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vibrating coil gradiometer

G

George Herold

Jan 1, 1970
0
Hi all, first off I may never build this.
At the moment I'm only thinking about it, with your 'most excellent'
help.

So from a thread started by Tim W. about frig magnets,
I got onto measuring B field gradients,
In particular the gradients, in the Earth's (~50uT ) field,
on the order of 1 uT/m.
Phil H started me thinking about wiggling a coil back and forth to
measure the gradient.
The emf will go as the field gradient, times the velocity, times the
area of the coil.
(** = ^ , 10**2 = 10^2 = 100)
For a 1cm**2 coil at 1m/s that's 10**-10 V. kinda depressing.

But I can add turns, increase area, (increase velocity?)
(Can I shake something at 1m/s? non-magnetically.)

To get to a microvolt I need a 10**4 increase.
1000 turns and 10cm**2 or 100 turns and 100 cm**2?
(or something else)

I also think the coil will have to sit in a thick walled (aluminum?)
shield
that will keep out the AC magnetic fields at (and above) the shake
frequency,
but allow the DC field gradients through. So a higher frequencies
will mean a thinner shield. (my total cost of 'gizmo' ~$100-200)
(How many skin depths will I need?)

And then how do I wiggle it? I'm thinking a coil on the end of a long
I-Beam.
Maybe I can drive it resonantly with a 'upstream' peizo?

I'm looking for some crazy ideas (or maybe this just doesn't work.)

George H.
 
R

Robert Macy

Jan 1, 1970
0
Hi all, first off I may never build this.
At the moment I'm only thinking about it, with your 'most excellent'
help.

So from a thread started by Tim W. about frig magnets,
I got onto measuring B field gradients,
In particular the gradients, in the Earth's (~50uT ) field,
on the order of 1 uT/m.
Phil H started me thinking about wiggling a coil back and forth to
measure the gradient.
The emf will go as the field gradient, times the velocity, times the
area of the coil.
(** = ^ , 10**2 = 10^2 = 100)
For a 1cm**2 coil at 1m/s that's 10**-10 V.  kinda depressing.

But I can add turns, increase area, (increase velocity?)
(Can I shake something at 1m/s?  non-magnetically.)

To get to a microvolt I need  a 10**4 increase.
1000 turns and 10cm**2 or 100 turns and 100 cm**2?
(or something else)

I also think the coil will have to sit in a thick walled (aluminum?)
shield
that will keep out the AC magnetic fields at (and above) the shake
frequency,
but allow the DC field gradients through.  So a higher frequencies
will mean a thinner shield.  (my total cost of 'gizmo' ~$100-200)
(How many skin depths will I need?)

And then how do I wiggle it?  I'm thinking a coil on the end of a long
I-Beam.
Maybe I can drive it resonantly with a 'upstream' peizo?

I'm looking for some crazy ideas  (or maybe this just doesn't work.)

George H.

When working with magnetic fields, remember they are LOW impedance.
Take advantage of that. It's not like you're making a voltmeter that
must have high impedance.


You could try 'chopping' Earth's DC field by using a spinning loop
coupled to a 'slip ring' transformer for non-contact transfer of
signal. [make the loop AND the coupling transformer, use air core]
Spin the loop at some 'proper' harmonic that is synchronous to AC
mains.

Speed of the spin determines the number of turns and knowing the
spinning speed you can synchronously detect the signal and really
lower the noise floor. maybe a cheap optical coupler/interrupter [take
apart an old mouse] as the loop is spun to provide the sampling pulse.

To analyze what's going on in the fields, get a copy of femm 4.2 If
you need help to start on the learning curve, let me know.

PS: Circa 1989, I made a portable 4 inch diameter AC magnetometer,
that had a 'flat' sensitivity of 1V per uT over spectrum from 5 Hz to
1MHz.[max 2uT] Gently rocking the meter would peg the output as it
moved through that 50uT field. The noise floor was in the range of 5nT
[on a scope the output signal was a crisp line and not thickened by
noise] and you could 'see' vehicles go by on the street adjacent to
our lab as the metal deflected the Earth's field.
 
