting that you may need to go
to Litz wire, but the skin depth in copper is 0.2mm at 100kHz,
so you'd need to be using fairly thick wire for this to be a problem
I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
On 5/17/2012 11:39 AM, jsscsha...@gmail.com wrote:
> Hi,
> I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
> I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
> jess
Hi, Jess -
First, I suspect that the DCR of the coil is less than .2 ohm from the link to the picture you posted. Have you checked the ohmmeter reading with the probes shorted?
If the coil is really .2 ohm, then the power dissipated in that much resistance would be about 16*.2 or about 3.2 watts if your current measurement you posted earlier is correct.
Third, the current through the coil is not determined by the DCR, it is set by the inductance, voltage, and frequency. That is, the current is mostly determined by the inductor's reactance.
> On 5/17/2012 11:39 AM, jsscsha...@gmail.com wrote:
>> Hi,
>> I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V
>> peak to peak is appearing across the coil than current in the coil is
>> 200A. How could it be true?
>> I am not using any core for Helmholtz coil pair. I am trying to power
>> up the secondary coil ( LC circuit) using Helmholtz coil pair. The
>> secondary coil is set up with the magnetic ferrite material.
>> jess
> Hi, Jess -
> First, I suspect that the DCR of the coil is less than .2 ohm from the
> link to the picture you posted. Have you checked the ohmmeter reading
> with the probes shorted?
> If the coil is really .2 ohm, then the power dissipated in that much
> resistance would be about 16*.2 or about 3.2 watts if your current
> measurement you posted earlier is correct.
My mistake. I just re-read your original post and you say that that is the current through the Helmholtz coil. However, based on your schematic, the current through the 22uH coil should result in only about .26W. It shouldn't be getting hot.
> Third, the current through the coil is not determined by the DCR, it is
> set by the inductance, voltage, and frequency. That is, the current is
> mostly determined by the inductor's reactance.
> I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
> I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
> jess
it is not unusual for the impedance to skyrocket in a coil caused by
skin depth in magnetic fields. increase of 4 to 10 times is highly
likely even at low frequencies of 100kHz.
Check. Get a copy of FREE femm 4.2 and analyze your coil in
axisymmetric mode [accurate 3D]
the program will calculate the fields, the impedance
>> I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
>> I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
>> jess
>it is not unusual for the impedance to skyrocket in a coil caused by
>skin depth in magnetic fields. increase of 4 to 10 times is highly
>likely even at low frequencies of 100kHz.
Clear back in the late '70's I was making switchers running at only
20kHz. The high temperatures of the transformers made me look into
skin effect... all the way up to the 9th harmonic produced significant
dissipation. So I changed to Litz.
>Check. Get a copy of FREE femm 4.2 and analyze your coil in
>axisymmetric mode [accurate 3D]
>the program will calculate the fields, the impedance
...Jim Thompson
-- | James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
I do not know how to use FEMM4.2. I also need to simulate the magnetic field generated by Helmholtz coils . I will appreciate if you can help with both issues.
> I do not know how to use FEMM4.2. I also need to simulate the magnetic field generated by Helmholtz coils . I will appreciate if you can help with both issues.
> jess
Get a copy, install. Join users group.
You can import/export .dxf files, too. Best way to learn is to jump
in, give me an email address to send a 'sample' Helmhotlz coil to. It
is also a text file and can be opened with any editor.
To run femm 4.2:
Exercise the original to learn terms/features.
Then copy and rename, and make appropriate changes:
Start with 'Problem' dimensions, frequency
re-Draw geometry.
Include a line around the hwole problem to 'terminate' the
calculations.
Define the BLOCKS - AIR, COPPER
Define the CIRCUIT - DRIVE, 1A makes it easy to get constants.
Define the BOUNDARY [set to infinite space - read manual, 1/(uo*r)]
Calculate, look at results
you can click on circuit symbol and get inductance impedance etc.
Explore capabilities of this tool.
femm deefaults to very coarse, and therefore useless for any accuracy,
mesh, but calculates fast and is still educational. To get useful
results, you need to make the 'mesh' fine enough. I always calculate
skin depth and then make certain the mesh has at least three nodes
within that dimension. Mesh can be 'adjusted' within any BLOCK
material, or along any contour line. I usually use both with the
finest mesh along contours, after all, all the interesting things
happen near boundaries.
