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Old 6th March 2006, 01:47 AM   #1
Optical is offline Optical  New Zealand
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Default Best way to get 12V from 17V @5A

Hi everyone
I have a transformer outputting a rectified and filtered 17VDC, and want 12VDC to power a mains powered fast battery charger.. is the best way to experiment with a series power resistor to drop teh voltage down? although i need all the current i can get..

otherwise would using something like the opa549 be viable?
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Old 6th March 2006, 03:14 AM   #2
BWRX is offline BWRX  United States
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I'd use a bunch of 5A diodes in series or an npn pass transistor with a 12.7V zener reference.
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Old 6th March 2006, 05:36 PM   #3
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Default SimpleSwitcher

Or- you can use a switching regulator. These are alot easier to design and build that one might think. Look at National Semiconductor's website at the SimpleSwitcher series of integrated switching regulators. The simplest I can think of would take only one diode, one coil and two caps.

If you're stuck on doing a linear regulator, please keep in mind the power that will be dissipated: (17-12)V x 5A = 25W dissipated as heat.

Hope this helps!

Steve
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Old 6th March 2006, 07:46 PM   #4
Optical is offline Optical  New Zealand
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yes id much rather hold onto that 25W..
thanks i'll have a browse through that website
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Old 6th March 2006, 08:17 PM   #5
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Cool! Just holler if you need any help. That's why we're here......................
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Old 8th March 2006, 12:47 AM   #6
BWRX is offline BWRX  United States
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A buck regulator would be better but it will also cost more.

The diodes, resistor, and transistor (darlington in a TO220 case would be good) will be simpler, easier, and cost less. You could use 3 diodes before the regulator (you coud also use a higher value zener and put one of the diodes after the transistor for reverse polarity protection on the output) to drop 2.1V and give you 14.9V at the collector of the transistor. Now the transistor only has to disspiate (14.9-12)*5=14.5W. You'll only need a total of 6 parts (3 diodes, 1 zener, 1 resistor, 1 transistor) plus a heatsink and can build the circuit easily (point to point assembly) by hand. Plus you can get each part for about a dollar

For the buck regulator you can either buy an off the shelf product or design your own pcb, order and buy all the parts, and build it. Don't forget the time it will take you to do all this.

In the end it's still your choice! Good luck
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Old 8th March 2006, 06:35 PM   #7
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I think the point he was trying to make was that he would rather "hold onto those 25Watts" and not dissipate them as waste. If one simply looks at any of the SimpleSwitcher Datasheets, they will not only see what the chips can do, but also schematics, design procedures, and suggested parts vendors. The design procedures are very cut-n-dried, nuts-n-bolts, rahter than alot of useless theory. They get right to the point.

I have constructed several buck, boost and inverting SMPSs using these chips and follwoing their design procedures. I have found them to be very helpful in understanding how they operate and how changing components affects the circuit operation as a whole.
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Old 8th March 2006, 08:38 PM   #8
Eva is offline Eva  Spain
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Furthermore, another disadvantage of the linear regulator is that those 25W are wasted power from the transformer, thus the transformer VA rating available for powering the load is actually reduced. The buck regulator does not suffer from that drawback as the input current is *lower* than the output current, thus the VA rating of the transformer is kept.
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Old 8th March 2006, 08:50 PM   #9
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hmmm, fast battery charger -- go to the Texas Instruments website -- depends upon the type of battery etc. For lead-acid batteries there some good application notes on using the UC3906 -- this is a $5.00 chip programmed specifically for this application.
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Old 8th March 2006, 11:50 PM   #10
BWRX is offline BWRX  United States
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Quote:
Originally posted by N-Channel
I think the point he was trying to make was that he would rather "hold onto those 25Watts" and not dissipate them as waste. The design procedures are very cut-n-dried, nuts-n-bolts, rahter than alot of useless theory.

I have constructed several buck, boost and inverting SMPSs using these chips and follwoing their design procedures. I have found them to be very helpful in understanding how they operate and how changing components affects the circuit operation as a whole.
I realize that a buck converter would be the best solution but I'm saying that a simple linear supply (even though it wastes a fair amount of energy) would be easier to build and less expensive. It would also seem to be a more suitable project for Optical who doesn't sound like he has a lot of experience with power supply design What better place to start than with one of the simplest linear supply designs? Unless they give you a proven pcb layout chances are good that most people without any experience won't come up with a very good layout. The whole make your own SMPS route is still more complicated than you're making it sound Don't forget that you've probably done this quite a few times already!

Alternatively, jack's solution seems like the way to go.
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