G

George Herold

Jan 1, 1970
0
Hi all, first off I may never build this.
At the moment I'm only thinking about it, with your 'most excellent'
help.
So from a thread started by Tim W. about frig magnets,
I got onto measuring B field gradients,
In particular the gradients, in the Earth's (~50uT ) field,
on the order of 1 uT/m.

That's very simple. Strong field, no problem at all.
Phil H started me thinking about wiggling a coil back and forth to
measure the gradient.

Not a good idea.

[...]
I'm looking for some crazy ideas  (or maybe this just doesn't work.)

http://www.magneticsensors.com/

Use GMR sensor ICs from Honeywell. With two sensors, few analog parts and
proper callibration, you will be able to resolve the gradients at the order
of 0.001 of the Earth field. GMR is as simple as it gets; there are of
course better sensors and technologies.

Oh, those might work, thanks (HMC1001, HMC1002) they're not GMR
though, but some Anisotropic MagnetoResistive (AMR) technology. Not
cheap (~$15 each) Or were you suggesting a different sensor?

We've got some hall sensors from Allegro (A1324), but they are way too
noisy.

George H.
 
G

George Herold

Jan 1, 1970
0
Hi all, first off I may never build this.
At the moment I'm only thinking about it, with your 'most excellent'
help.
So from a thread started by Tim W. about frig magnets,
I got onto measuring B field gradients,
In particular the gradients, in the Earth's (~50uT ) field,
on the order of 1 uT/m.
Phil H started me thinking about wiggling a coil back and forth to
measure the gradient.
The emf will go as the field gradient, times the velocity, times the
area of the coil.
(** = ^ , 10**2 = 10^2 = 100)
For a 1cm**2 coil at 1m/s that's 10**-10 V.  kinda depressing.
But I can add turns, increase area, (increase velocity?)
(Can I shake something at 1m/s?  non-magnetically.)
To get to a microvolt I need  a 10**4 increase.
1000 turns and 10cm**2 or 100 turns and 100 cm**2?
(or something else)
I also think the coil will have to sit in a thick walled (aluminum?)
shield
that will keep out the AC magnetic fields at (and above) the shake
frequency,
but allow the DC field gradients through.  So a higher frequencies
will mean a thinner shield.  (my total cost of 'gizmo' ~$100-200)
(How many skin depths will I need?)
And then how do I wiggle it?  I'm thinking a coil on the end of a long
I-Beam.
Maybe I can drive it resonantly with a 'upstream' peizo?
I'm looking for some crazy ideas  (or maybe this just doesn't work.)
George H.

When working with magnetic fields, remember they are LOW impedance.
Take advantage of that.  It's not like you're making a voltmeter that
must have high impedance.

You could try 'chopping' Earth's DC field by using a spinning loop
coupled to a 'slip ring' transformer for non-contact transfer of
signal. [make the loop AND the coupling transformer, use air core]
Spin the loop at some 'proper' harmonic that is synchronous to AC
mains.

Speed of the spin determines the number of turns and knowing the
spinning speed you can synchronously detect the signal and really
lower the noise floor. maybe a cheap optical coupler/interrupter [take
apart an old mouse] as the loop is spun to provide the sampling pulse.

To analyze what's going on in the fields, get a copy of femm 4.2  If
you need help to start on the learning curve, let me know.

PS: Circa 1989, I made a portable 4 inch diameter AC magnetometer,
that had a 'flat' sensitivity of 1V per uT over spectrum from 5 Hz to
1MHz.[max 2uT]  Gently rocking the meter would peg the output as it
moved through that 50uT field. The noise floor was in the range of 5nT
[on a scope the output signal was a crisp line and not thickened by
noise] and you could 'see' vehicles go by on the street adjacent to
our lab as the metal deflected the Earth's field.- Hide quoted text -

- Show quoted text -

Hi Robert, thanks for that. I bet your 1989 magnetometer cost more
than a few hundred dollars.

