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olddawgsrule
Advanced Member


USA
1434 Posts

Posted - October 08 2012 :  10:39:08  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Looking to make up a 125mA dis-charge circuit for my 14.4v drill batteries.

First; I believe 125mA is the C20 rate for these (2.2-2.8 ah batteries)
Second; I believe I have the circuit correct
Lastly; getting that LDR in the circuit for future projects



LDR circuits on page are my attempts at seeing and understanding what it may take to do this.

Old in age, not in mind, so
'Teach me something new'!

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USA
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Anothercoilgun
Junior Member



140 Posts

Posted - October 08 2012 :  16:03:53  Show Profile Send Anothercoilgun a Private Message  Reply with Quote
Looks nice. Question, is the C20 rating important or just a ballpark for make a light source?

As the battery voltage drops, so to will the current drawn from the LEDs. You may end up with a much longer discharge time than anticipated. 3.2v LEDs will still light well below 3.2v. But this also depends on battery type. Lithiums will (may) self cut off at their discharge voltage. Modern ones have internal circuits to do this. NiMH and NiCd will continue to drop below discharge voltage.

This is why I like LEDs for lighting. They are natural joule thieves.

OLD Boy's Bench?
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - October 08 2012 :  16:48:48  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
ACG, thanks for the reply!

The idea is 2 fold..
One I need a 125ma draw for a few batteries I hope to re-claim.
Reason for the C20.
These are the 'old' style re-chargeable'.
Not NiCad or Nim or whatever else is out there right now.


Secondly, I can see crap in the garage when I leave in morning!
Tired of tripping over whatever I left out...

Understand the current drop and why I went where I did.
These are 14.4v batteries I'm working with.
They will hold 16v and hopefully the lights will drop out somewhere around 11v.

Time will tell or maybe you will???

I'm trying to teach an old battery a new trick!
Ya and trying to stop this 'Olddawg' from tripping over whatever he can't see in the morning (which can be quite a bit)..

Old in age, not in mind, so
'Teach me something new'!


Edited by - olddawgsrule on October 08 2012 16:55:11
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - October 08 2012 :  16:59:58  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
quote:
OLD Boy's Bench?

Of course!
Now talk to me ACG!!
Another coil gun???

We need to talk some...
What you been doin'??

Old in age, not in mind, so
'Teach me something new'!


Edited by - olddawgsrule on October 08 2012 17:01:45
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - October 09 2012 :  16:45:48  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Olddawgsrule
the discharge circuit shown in the link below might be of use to you ( it also overcomes the "dischage rate current tail-off" that Anothercoilgun pointed out )

You must be logged in to see this link.

ron
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - October 09 2012 :  17:42:40  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Thank you RonO!
Again need time to digest and adjust.

Scary thing about learning is the time it takes and what we do..

Age has taught me to take the time.
Seems adjustment takes some also, LOL..

Old in age, not in mind, so
'Teach me something new'!

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kcarring
Moderator



Canada
1057 Posts

Posted - October 09 2012 :  23:33:40  Show Profile Send kcarring a Private Message  Reply with Quote
Olddawg...

I'm gonna step out on a limb and make an assumption:

you don't want to -just- light LED lamp, rather you wish to -also- discharge the battery, to se, if, in fact, you are "gaining" during desulphation.

Your circuit will work, sort of... but, keep in mind...

As the the battery voltage drop, so does the current, and so really, running it for a period of time... doesn't really say a lot because if you graphed the energy, the wattage would start at a peak and slowly decline, due to ohms law. so at best, it is an approximation.

What you'd really want is a constant current supply, staying well above the operating voltage of the LED array.

Having said that, this is not an efficient circuit, BUT, it would server as a better load tester!

Take this info for what it's worth:

The LM317 regulator gives out a constant voltage of 1,25 volts between ADJ and Vout, so by adding a resistor between these two outputs, you'll get a constant current.

Ohm's law says that U/I=R, which means that Voltage divided by Ampere makes resistance.

