How many Tesla is healthy for my Traco TME DC-DC converters? 
My concern is if the DC-DC converter may get saturated transformer cores.
My application is two 300 A, DC 10 x 20 mm copper bars 60 mm apart (50 mm space). Between the copper bars I’m going to place a H-bridge driver including three 12 V DC-Dc converters.
A forumla says 0.3 Tesla 1 cm from the conductor. This seems much. Anybody who have a theorectical insight in this?
The current flows is different directions.
My concern is if the DC-DC converter may get saturated transformer cores.My application is two 300 A, DC 10 x 20 mm copper bars 60 mm apart (50 mm space). Between the copper bars I’m going to place a H-bridge driver including three 12 V DC-Dc converters.
A forumla says 0.3 Tesla 1 cm from the conductor. This seems much. Anybody who have a theorectical insight in this?
The current flows is different directions.
peranders said:How many Tesla is healthy for my Traco TME DC-DC converters?
Hopefully, traco will specify the sensitivity of their supply design to external DC magnetic fields. There will be minimum as well as maximum orientation sensitivity, depending on the design of their ferrites.
There will also be a temperature component to that sensitivity as well, higher temps of course being worse for magnetic sensitivity.
As posed, that specific question is not answerable by anybody other than the manufacturer.
peranders said:
Go here:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Under Biot-Savart, there is a calculator for a single loop magnetic field.
Just looked at what the link gives..
Click on ele and magnetism...then biot-savart, scroll down to the pic of the single loop, click on it.
Cheers, John
Magnetic field near straight filamnetary conductor
For a long straight, filamentary conductor
B = mu0 2i / 4 pi R
where mu0 = 4 pi x 10^-7, i is the current in amps and R is the radius in meters.
cancelling the 4pi gives
B = 2 x 10^-7 i /R
For R = 1cm and i = 300A, this is 6x10^-3 T
For an extended conductor, you should really integrate over the cross section, but usually just measuring the radius from the centre is close enough, which reduces B a bit more.
I would be amazed if 6mT had much effect on the ferrites.
For a long straight, filamentary conductor
B = mu0 2i / 4 pi R
where mu0 = 4 pi x 10^-7, i is the current in amps and R is the radius in meters.
cancelling the 4pi gives
B = 2 x 10^-7 i /R
For R = 1cm and i = 300A, this is 6x10^-3 T
For an extended conductor, you should really integrate over the cross section, but usually just measuring the radius from the centre is close enough, which reduces B a bit more.
I would be amazed if 6mT had much effect on the ferrites.
Re: Magnetic field near straight filamnetary conductor
As you stated correctly, the formula is for a long straight conductor.
Outside of the actual conductor, assuming circular cross section, the field will be independent of the size of the conductor. If the conductor is not cylindrical, that will not be correct.
For the problem at hand, if the supply is between the two conductors, field will enhance, it will be necessary to double the strength calculation...a whopping 12 mT..
If the 300 amps forms a loop, it can be as much as pi times 6 mT. It was the loop case I linked to.
Sheesh, at those field strengths, shouldn't there be some kind of safety barrier???
Cheers, John
ps...it's great to see others using Max's stuff..
PigletsDad said:
As you stated correctly, the formula is for a long straight conductor.
Outside of the actual conductor, assuming circular cross section, the field will be independent of the size of the conductor. If the conductor is not cylindrical, that will not be correct.
For the problem at hand, if the supply is between the two conductors, field will enhance, it will be necessary to double the strength calculation...a whopping 12 mT..
If the 300 amps forms a loop, it can be as much as pi times 6 mT. It was the loop case I linked to.
Sheesh, at those field strengths, shouldn't there be some kind of safety barrier???
Cheers, John
ps...it's great to see others using Max's stuff..
oops, missed a decimal 12mT*500=6T
or way beyond saturation for any ferromagnetic material
for a 2-D problem (some free) FEM programs are available
http://www.diyaudio.com/forums/showthread.php?s=&postid=251006&highlight=#post251006
http://www.quickfield.com/free_soft.htm
or way beyond saturation for any ferromagnetic material
for a 2-D problem (some free) FEM programs are available
http://www.diyaudio.com/forums/showthread.php?s=&postid=251006&highlight=#post251006
http://www.quickfield.com/free_soft.htm
jcx said:
The mere presence of the iron will suck the field lines into new and wonderful patterns...never to your advantage of course...
jcx said:
Tisk, tisk..gotta watch those decimal points..shame on you
Iron typically begins saturation in the 1 to 2 tesla range, that's why we don't use iron as the dominant flux path for anything over about 2 tesla. It certainly helps even in the 5 to 10 tesla range, but we use some high falutin analysis programs which have the iron characteristics embedded within even at 4 kelvin.
peranders said:
Nah, 2 kilo is childs play. Let me know when you start woikin wit some real current..
Your test was very nice, thanks for tellin us bout it, I enjoyed it.
Cheers, John
peranders said:
A tad more at times. 2kA is only a #18awg sized conductor. And at that, I have to put the magnet into an 8 to 10 tesla background field to limit the conductor capability...the leads can't handle the conductor without the external field to supress the wire's inherent ability.
30k is the limit for the highest current supply we use, but honestly, we don't use that one too much lately. We got's lots of supplies in the 5k range, and scads of em in the 300 or so range.
Connections are typically 200 microinch surface finish, silver plated, and many times we have to plop a 3 mil layer of indium foil on the contact area to keep resistance down. (soldering isn't an option for contacts which are 6 inch by 12 inch in area and made of 1 inch thick copper). Joints of that size will normally have 8 to 10 pieces of 1/2 inch diameter hardened steel or titanium bolts with bellvilles to maintain contact force.
Oh, a good rule is a thousand amperes per square inch wire section. It keeps the insulation from frying. 500 mcm is 500 amps.
Cheers, John
Umm, eva?
He did say...
""My application is two 300 A, DC 10 x 20 mm copper bars 60 mm apart (50 mm space). Between the copper bars I’m going to place a H-bridge driver including three 12 V DC-Dc converters."".
The ferrites in the converters could possibly be affected by the external magnetic field. That was the OP question.
Cheers, John
He did say...
""My application is two 300 A, DC 10 x 20 mm copper bars 60 mm apart (50 mm space). Between the copper bars I’m going to place a H-bridge driver including three 12 V DC-Dc converters."".
The ferrites in the converters could possibly be affected by the external magnetic field. That was the OP question.
Cheers, John
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