A colleague has done the field gradient measurment, by weighting a
permenant magnet. (First with the magnetic moment pointing up and
then down.) He claims that a gradient of 1uT/m gives a weight
difference of ~ 1 milli gram (out of a 100 gram magnet) Which can be
done with some of these cheap balances. This doesn't give all the
components of the gradient, but it might be enough.

George H.
 
Hi all, first off I may never build this.
At the moment I'm only thinking about it, with your 'most excellent'
help.

So from a thread started by Tim W. about frig magnets,
I got onto measuring B field gradients,
In particular the gradients, in the Earth's (~50uT ) field,
on the order of 1 uT/m.
Phil H started me thinking about wiggling a coil back and forth to
measure the gradient.
The emf will go as the field gradient, times the velocity, times the
area of the coil.
(** = ^ , 10**2 = 10^2 = 100)
For a 1cm**2 coil at 1m/s that's 10**-10 V.  kinda depressing.

But I can add turns, increase area, (increase velocity?)
(Can I shake something at 1m/s?  non-magnetically.)

To get to a microvolt I need  a 10**4 increase.
1000 turns and 10cm**2 or 100 turns and 100 cm**2?
(or something else)

I also think the coil will have to sit in a thick walled (aluminum?)
shield
that will keep out the AC magnetic fields at (and above) the shake
frequency,
but allow the DC field gradients through.  So a higher frequencies
will mean a thinner shield.  (my total cost of 'gizmo' ~$100-200)
(How many skin depths will I need?)

And then how do I wiggle it?  I'm thinking a coil on the end of a long
I-Beam.
Maybe I can drive it resonantly with a 'upstream' peizo?

I'm looking for some crazy ideas  (or maybe this just doesn't work.)

George H.

http://en.wikipedia.org/wiki/Earth's_field_NMR

-Lasse
 
G

George Herold

Jan 1, 1970
0
There is really no problem to measure weak magnetic fields. The problem is
to measure weak gradients on the background of uniform but strong magnetic
field of the Earth. What your colleague measured is probably caused by
misalignment of the magnet axis wrt Earth magnetic field. So the magnet
essentially acted like a compass. Compassing is common problem for all
mechanical magnetic gizmos,
so sensitivity to gradient is limited by mechanical misalignment.

Vladimir Vassilevsky
DSP and Mixed Signal Consultantwww.abvolt.com

Hi Vlad, I hope it's OK if I disagree with you. There is certainly
misalignment between the B field and magnetic moment. But a uniform B
field can only cause a torque on the magnet. (So there is certainly
some 'compassing' when one is weighing the magnet, but that pushes
down more on one side and less on the other... no net force.) The
force on the magnet is caused by the gradient in the field. You can
see this pretty easily with a spring, magnet and rather strong field
gradients.
http://www.teachspin.com/instruments/magnetic_force/experiments.shtml

Figure 2 down the bottom of the page is for a single coil, along the
axis. It's fun to write down the field as a function of z,
differentiate to get dB/dz and then see how that fits the data.

(I've ordered some of the honeywell sensors, it be great to be able to
measure both the field and the gradient... but it's a bit of a pain
getting two sensors zero'd I've got a little mu metal tube that is
maybe close to zero field inside.)

George H.
 
F

Fred Bartoli

Jan 1, 1970
0
George Herold a écrit :
Grin, Thanks Lasse, I'd like the gradiometer so our users would know
where NOT to put this,
http://www.teachspin.com/instruments/efnmr/index.shtml

George H.

One way to do rotating coil would be with two coils:
One fixed and external one to sense the varying field and the other one
being a rotating short circuited ring.
The rotating ring will see an induced emf=w.B.s.sin(theta) which shorted
will induce a current in the loop, which in turn will produce a rotating
B component, which will be sensed by the enclosing sense coil.

I did not run the figures, you know how to do that, so I don't know if
the figures are practical or not.