> I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
I hope you used a Kelvin connection - contact resistance is usually
around 0.1 Ohm. If you know the diameter of the copper wire used to
wind the coil, the diameter of the coil and the number of turns you
can look up the resistance of the wire per metre and calculate the
length of wire and its resistance, which can get you a good-enough
resistance rather more easily.
> I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
That core may concentrate the field generated by the Helmholtz coils,
but not much - the permeability Manganese/Zinc ferrites is of the
order of 1000, so it represents something close to a dead short in the
flux path, but it's not going to represent much of the flux path nor
snaffle a large proportion of the total flux generated.
The 40V peak to peak is being generated across the impedance of the
coil, which - as John S. has pointed out - is the vector sum of the
resistance of the coil and its inductive reactance (which is
presumably rather higher at the frequency of interest).
As Jim Thompson has pointed out, if you are exciting the secondary
with a square wave you have to take into account that a square wave
includes all the odd harmonics of the fundamental, and while a skin
depth of 0.2mm at 100kHz isn't likely to be a problem, the skin depth
at 900kHz - the nineth harmonic - is down to 0.022mm.
> > I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
> I hope you used a Kelvin connection - contact resistance is usually
> around 0.1 Ohm. If you know the diameter of the copper wire used to
> wind the coil, the diameter of the coil and the number of turns you
> can look up the resistance of the wire per metre and calculate the
> length of wire and its resistance, which can get you a good-enough
> resistance rather more easily.
> > I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
> That core may concentrate the field generated by the Helmholtz coils,
> but not much - the permeability Manganese/Zinc ferrites is of the
> order of 1000, so it represents something close to a dead short in the
> flux path, but it's not going to represent much of the flux path nor
> snaffle a large proportion of the total flux generated.
> The 40V peak to peak is being generated across the impedance of the
> coil, which - as John S. has pointed out - is the vector sum of the
> resistance of the coil and its inductive reactance (which is
> presumably rather higher at the frequency of interest).
> As Jim Thompson has pointed out, if you are exciting the secondary
> with a square wave you have to take into account that a square wave
> includes all the odd harmonics of the fundamental, and while a skin
> depth of 0.2mm at 100kHz isn't likely to be a problem, the skin depth
> at 900kHz - the nineth harmonic - is down to 0.022mm.
> --
> Bill Sloman, Nijmegen
Estimating resistance by calculating diameter times length is just
that. From experience winding coils, I've found that estimation is
always low. Just accepted it as cheap manufacturers were cutting the
wire size down, until I carefully wrapped a coil as tight as I could,
layering it into the windings for maximum stacking factor. When I
measured that coil's resistance it was one of those 'duh!' moments,
because the resistance was 45% higher than expected. Obviously
stretching the wire, eh?
In a coil skin depth effect 'lumps' the current into a tiny side
portion of the wire, thus the resistance is way high compared to the
resistance if the wire is NOT in a field. And, you guessed it, the
stronger the field, the more that skin effect lumps those conductors -
and the higher the resistance.
> On May 18, 3:18 am, Bill Sloman <bill.slo...@ieee.org> wrote:
> > On May 17, 6:39 pm, jsscsha...@gmail.com wrote:
> > > Hi,
> > > I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
> > I hope you used a Kelvin connection - contact resistance is usually
> > around 0.1 Ohm. If you know the diameter of the copper wire used to
> > wind the coil, the diameter of the coil and the number of turns you
> > can look up the resistance of the wire per metre and calculate the
> > length of wire and its resistance, which can get you a good-enough
> > resistance rather more easily.
> > > I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
> > That core may concentrate the field generated by the Helmholtz coils,
> > but not much - the permeability Manganese/Zinc ferrites is of the
> > order of 1000, so it represents something close to a dead short in the
> > flux path, but it's not going to represent much of the flux path nor
> > snaffle a large proportion of the total flux generated.
> > The 40V peak to peak is being generated across the impedance of the
> > coil, which - as John S. has pointed out - is the vector sum of the
> > resistance of the coil and its inductive reactance (which is
> > presumably rather higher at the frequency of interest).
> > As Jim Thompson has pointed out, if you are exciting the secondary
> > with a square wave you have to take into account that a square wave
> > includes all the odd harmonics of the fundamental, and while a skin
> > depth of 0.2mm at 100kHz isn't likely to be a problem, the skin depth
> > at 900kHz - the nineth harmonic - is down to 0.022mm.