If you want to connect one or more luxeon 1W LEDs, (power consumption of 350mA), the calculation should look like this: 1.25 (the constant reference voltage of the LM317) divided by 0.350 (the LEDs power consumption) makes 3.57. So if the resistor is 3.57ohmd, constant current will be 350mA. The closest E12 value is 3.9 ohms, it will give you a constant current of 321mA.

This idealogy can, of course be modified to different loads & their requirements.

Cheers

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - October 10 2012 :  16:02:52  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
You got it KC.
Exactly what I'm attempting.

Dis-charge the battery at rate and have a bonus LED light.

I will look into this LM317 regulator.
By what your saying, it's what I need to not only accomplish the task in mind, but needed if I proceed (as I'm thinking) to build the LED workbench lighting I'm contemplating.

Thank you!


quote:
Originally posted by kcarring

Olddawg...

I'm gonna step out on a limb and make an assumption:

you don't want to -just- light LED lamp, rather you wish to -also- discharge the battery, to se, if, in fact, you are "gaining" during desulphation.

Your circuit will work, sort of... but, keep in mind...

As the the battery voltage drop, so does the current, and so really, running it for a period of time... doesn't really say a lot because if you graphed the energy, the wattage would start at a peak and slowly decline, due to ohms law. so at best, it is an approximation.

What you'd really want is a constant current supply, staying well above the operating voltage of the LED array.

Having said that, this is not an efficient circuit, BUT, it would server as a better load tester!

Take this info for what it's worth:

The LM317 regulator gives out a constant voltage of 1,25 volts between ADJ and Vout, so by adding a resistor between these two outputs, you'll get a constant current.

Ohm's law says that U/I=R, which means that Voltage divided by Ampere makes resistance.

If you want to connect one or more luxeon 1W LEDs, (power consumption of 350mA), the calculation should look like this: 1.25 (the constant reference voltage of the LM317) divided by 0.350 (the LEDs power consumption) makes 3.57. So if the resistor is 3.57ohmd, constant current will be 350mA. The closest E12 value is 3.9 ohms, it will give you a constant current of 321mA.

This idealogy can, of course be modified to different loads & their requirements.

Cheers

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!



Old in age, not in mind, so
'Teach me something new'!

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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - October 10 2012 :  16:59:28  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Okay, let's make an attempt at this.
Here's the 14.4 LED circuit with the LM317 added.

I'm hoping I have the connections and resistors right..



Old in age, not in mind, so
'Teach me something new'!

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kcarring
Moderator



Canada
1057 Posts

Posted - October 10 2012 :  22:13:49  Show Profile Send kcarring a Private Message  Reply with Quote
You'll know if it it's right , if:

1. the current remains the same as the voltage drops
2. the LEDs are not overdriven.

I see you have 4 LEDs in series.
At 3.2V per LED, that's a 12.8V drive - not including the added resistor on the tail end of the 4 LEDs.

Do you plan on taking the battery no lower than 12.8V?

You want to make sure that the series leg, will still function properly at the lowest voltage point you intend to take the battery "down to".

On the other hand, you want to not be overdriving the LEDs at the voltage peak of the battery, either.

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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Anothercoilgun
Junior Member



140 Posts

Posted - October 10 2012 :  23:22:42  Show Profile Send Anothercoilgun a Private Message  Reply with Quote
Let the LEDs charge a small solar panel along with lighting the room. Loop that panel back to battery pack. Send me all royalties when you make it big :)

OLD Boy's Bench?
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - October 11 2012 :  16:18:28  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Theory was that is a 14.4v circuit...
I was over-driving the LED's at start (being the battery will hold 16+v). 20% over voltage to 20% under and time it.

Well, I'm not around to see the lights fade or stop, so the theory lacked..

The LM317 seems to be the ticket item I am missing.
Still looking for that full charge (16v), then dis-charge to 12v.

This may not be the best for recovery of some the older batteries, but seems to be a good usage cycle for everyday running.

For the batteries I'm trying to recover, I hope to build a circuit that will run them down to 9v.

Thank you KC for the input!
This is really helping!
Just disappointed I missed the circuit by that much...



Old in age, not in mind, so
'Teach me something new'!