A gradiometer could be build from this with two rotating rings and one
big enclosing coil, with turns in one sense for half its length, then
the same turns count in the other sense for the second half length.
You sense the field gradient emf across the whole coil...

Lock-ins should give you more than enough sensitivity, but again I did
not run the figures.
 
G

George Herold

Jan 1, 1970
0
George Herold a crit :








One way to do rotating coil would be with two coils:
One fixed and external one to sense the varying field and the other one
being a rotating short circuited ring.
The rotating ring will see an induced emf=w.B.s.sin(theta) which shorted
will induce a current in the loop, which in turn will produce a rotating
B component, which will be sensed by the enclosing sense coil.

Wow! I think I see it. Does this really work? (I get baffled by
changing B fields sometimes.)

I'll have to put some real numbers in for the rotating coil (wire size/
resistance). If I'm not mistaken, it will be the current in the
rotating coil that I need to calculate.
I did not run the figures, you know how to do that, so I don't know if
the figures are practical or not.

A gradiometer could be build from this with two rotating rings and one
big enclosing coil, with turns in one sense for half its length, then
the same turns count in the other sense for the second  half length.
You sense the field gradient emf across the whole coil...

Cute, but maybe instead of changing the 'sign' of the stationary sense
coil, I could have the rotating coils wired in the opposite
direction.
(Still the rotating coils are going to have to be pretty well
matched.)
Lock-ins should give you more than enough sensitivity, but again I did
not run the figures.

OK here's another idea.
I was thinking about using Vlad's field sensing chips to measure the
gradient.
If I use two of them, the problem is always getting them 'matched'.
So how about if I just shake one chip around and look at the AC signal
at the shake frequency? I'll want to keep it nice and 'flat' as I
shake it. It's not clear to me if unintended rotations (as I shake
it) will give signals at f or 2f....

(Hmm looks like f unless I get pointed along the field direction)

George H.
 
F

Fred Bartoli

Jan 1, 1970
0
George Herold a écrit :
On Oct 31, 7:37 pm, Fred Bartoli <" "> wrote: ......


Wow! I think I see it. Does this really work? (I get baffled by
changing B fields sometimes.)

I'll have to put some real numbers in for the rotating coil (wire size/
resistance). If I'm not mistaken, it will be the current in the
rotating coil that I need to calculate.

That one is easy...
The interesting one is the total flux seen by the enclosing coil which
is, well... better measured I think :)

Cute, but maybe instead of changing the 'sign' of the stationary sense
coil, I could have the rotating coils wired in the opposite
direction.

Wired in the opposite direction? I'd sure be interested to have a
picture of this: just look at what it is with the one turn coils :)

I also think that rotating them in opposite direction won't make a
difference.

(Still the rotating coils are going to have to be pretty well
matched.)

One turn rings lathed from copper tubing?
 
R

Robert Macy

Jan 1, 1970
0
...snip...
Hi Robert, thanks for that.  I bet your 1989 magnetometer cost more
than a few hundred dollars.

A colleague has done the field gradient measurment, by weighting a
permenant magnet.  (First with the magnetic moment pointing up and
then down.)  He claims that a gradient of 1uT/m gives a weight
difference of ~ 1 milli gram (out of a 100 gram magnet) Which can be
done with some of these cheap balances.  This doesn't give all the
components of the gradient, but it might be enough.

George H.

Don't tell the customers, more like $24 Copper is EXPENSIVE! I even
got away with cannibalizing a cheap meter to mount it on, so I didn't
have to spring $$ for a display..

Weight difference? I wonder if that could be done using MEMS?
Combine with stress sensor and voila!
 
G

George Herold

Jan 1, 1970
0
George Herold a écrit :> On Oct 31, 7:37 pm, Fred Bartoli <" "> wrote:

.....