> > --
> > Bill Sloman, Nijmegen
> Estimating resistance by calculating diameter times length is just
> that. From experience winding coils, I've found that estimation is
> always low. Just accepted it as cheap manufacturers were cutting the
> wire size down, until I carefully wrapped a coil as tight as I could,
> layering it into the windings for maximum stacking factor. When I
> measured that coil's resistance it was one of those 'duh!' moments,
> because the resistance was 45% higher than expected. Obviously
> stretching the wire, eh?
Seems likely. It still beats measuring 0.2 ohms with anything short of
a proper Ohm-meter with Kelvin (four terminal) connections.
> In a coil skin depth effect 'lumps' the current into a tiny side
> portion of the wire, thus the resistance is way high compared to the
> resistance if the wire is NOT in a field. And, you guessed it, the
> stronger the field, the more that skin effect lumps those conductors -
> and the higher the resistance.
The guesswork is all yours. Skin effect depth is purely frequency
dependent.
On Fri, 18 May 2012 08:44:34 -0700 (PDT), Robert Macy
<robert.a.m...@gmail.com> wrote:
>Just accepted it as cheap manufacturers were cutting the
>wire size down, until I carefully wrapped a coil as tight as I could,
>layering it into the windings for maximum stacking factor. When I
>measured that coil's resistance it was one of those 'duh!' moments,
>because the resistance was 45% higher than expected. Obviously
>stretching the wire, eh?
Maybe. I might expect magnet wire to be on the high side of the
tolerance band- it's sold by weight and they can sell more rolls if
they make it a bit thicker- since it's used (pretty much) by linear
measure (in fact the linear use goes up with thickness as you fill the
bobbin).
>>>I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
>>I hope you used a Kelvin connection - contact resistance is usually
>>around 0.1 Ohm. If you know the diameter of the copper wire used to
>>wind the coil, the diameter of the coil and the number of turns you
>>can look up the resistance of the wire per metre and calculate the
>>length of wire and its resistance, which can get you a good-enough
>>resistance rather more easily.
>>>I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
>>That core may concentrate the field generated by the Helmholtz coils,
>>but not much - the permeability Manganese/Zinc ferrites is of the
>>order of 1000, so it represents something close to a dead short in the
>>flux path, but it's not going to represent much of the flux path nor
>>snaffle a large proportion of the total flux generated.
>>The 40V peak to peak is being generated across the impedance of the
>>coil, which - as John S. has pointed out - is the vector sum of the
>>resistance of the coil and its inductive reactance (which is
>>presumably rather higher at the frequency of interest).
>>As Jim Thompson has pointed out, if you are exciting the secondary
>>with a square wave you have to take into account that a square wave
>>includes all the odd harmonics of the fundamental, and while a skin
>>depth of 0.2mm at 100kHz isn't likely to be a problem, the skin depth
>>at 900kHz - the nineth harmonic - is down to 0.022mm.
>>--
>>Bill Sloman, Nijmegen
> Estimating resistance by calculating diameter times length is just
> that. From experience winding coils, I've found that estimation is
> always low. Just accepted it as cheap manufacturers were cutting the
> wire size down, until I carefully wrapped a coil as tight as I could,
> layering it into the windings for maximum stacking factor. When I
> measured that coil's resistance it was one of those 'duh!' moments,
> because the resistance was 45% higher than expected. Obviously
> stretching the wire, eh?
> In a coil skin depth effect 'lumps' the current into a tiny side
> portion of the wire, thus the resistance is way high compared to the
> resistance if the wire is NOT in a field. And, you guessed it, the
> stronger the field, the more that skin effect lumps those conductors -
> and the higher the resistance.
It is always advisable to be micro measuring your wire if you plan on using it for specifics, do to its expected characteristics.
I know this because I happen to work for a company that makes wire.. Copper reduction is getting so bad now that instead of the OD being slightly near the odd size on the min side, it has been reaching over towards the next smaller gauge.
> On May 18, 5:44 pm, Robert Macy <robert.a.m...@gmail.com> wrote:
> > On May 18, 3:18 am, Bill Sloman <bill.slo...@ieee.org> wrote:
> > > On May 17, 6:39 pm, jsscsha...@gmail.com wrote:
> > > > Hi,
> > > > I measured the DC resistance of the coil. Its 0.2 Ohm. So, if 40 V peak to peak is appearing across the coil than current in the coil is 200A. How could it be true?
