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stube40

Feb 9, 2010
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If your load has zero resistance then there is no need to have a 150V power supply. All of that will be dropped across your current limiting resistor.

You are better off using a single 12V lead acid battery (it will be more than capable of supplying 40A).

If your load really has a zero resistance then all transient conditions can do is vary the magnetic field surrounding the conductor as far as I can tell. That may be an issue if your device is superconducting or if it is constructed in such a way that excess magnetic fields may affect it in some mechanical way.

<shrug>

Still, you could use a similar arrangement to introduce transients if you replace the battery with your battery and current limiting resistor combination.

OK Steve, thanks for that. Yes I appreciate your thoughts about the zero-resistance and the total voltage drop across the current limiting resistors, however we are dealing with some pretty weird loads that dont behave like you might expect - this is the crux of the experiment.
 

55pilot

Feb 23, 2010
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I have been thinking about this and frankly I keep coming back to the point that Steve made: If the resistance is "zero" then why waste so much effort on doing it at 150V?

Is your real goal to have a 40A current supply? Are you dealing with a negative resistance device that requires a large voltage to overcome an initial large resistance then enters a negative resistance zone? If so, some of the concepts from CCFL lamp control can be applicable after a huge scaling up in current and scaling down in voltage.

Are you testing SCRs? The voltage is too high for a superconductor and too low for any kind of gas/plasma conduction.

Bottom line is that there may be other and better ways of coming at this, but my ability to help you is hitting a brick wall without knowing the parameters of the load you are dealing with and a better understanding of what you are trying to do with it.

I am not trying to be an ass about it (which I can be at times). This is really a case of not being able to help without better understanding the problem. The only avenue you have left open is to brute force it. My gut tells me that this is a problem that can be solved with a scalpel, but the only tool you are allowing is bulldozer.

---55p
 

stube40

Feb 9, 2010
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OK, this is where it gets tricky. I appreciate that this is an open public forum and it was me that came here looking for help, not vice versa. However, we are a research organisation who might well be on the verge of discovering something that means re-writing some of the basic electrical laws that we all learnt at school.

What you have both said makes perfect sense based on the simple V=IR laws that we are all familiar with, but I can tell you categorically right now that these laws break down in certain situations.

However, I'm afraid I have to end it there. I'd LOVE to tell you guys more, but it doesnt work like that for obvious reasons. I appreciate it if that's the end of the thread for now, but it's more than my job is worth to discuss it further.
 

55pilot

Feb 23, 2010
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I understand that you can not post proprietary information. I also understand there are many places where V=IR appears to not work. In all those cases, as you dig deeper into the physics, you eventually discover that the reason it does not work is because R happens to be a function of V or I and not a constant as the equation requires. When express R in those terms, the re-written equation starts working again. Look closely at your physics. Don't embarrass yourself being the next cold fusion scientists!

Finally, it is obvious that you are approaching your "need to vary the current through the load" problem with brute force rather than finesse. If you get someone involved that really knows what they are doing in the area of power supplies, you will be able to open the door with a key rather than bulldozing the house, like you are planning to do.

Good luck.

---55p
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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I also understand there are many places where V=IR appears to not work

From an early comment, it could be that the load is highly inductive.

On of my fanciful guesses was that it was a superconducting electromagnet and this question is coming from someone at the LHC :-D
 

stube40

Feb 9, 2010
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From an early comment, it could be that the load is highly inductive.

On of my fanciful guesses was that it was a superconducting electromagnet and this question is coming from someone at the LHC :-D

We could do with you on our team! When you next visiting Oz?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I realise my work takes me overseas a lot, but this is the first time that being home was termed as "visiting" :D

I live in Perth.
 

55pilot

Feb 23, 2010
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I have to admit that I had not considered the highly inductive angle. That is an even simpler problem that is encountered in high performance solenoid and relay drive electronics where you have to turn on a highly inductive coil by very quickly getting it up to some high current A and then bring the current down to a lower holding current B.

---55p
 

stube40

Feb 9, 2010
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I realise my work takes me overseas a lot, but this is the first time that being home was termed as "visiting" :D

I live in Perth.

I just left Perth in November to move to Queensland. We lived in Carramar for 5 years, so I know Perth fairly well. Ever heard of companies called Telvent or Almos?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I've heard of Almos.

I had some peripheral dealings with them sometime around the mid 90s I guess.
 

trobbins

Jun 15, 2010
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Stube40,

Have you been able to identify the magnitude of the droop in battery source voltage that is effectively causing your output voltage regulation problems?

You could relatively easily put together an active linear regulator that would 'span' the battery voltage change over the time of the test - ie. the regulator would start with say 10V across it, which would reduce to <1V at end of test. The peak dissipation would be circa 400W - not too unwieldy with a row of FETs each dissipating 20Wpk. You need some minimal source ballast resistance for each FET, and separate gate resistors - and you can tweak even more if needed for current sharing. Simple protection to turn off if voltage drop is above say 10V. Use a shunt for current limit protection. Or go the whole hog and use a multiplier for over-power protection.

I've used that form of active drop in series (and parallel) with load resistors to achieve constant current and constant power test loads for VRLA at over 100A.

Ciao, Tim
 
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