That one is easy...
The interesting one is the total flux seen by the enclosing coil which
is, well... better measured I think :)

OK I'm still trying to get my head around how this works. But I think
I'm starting to see it. Even if this isn't used to measure field
gradients it might be fun to build one and see it work. "I'm not as
smart as other's and it's nice to have some data to help guide my
thinking"*
Wired in the opposite direction? I'd sure be interested to have a
picture of this: just look at what it is with the one turn coils :)

Yeah, sorry about that. Sticking my foot in my mouth again.
(But I have no shame and will certainly do it again.)
I also think that rotating them in opposite direction won't make a
difference.


One turn rings lathed from copper tubing?

Do I gain anything by having multiple turns? If not then how about a
copper disk?

I was first thinking I wanted the stationary pickup coil with it's
axis along the field direction, but now I'm thinking it needs to be at
right angles?


And thanks for the 'crazy' idea Fred.
Do you know if this has ever been made?

George H.
 
G

George Herold

Jan 1, 1970
0
Use a PIC. Match the sensors digitally. That works very well; to the
accuracy limited by self noise.

OK then I've got to calibrate each sensor separately. And then worry
about DC offset drift with temperature. I'm not saying it's not
possible.....but I need to measure differences of ~1-2 parts per
thousand in the field.
Unintended rotations will result in compassing.

It's worse than that, it gives me a signal. Roughly proportional to
the rotation angle. One good thing about vibrating the sensor is that
it moves the signal up in frequency, away from all the 1/f crud.

George H.
 
W

whit3rd

Jan 1, 1970
0
(I've ordered some of the honeywell sensors, it be great to be able to
measure both the field and the gradient... but it's a bit of a pain
getting two sensors zero'd I've got a little mu metal tube that is
maybe close to zero field inside.)

It doesn't take a zero-field region to make the adjustment; just
a way to reverse the sensor orientation, and make sure if
it reads +0.457 gauss at zero degrees, it reads -0.457 gauss
at 180 degrees.

Nonmagnetic materials are important, too; switches, resistor leads,
nuts 'n bolts, all will have to be degaussed if they are near the sensor.
 
G

George Herold

Jan 1, 1970
0
It doesn't take a zero-field region to make the adjustment; just
a way to reverse the sensor orientation, and make sure if
it reads +0.457 gauss at zero degrees, it reads -0.457 gauss
at 180 degrees.

Hi whit3rd, I've done that for a rough zero. Plenty good most of the
time when I only care about the field to a few percent or so. But I'm
contemplating a gradient measurement. I'd like to measure the
gradient over a length about equal to the sample size ~10 cm. And I
need to measure the gradients at the ~ 1uT/m (10mG/m... 1mG/10cm)
level. So something better than 1 mG. Now flipping over is fine...
(as long as you think carefully about how it's flipped) But a further
issue is that the field in the old building that I work in jumps
around at the few mG level... kinda randomly as elevators and fork
lifts move around. (I guess I could take the sensors to my house in
the country for zero adjustment.... then one worries about temperature
changes...) Anyway the mu-metal tube reduces the effects of a
changing local field.
Nonmagnetic materials are important, too; switches, resistor leads,
nuts 'n bolts, all will have to be degaussed if they are near the sensor.

Yeah been there, all the brass bolts are screened for magnetic
effects beofre being used in the 'non-magnetic' instuments. (Old
carbon comp resistors are non-magnetic BTW)

George H.
 
F

Fred Bartoli

Jan 1, 1970
0
George Herold a écrit :
OK I'm still trying to get my head around how this works. But I think
I'm starting to see it. Even if this isn't used to measure field
gradients it might be fun to build one and see it work. "I'm not as
smart as other's and it's nice to have some data to help guide my
thinking"*

Yeah, sorry about that. Sticking my foot in my mouth again.
(But I have no shame and will certainly do it again.)

Do I gain anything by having multiple turns? If not then how about a
copper disk?

I didn't think of copper disks but it should work. Anyway field
calculations aren't trivial anymore...

That and multi-turns might be a win... or not:

Emf goes with the number of turns. Then :

1) as well as the resistance (for a constant wire gauge). Thus current
is constant with turns and the generated field goes with the number of
turns.

