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Registered Member #1960
Joined: Thu Feb 05 2009, 11:57PM
Location:
Posts: 48
Hmm Kluge, I'm not sure I quite follow your explanation. If a toroid encapsulates a current carrying wire - won't it just keep the predicted low magnetic field somewhat more stable than if it was free air?
I pictured threading a toroid on a 10mm dia wire. Then in the gap I install a A1362 hall effect sensor. This has a sensitivity of 4,5mV/G (or 16mV/g programmable) - so if I have a current flowing of 1000A(400G @5mm from conductor), it will swing 1,8V either way. In bidirectional-mode, the Vout is Vin/2.
Some calibration is probably needed, but I can do that with my very accurate 200A IC's.
A ferrite would take 400Gauss day in and day out, but does it work the way I think it does - or are there some other things I have to calculate in?
Allegro's site featured a nice drawing, I'm adding this here:
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Storken wrote ... Hmm Kluge, I'm not sure I quite follow your explanation. If a toroid encapsulates a current carrying wire - won't it just keep the predicted low magnetic field somewhat more stable than if it was free air?
No no no, that's the bit you're missing. (Nice picture, by the way.) Sorry I don't have time this morning to write or point to a good tutorial about B, H, and their scalar ratio which is the material permeability "mu".
Yes, for any path encircling a 1000 A current, "H" integrated all the way around is 1000 A (1256 Oe-cm in cgs units) for any distribution of material. If the material is uniform along whole path, |H| is uniform: 15900 A/m (200 Oe) on circle at r=10mm. In air the resulting |B| is 0.02 T (200 G).
In a solid ferrite or iron toroidal core, |B| will be greater according to the relative permeability, but in this case would hit saturation if mu/mu0 exceeds about 20 for ferrite or 100 for iron.
In a gapped core as normally applied, |B| is continuous around the loop while |H| is very high inside the air gap and very low in the permeable material. Can you take it from there? Good luck.
Registered Member #56
Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445
Can you use something like the allegro sensor (200A) with parallel resistor (read as 'piece of copper') across it with 1/5 the resistance? A bit of a hack, but at least for DIY it is probably a lot easier than making your own module from scratch...
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
... wrote ... Can you use something like the allegro sensor (200A) with parallel resistor (read as 'piece of copper') across it with 1/5 the resistance?
I think that's a spectacularly excellent idea, tho' the shunt should have 1/4 of the resistance to carry 800 A.
Try to make the temperature of your copper shunt resistor track the temperature of the copper path through the Allegro device. Perhaps solder the Allegro part on one edge of a thick copper strip. Then trim the ratio by progressively slotting or drilling holes in the strip. I bet the 200 amp conductor in Allegro part depends on cooling by heat conduction to much heavier external conductors.
Registered Member #1960
Joined: Thu Feb 05 2009, 11:57PM
Location:
Posts: 48
klugesmith wrote ...
... wrote ... Can you use something like the allegro sensor (200A) with parallel resistor (read as 'piece of copper') across it with 1/5 the resistance?
I think that's a spectacularly excellent idea, tho' the shunt should have 1/4 of the resistance to carry 800 A.
Try to make the temperature of your copper shunt resistor track the temperature of the copper path through the Allegro device. Perhaps solder the Allegro part on one edge of a thick copper strip. Then trim the ratio by progressively slotting or drilling holes in the strip. I bet the 200 amp conductor in Allegro part depends on cooling by heat conduction to much heavier external conductors.
The 200A IC package features a 6mm^2 conductor, so the total resistance over the conductor is about 130microohms. Not much. It can handle spikes of 1000A, but it will not give any form of feedback ofcource. A heavy ass shunt over the leads of this sensor has struck my mind more than once, but I want this to be more of a production thing - so I've been reluctant to DIY it like this. Sanding/drilling to calibrate it is probably a good way to go imo, hmm... The leads are spaced 6mms apart, that's copper shunt territory :P a ~3mm^2 copper conductor between these (awg ~12) could be tested. Still, handling 1000A seems like crazy..
Still plundering about what you said earlier, I'll post some once my understanding of it is better :)
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Storken wrote ... The 200A IC package features a 6mm^2 conductor, so the total resistance over the conductor is about 130microohms. Not much. It can handle spikes of 1000A, but it will not give any form of feedback ofcource. A heavy ass shunt over the leads of this sensor has struck my mind more than once, but I want this to be more of a production thing - so I've been reluctant to DIY it like this. Sanding/drilling to calibrate it is probably a good way to go imo, hmm... The leads are spaced 6mms apart, that's copper shunt territory :P a ~3mm^2 copper conductor between these (awg ~12) could be tested. Still, handling 1000A seems like crazy...
Yes to what I said about the Hall module conductor being much to thin to carry 200 A without heat sinking. At 130 microohms it would dissipate over 5 watts! The current density is 33 A/mm^2. Look at the board layout recommendation, designed to spread the current and heat radially.
Copper busbars 5 mm thick can carry about 4 A/mm^2 at 65 degree temperature rise. A 5 x 50 mm strip, passively cooled, could carry 1000 A all day (but would be hot enough to burn you). Just make it thin (or perforated) in a very short section that's straddled by your Hall device. That high-resistance section will be cooled by thermal conduction in the direction of electrical current. Hint: the thinning will be easier if you choose a Hall device rated for much less than 200 A.
Registered Member #103
Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845
Just my thoughts, but $80 for those LEM sensors EVR linked to seems a bargain. We use similar parts here but they're more like $350 for a similar spec. Your actual hall effect device might be cheap, but when you add up all the costs of toroid, control electronics etc that you'll need to develop the full sensor you probably won't save much in the end... however if it's a project in it's own right then it sounds interesting
Regarding shunts, if you want any kind of accuracy avoid copper and use manganin instead! You should be able to buy it in sheets off ebay. Copper is almost useless for accurate current measurement because of the temperature coefficient, but manganin is very nice material.
I've built a similar thing to what you seem to be wanting, using manganin shunts and the IR2175 shunt interface IC. The IR2175 converts the shunt voltage to a PWM signal for isolation, which you RC filter to get an analogue voltage. We needed decent bandwidth down to DC for torque at zero speed on a BLAC motor. The IR2175 will give about 15kHz bandwidth, and has about a +/-10mV offset on the shunt input. If you need better accuracy, then I would build a differential amp using a nice low offset opamp (like the OP07?) but then you will need to sort out your own power and signal isolation.
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