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kcarring
Moderator



Canada
1057 Posts

Posted - October 11 2012 :  23:50:34  Show Profile Send kcarring a Private Message  Reply with Quote
if you wish to run the bat down to 9V, consider that the LM317 will impose a voltage drop, within itself. Using a constant current, you KNOW you will get X amps per hour, for example.

We don't really care where the voltage of the battery is at, at any point, all we care is the circuit is designed such that it will maintain our goal of current, but...

If the bats start around 15-16V, yet we want to be running the light, still, at 9V (and the LM317 itself imparts a loss of voltage) THEN, in a very inefficient (but more accurate) fashio, we'd set the string of LEDs in two per series string.

That way, no matter whether the voltage is 15-16V, or only 9... we can still produce the desired 3.2 X 2 = 6.4V. The job of the circuit is to disregard the input voltage and maintain a specific current.

When the battery drops to 9V, despite the volt drop across the LM317, we'll still have our needed 6.4V.

You'll have to put a heat sink on the LM317, a nice large one will help. Like I said, this circuit is designed to maintain an AMP flow, not be efficient; in fact, it's pretty darn inefficient!

But it will serve as a constant load for battery capacity checking.

Cheers

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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kcarring
Moderator



Canada
1057 Posts

Posted - October 12 2012 :  01:18:30  Show Profile Send kcarring a Private Message  Reply with Quote
The problem, is, though... the output voltage will change! (Because you are setting the circuit to control current, not voltage!)

So...

you'd have to build the array to handle the highest voltage it would see (i.e. if the bats are at 16V, you'll see about 2V drop, so 14V) - done by adding current limiting resistors...

End result, very dim lamps by the time you reach 9.25V (the theoretical lowest you'd be able to run the circuit, properly - i think...)

A better all round approach might be to use something like this:

You must be logged in to see this link.

It's all good learning and experimenting tho...

Myself, I use a much less scientific approach. But I have thought about the same thing... getting a good lamp to be constant load source. There are devices you can buy, to test batteries, but all that i have seen, essentially waste the energy.

My tests have been more or less "qualitative", i.e. can it load THE SAID lamp, for ... 3 hrs? 4 hrs? What is the final voltage after doing so? What is the levelling plateau of operation.

I think if you got 49'ers input... he also really knows and like monitering voltage...

If a battery improves (from being sub standard...) so does it's operational voltage... how (at what voltage) does it actually maintain it's C20 load? At what voltage is it leveling off at?

It's not real scientific, but in a way, it can be very indicative and more than accurate enough to see (or not see) improvement!

cheers



~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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kcarring
Moderator



Canada
1057 Posts

Posted - October 12 2012 :  01:35:26  Show Profile Send kcarring a Private Message  Reply with Quote
This is interesting too. (altho, another wasteful circuit...)

You must be logged in to see this link.

BTW:

If, you create a constant current source, and just use incandescent lamp... then you need not worry about the semantics of LEDs, you get your constant current (for amp hours) and the incandescent will handle the voltage ups and downs (albeit dimmed). A 10 watt halogen, for example. Or in your case, perhaps a 5 watt auto bulb.

~~~~~~~~~~~~~
You wouldn't laugh at my igloo if you knew how cold my beer is!
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olddawgsrule
Advanced Member



USA
1434 Posts

Posted - November 11 2012 :  09:17:31  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I did buy the LM317 and received then a week or so back.
With the recent weather and events, I've just returned to them to learn and experiment.

Quite a nice experience.
Still a bit confusing when used as voltage regulation..
Just looks like shorts all over the place, but that's for me to come back to and learn.
But I can see how they regulate current much easier.
Wiring just makes sense to me.

I also saw when used as a voltage regulator, I could expect a 4watt loss.
But when used as a current regulator, talking mW loss.
This direction seems to make the LM317 practical for my use.

The attached image is my continuation in this learning experience.
Ignore the lower LED circuit and LDR pic's for now.
They're notes for me.

Oh yes, I also purchased a 12v 10watt LED lamp to play with and is now in the diagram.



Old in age, not in mind, so
'Teach me something new'!

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