My suspicion is that the increased burden in the primary circuit from adding a Hall effect sensor is probably unmeasurable, which is why I asked about a theoretical solution. And it is just out of curiosity, but I thought that somebody else with better analytical chops than me and better knowledge of Hall effect device design might take up the challenge.
o 160A for 10msec current surges from both AC and DC sources... I'm after t he best resolution I can get... I don't know if it's possible to do this fo r both AC and DC off the same current sense circuit... I was thinking a shu nt through a current sense amplifier then to an RMS to DC converter IC... b ut I'm not sure if this is the best approach... any suggestions?
an isolated sigma delta converter right now....
why not get it all in one package:
On a sunny day (Wed, 16 Apr 2014 11:52:17 -0700 (PDT)) it happened whit3rd wrote in :
Yes, and there is an other way to use a coil with core, as both AC and DC sensor, Here is a little inductor setup in a FET oscillator:
Now with a bit of DC (say permanent magnet for demo):
As the core saturates the L decreases, and frequency goes way up. If f resonance is much higher than the frequency of the measured 'signal' (AC with DC component), then it should follow that, It sort of de-magnetizes itself...
That is actually a several hundred mH coil.
I'm pretty sure the Hall effect is due to electron drift velocity and deflection of the moving electrons by a magnetic field. There's no work whatever done by the magnetic field on such a moving charge. No work, zero power, zero resistive-like energy losses.
Magnetic force on a moving charge is perpendicular to velocity, the power is zero because F-vector and V-vector are orthogonal.
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IIRR Hall-effect sensors depend on having roughly equal currents being carr ied by positive and negative charge-carriers.
The work being done by those charge carriers is supplied by your measuring circuit and has no effect on the load.
Getting a magnetic field for the Hall effect sensor to detect could involve forming the current carrying conductor into a loop, which would add induct ance to the circuit being measured, but you can just measure the magnetic f ield being created around a straight wire, and the inductance of a straight wire is around 5nH per cm, which isn't much.
-- Bill Sloman, Sydney
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rried by positive and negative charge-carriers.
I don't think so.. OK I'm not hall effect expert. But I have been measurin g it recently in Si and Ge wafers. (I need to start a thread about that... later.) So if you had equal pos. and neg. charge carriers (each with the same mobil ity*) then in theory you'd get no Hall voltage. The sign of the Hall voltage giv es you the sign of the charge carriers.
As far as the load of a Hall effect on the line... If I'm allowed to specul ate wildly. I'd guess the load will only be on changing magnetic fileds... Any nearby conductor will have induced eddy current, and that will be some sort of "load" (loss). Since Hall effect devices are pretty darn small, an d also will be fairly poor conductors**, I'm thinking this is a pretty smal l effect.
George H.
*the Hall mobility is different from the normal electric field mobility. by somethng like a factor of 1.2 to a factor of almost 2. (Sze sec 1.5.2) ** the Hall signal is proportional to the carrier velocity, so all other th ings being equal you get a bigger signal with fewer carriers.g circuit and has no effect on the load.
ve forming the current carrying conductor into a loop, which would add indu ctance to the circuit being measured, but you can just measure the magnetic field being created around a straight wire, and the inductance of a straig ht wire is around 5nH per cm, which isn't much.
Only if the applied force is in the direction of the motion, which is never the case in Hall-effect cells.
-- umop apisdn --- news://freenews.netfront.net/ - complaints: news@netfront.net ---
asked
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Stuff and nonsense. If you change the path of a particle you have accelerated it. That takes work. However the amount involved in a hall effect sensor is mighty low.
?-)
Like magnetic amplifiers. I used them back in the early '60s.
On a sunny day (Fri, 18 Apr 2014 09:05:04 -0500) it happened John S wrote in :
Yep, same here, company I worked made huge ones, hundreds of amps.. I designed the controller..
Sounds like you need some remedial work in vector mathematics (and the "right-hand rule" :-) ...Jim Thompson
--
| James E.Thompson | mens |
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| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
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I love to cook with wine. Sometimes I even put it in the food.
Rethink that. The moon's straight path has been 'changed' into an ellipse by Earth's gravity, but the continual acceleration does NO net work on the moon. Similarly, a permanent magnet placed near a Hall sensor will give a positive Hall effect indication for an indefinite period of time.
The magnet won't go flat. And, the moon isn't falling from the sky. The word 'work' in physics has a specific energy-is-transferred meaning.
Well, so much for my intuitive conclusion that Hall effect sensors have burden. I suppose if that were true, one could cool a permanent magnet by surrounding it with Hall effect sensors. Thanks all. That's one more small piece of ignorance removed.
How do measurements off a shunt work for AC, are there OpAmps that can handle that kind of common mode voltage at their inputs? Or is Hall Effect pretty much the way to go for these kinds of things?
No, he's quite right. Work is a dot product, viz. force dot distance, whereas Lorentz (magnetic) force goes as v cross B.
The dot product v dot (v cross B) is identically zero.
Otherwise, the Sun would be doing work on the Earth by bending its path into an ellipse.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
So you get something for nothing? ...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 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.
Unless the acceleration on a charged particle radiates a photon.
If it does, the particle must slow down. Is that the reaction to the momentum of the photon? Must be. So that is why the photon is emitted in the direction of the particle motion?
-- John Larkin Highland Technology, Inc jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Which is why photons must have momentum.
-- John Larkin Highland Technology, Inc jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
The emitted photon (called synchrotron radiation, as if the charged particle is traveling in a circle in a cyclotron) is polarized, and there is no way to emit such a polarized photon in a radial direction because the E of a photon is always perpendicular to its momentum/travel direction, and that E direction is parallel to the force applied (radial).
It's a very rare photon emission except in the case of ultrarelativistic beams, where there's a 'headlight effect'. This makes a narrow forward beam from what (in the electron's rest frame) looks like a dipole-radiation 'doughnut'.
No. If you try drawing current from a Hall sensor, it does work because there's a component of the current in the direction of the E field.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
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