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How to configure LM317T to deliver 5 amps?

Hi guys, I'm new to the site.
I wonder if someone could help me resolve my problem.

Well, I'm trying to build a Laptop AC Adapter, but I can't get these components right now:
-A Transformer
-A voltage regulator other than LM317T and I have no experience with other ICs.

So I'm using a DC to DC converter.
The input power supply is a Z12 Lexmark printer AC adapter. It gives a DC output current of 830mA under 30 volts (actually it's 37.5 volts, I've checked this with my multimeter).
However, the output I would like to get is 4-5 amps under 19.5 volts.

I would like to mention that I've already given it a try. But the output current I've got wasn't enough to feed the laptop.
I've used an LM317T and adjusted the voltage with a 6.6k resistor and a 47k potentiometer down to 19.5 volts.

If there is a way to boost up the current with a joulethief or using a set of transistors or parallels LM317Ts, please tell me.
Also, I hope that you give me the part number of all the components to match up with the LM317T because I must use this type of regulator.

Thanks for your help and please consider my situation.

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junkrecycler (author) 4 years ago
Okay, may be I'm strange or crazy, but there is something I believe in:
"If you're looking for small something in this huge world, only time would seperate you. No distance, no other dimensions as well. Always keep looking, and someday time you've spent will end up into past."

Thanks guys for trying to help me, but to tell you the truth...

YOU LOSE!

No one has succeded afterall in helping me.
I thought that someone would have some brilliant ideas and could bring some new concepts.
Yet...

CONCLUDED.
Make what you believe.
Make it work.
   Annnnnd
We each who have over
10,000 Hours circuit D&D under our respective belts.
Will be kneeling at the Door to your work shack.
    Buuuuut
Only if the power output is more then that coming in !

D&D == Design and Development

There is a way to increase current using an inductor because it has the propensity of storing energy in a field that surrounds it.
Alas the voltage is conversely reduced..

    Also
There is a way to use an array of condensers to increase current by charging them in series and discharging them in parallel.
Alas the voltage is conversely reduced..


Until then, in my case 20 years is all you have,  I HAVE LOST
to the intuitive tunnels in your mind.



junkrecycler (author)  iceng4 years ago
Sorry for being too rude. I was too excited in making a new experiment, and suddenly words have flown out away of me. Please I'm trying to be so nice and clear with you guys, I don't want any trouble. I did understand the relation that joins volts and amps at least.(Only at 98.382% for now LOL) Anyway, this was my first topic on the site! Of course, thanks to me and ICENG, you didn't get bored for about two days.
All is well :-)
junkrecycler (author) 4 years ago
Well for now I've got a better idea and I will use it on my own risk. I will post a new topic when ready and please try to understand me this time because I'm already understanding and respecting all of you BRO and PRO.
junkrecycler (author) 4 years ago
Anybody knows a way of how to step down AC voltage without using a transformer. I have a bunch of different capacitors, resistors, diodes, transistors and bridge rectifiers.
Please help me, sorry for being too much annoying! :'(
No, you absolutely HAVE to use a transformer to reduce AC mains to safe levels and isolation.
iceng4 years ago
This is what you must construct 
IF your source can Deliver.

A
lm317-5 amps-power-circuit.jpg
junkrecycler (author)  iceng4 years ago
Well, this might be helpful but please could you provide a short description of the sub-circuits functions or at least post a link to the official page. The input source I have can deliver up to 830mA under 30 volts.
Thank you in advance!

" IF your source can Deliver. "  which it cannot !!

is what I said

and everyone else said

Unhappily you cannot get more out then you  put in

Also until you grasp this one understanding
Explaining the Real details will be lost in the windmills of your mind
iceng iceng4 years ago
You can't get more power out (19.5 x 5= 97W) than you put in (0.83 x 30= 25W)

You can't do what you want with the parts you have.

Sorry.
junkrecycler (author)  steveastrouk4 years ago
Thanks for answering me sir, but I thought that there was a way by storing energy using capacitors or inductors and then delivering it back to the circuit again. That would solve the problem of not having enough power but I didn't try it. I'm always looking for an answer.
No, a joule thief does the opposite, increasing voltage at the cost of more current. (same watts)...going from less than a volt up to about 4 volts for very short pulses, at 5-10x the current draw.

You can increase current, but only at the cost of dropping more voltage. (same watts)

Watts are total power, volts multiplied by amps. Watts stay constant while volts and amps can be interchanged with different transformations.
junkrecycler (author)  frollard4 years ago
So there is no hope making this converter.
Any better ideas, at low cost?
buy a laptop power supply. I wouldn't trust powering my laptop from anything but the original article.

Device worth upwards of several thousand dollars...
...to cheap out on the power supply that drives everything in that device.
No it WOULDN'T

YOU CAN'T GET MORE OUT THAN YOU PUT IN.....
Only for a very short period.

Essentially over time power in = slightly less power out.

if it didn't we could easily build over unity machines. (perpetual motion)

For what they cost it's lot easier and possibly not a lot more costly to buy one & get a neat package.