> > > I hope you used a Kelvin connection - contact resistance is usually
> > > around 0.1 Ohm. If you know the diameter of the copper wire used to
> > > wind the coil, the diameter of the coil and the number of turns you
> > > can look up the resistance of the wire per metre and calculate the
> > > length of wire and its resistance, which can get you a good-enough
> > > resistance rather more easily.
> > > > I am not using any core for Helmholtz coil pair. I am trying to power up the secondary coil ( LC circuit) using Helmholtz coil pair. The secondary coil is set up with the magnetic ferrite material.
> > > That core may concentrate the field generated by the Helmholtz coils,
> > > but not much - the permeability Manganese/Zinc ferrites is of the
> > > order of 1000, so it represents something close to a dead short in the
> > > flux path, but it's not going to represent much of the flux path nor
> > > snaffle a large proportion of the total flux generated.
> > > The 40V peak to peak is being generated across the impedance of the
> > > coil, which - as John S. has pointed out - is the vector sum of the
> > > resistance of the coil and its inductive reactance (which is
> > > presumably rather higher at the frequency of interest).
> > > As Jim Thompson has pointed out, if you are exciting the secondary
> > > with a square wave you have to take into account that a square wave
> > > includes all the odd harmonics of the fundamental, and while a skin
> > > depth of 0.2mm at 100kHz isn't likely to be a problem, the skin depth
> > > at 900kHz - the nineth harmonic - is down to 0.022mm.
> > > --
> > > Bill Sloman, Nijmegen
> > Estimating resistance by calculating diameter times length is just
> > that. From experience winding coils, I've found that estimation is
> > always low. Just accepted it as cheap manufacturers were cutting the
> > wire size down, until I carefully wrapped a coil as tight as I could,
> > layering it into the windings for maximum stacking factor. When I
> > measured that coil's resistance it was one of those 'duh!' moments,
> > because the resistance was 45% higher than expected. Obviously
> > stretching the wire, eh?
> Seems likely. It still beats measuring 0.2 ohms with anything short of
> a proper Ohm-meter with Kelvin (four terminal) connections.
> > In a coil skin depth effect 'lumps' the current into a tiny side
> > portion of the wire, thus the resistance is way high compared to the
> > resistance if the wire is NOT in a field. And, you guessed it, the
> > stronger the field, the more that skin effect lumps those conductors -
> > and the higher the resistance.
> The guesswork is all yours. Skin effect depth is purely frequency
> dependent.
> --
> Bill Sloman, Nijmegen
What? that's combining a lot here
skin effect is an effect caused by current carriers doing something
different because they're in a field. Could even be their own field.
from memory: skin depth is a concept that is useful only and is
defined as where the current density has dropped to 1/e within a
PLANAR surface with a PLANAR field impinging upon it. There in lies
some key information: planar this and planar that are some kind of
restrictions.
Now more confusion with skin depth being planar and people refer to
skin depth of a wire. Are they the same dimension? No. Same origin.
From the same effect, but different values.
The reason I cannot recall the EXACT definition? It's because skin
depth has too many assumptions applied to the calculation which pretty
much renders it useless for anything of value. However, it is a good
staritng point.
Plus, the equation is super easy to remember in MKS units
skin depth = sqrt ( 2 / (w*perm*cond) ), where
w is radians per second
perm is absolute permeability
cond is in S/m which for copper is around 58e6 S/m
Of major importance, if you have ANY gradient in the field, that
magnetic field will 'punch' right through. I've seen a lot of people
calculate skin depth through a shield, and then wonder why the shield
looks transparent! the magnetic field punches right through. Again,
skin depth is a PLANAR concept, and should be used carefully.
In a plain wire in free space skin effect causes the carriers to go
towards the outside of the wire, uniformly distributiing themselves
about the wire. For that situation a good estimation of conducting
cross sectional area is skin depth times pi times diameter. HOWEVER,
that same wire coiled upon itself suddenly has its own field pushing
the conductors around even more and REALLY bunches them, rather
tightly, and with more turns it gets worse. They really end up so
tight, it makes it look like your wire is 54 Awg, no longer the 20 Awg
you started with.
