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| The Projects Forum Working on an electronics project and would like some suggestions, help or critiques? If you would like to comment or assist others with their projects, this is the place to do it. |
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#1
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Hi,
I am working on a project and I was trying to find some information regarding how to power devices inductively off of a power line. The lines will be 100A +- 50A and anywhere between 4kV to 12kV. What I am building will be clamped around the line. I am looking to make a 9VDC power supply to power a device and also to charge a battery. I was able to find this post regarding a similar idea. http://forum.allaboutcircuits.com/showthread.php?t=2688 My first thoughts were to use some coils placed near the line to induce some voltage a la Faraday's law. However, a lot of turns were required to do that. I am now considering using a CT like device around it. I am just looking for some insight into this. A CT has a burden or some type of load across it, never open circuited so to prevent huge voltage differences. I guess what I'm wondering is how to get the voltage I want out of the secondary. A 100:5 CT will give me 5A at normal current, is this regardless of the line voltage? If I put the same CT on a line with half as much voltage, will it still deliver 5A on the secondary with a 100A primary? The reason I ask is because the power transfer doesn't quite make sense to me. 12kV 100A primary put through a 100:5 CT with a 1 ohm load will have 5A and 10V across the secondary. Why is the voltage not 240kV as expected from a 100:5 ratio? This seems to ignore power transfer equation for transformers. Am I to model the secondary as a perfect dependent current source? In reference to the link I posted, the poster mentioned putting a diode bridge with cap and zener to create his power supply. My only question is how would he determine the voltage input to the bridge? If he needed a 3VDC output, he'd need around 5-6 VAC peak to generate it. If the power supply is to load the secondary windings, I don't see how to set an input voltage. Sorry about the long post, but thanks in advance to everyone. |
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#2
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In order to be able to get useful power from inductance, very close-coupling is needed. I believe the effect is used to recharge batteries in electric toothbrushes.
The field around a single conductor is not dense enough to get power from it. That is why transformers use many turns and a flux-coupling core to be efficient. You probably know most of this, but here is apointer to the Wikipedia article on CT's - http://en.wikipedia.org/wiki/Current_transformer. |
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#3
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Thanks for the reply. If I attach the CT around the wire, I'm assuming that the coupling will be much greater. Even with better coupling though, the questions at the bottom of my first post still apply. If I connect a rectifier bridge, cap and LM317 regulator to the secondary of the CT, what will it rectify? IE: If it is a 100:5 CT and 100A is on the primary, will it be rectifying :V = 5A * load impedance of my circuit?
Beenthere: I think the CT will will help with coupling, if not, please tell me Last edited by denman_100; 10-28-2007 at 01:03 PM. |
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#4
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You're missing the point of a current transformer. It exists to provide an analog of the main power line condirions so those conditions may be safely monitored. It is not a device that may be used to produce power.
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#5
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contact me
<snip> i would be happy to help you!!! Last edited by beenthere; 11-02-2009 at 11:53 PM. Reason: email removed by moderator |
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#6
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If you are going to offer help, you need to post it here so others looking for similar info can have access to it if needed..... that is the whole point of these forums.....
__________________
There is no spoon -Neo- ![]() Disclaimer: The example programs,circuits, projects and information I provide or post on this web site are for educational purposes only. By copying anything from this site posted by me, you agree to the "as is" nature of the programs, circuits, information and to the statements listed in this disclaimer.No warranty or liability is expressed or implied. Working with AC /DC voltages can be dangerous and even deadly. Proceed at your own risk! Last edited by beenthere; 11-02-2009 at 11:54 PM. |
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#7
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and to warn one of impending death.
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#8
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You took the words right out of my mouth.
__________________
__________________________________________________ ___________________________ Disclaimer: Any information that I post is intended for educational purposes only. No warranty or liability is expressed or implied. Note: Working with household line voltages (Mains) can be dangerous. Proceed at your own risk! |
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