2) Now if you only have a defined section to wind your coil, it is
another matter as the total resistance goes with turns squared, then the
generated field stays constant.


I was first thinking I wanted the stationary pickup coil with it's
axis along the field direction, but now I'm thinking it needs to be at
right angles?

Say you have an X,Y,Z orthonormal axis system
Your pick up coil will have a Z axis.
Then draw your "chopping" coil in the X-Y plane and have it rotating on,
say X axis. That way you'll have generate a rotating field in the Y-Z
plane (well for the loop on axis component).
Now you just keep the Z axis component of that for the pick up coil...

And thanks for the 'crazy' idea Fred.
Do you know if this has ever been made?

No, I just dreamed that up. Hey, if you patent it, put my name on it :)
Ooops, to late...
 
G

George Herold

Jan 1, 1970
0
You just have to match one sensor against the other in the uniform field.

Ahh where am I getting this uniform field? I'd like to sell this to
users so they can find a (roughly) uniform field region in their
building.
Done that. It isn't very difficult.


That's why I prefer digital correction. Linear and nonlinear effects could
be taken care off.


IIRC those NMR sensors generate fairly uniform noise; at least the ones I
tried.

Hmm, I wonder if we are talking about the same sensors. HMC1001 or
(HMC10xx)?

http://www51.honeywell.com/aero/com...itions/HMC_1001-1002-1021-1022_Data_Sheet.pdf

(Or if that's too long find them at digikey)

The high sensitivity ones (HMC1001) have a gain that varies from 2.4
to 4.0 mV/V/G
and a sensitivity tempco of ~0.3%/C... (much less with a current
drive, which I don't quite get.)

So matching two of these across B and T to the 0.1% level looks like a
lot of work.

I still like the idea of wiggling one back and forth. That seems to
get rid of all this 'common mode' crap, and just give me the
difference that I want. But I'm not sure how to wiggle it 10 cm and
keep it flat.

(ohh there's a noise graph (for the less sensitive flavor) on page 4,
1/f corner at ~100Hz.)

George H.
 
G

George Herold

Jan 1, 1970
0
George Herold a écrit :









I didn't think of copper disks but it should work. Anyway field
calculations aren't trivial anymore...
That and multi-turns might be a win... or not:

Geesh you're no help at all. :^)
Emf goes with the number of turns. Then :

1) as well as the resistance (for a constant wire gauge). Thus current
is constant with turns and the generated field goes with the number of
turns.

2) Now if you only have a defined section to wind your coil, it is
another matter as the total resistance goes with turns squared, then the
generated field stays constant.


Say you have an X,Y,Z orthonormal axis system
Your pick up coil will have a Z axis.
Then draw your "chopping" coil in the X-Y plane and have it rotating on,
say X axis. That way you'll have generate a rotating field in the Y-Z
plane (well for the loop on axis component).
Now you just keep the Z axis component of that for the pick up coil...

OK I'll have to try and mock something up. I've got a copper disk on
a stick, but spinning it with a power drill was a no-go. (As you
might have guessed the power drill spits out all sorts of B field
stuff.)


No, I just dreamed that up. Hey, if you patent it, put my name on it :)
Ooops, to late...

We can't afford to patent anything.

George H.
 
F

Fred Bartoli

Jan 1, 1970
0
George Herold a écrit :
Geesh you're no help at all. :^)

OK I'll have to try and mock something up. I've got a copper disk on
a stick, but spinning it with a power drill was a no-go. (As you
might have guessed the power drill spits out all sorts of B field
stuff.)

BTW it occurred to me that the picked up voltage should be at 2 times
the rotating frequency, so you'll have to lock on 2nd harmonic.
We can't afford to patent anything.

Was joking, of course :)
 
G

George Herold

Jan 1, 1970
0
George Herold a écrit :








BTW it occurred to me that the picked up voltage should be at 2 times
the rotating frequency, so you'll have to lock on 2nd harmonic.

Good I somehow had the same picture.

George H.
 
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