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Old Feb 17, 2003, 01:02 AM
MX is offline
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Here's how mine looks all packaged.

MX
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Old Feb 17, 2003, 02:43 PM
jeffs555 is offline
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"Simplify, then add lightness"
Andy,
If the lamp is not bright enough across R3, you can always put 2 power diodes (1n4001) in series with the 12v supply to the charger. Then if you put a 1.5v bulb across the diodes, it will be bright when charging, and go out when it is charged. You could also put the base-emitter junction of a power transistor in series with the 12v supply, and drive an led with the collector. Then put a low value resistor across the base-emitter, so that at low current the transistor is turned off. Your L200 charger is so simple, it would seem a shame to have to add a complex circuit just to get a charging indicator.

Jeff
Old Feb 17, 2003, 04:12 PM
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Hornet Basher
Oops, I hadn't scrolled down far enough on that page.

Thanks for pointing that out.
Old Feb 17, 2003, 04:22 PM
Alfredo Rubio is offline
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Andy, I saw you design, but I do not know about electronic may I can get a friend to do it, I saw you recomend many options in R3, do you can tell me wath is to charge a 145 li-poly cell?
Old Feb 17, 2003, 07:01 PM
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Charg indication


Jeff,

The thing about detecting across the diode(s) is that there will still be as much residual voltage across the diode(s) at the end of charge as at the beginning (the voltage drop is largely independent of the current through it). I haven't thought enough about the process to say it definitely won't work, I will have to try some tests in the lab.

Alfredo,
You will need to keep the current down to 145mA or less, this will mean using 3.3 Ohm for R3, I use 4.7 Ohm which reduces the current to about 90mA. Lower charge current will preserve Li-poly cycle life.

Andy.
Old Feb 17, 2003, 07:25 PM
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"Simplify, then add lightness"
Andy,
The diodes are only to protect the lamp from excessive voltage. As current drops, the resistance of the lamp will cause the voltage across the lamp to drop, and the lamp will dim. As the current continues to drop, the lamp will go out.

Jeff
Old Feb 17, 2003, 08:04 PM
Alfredo Rubio is offline
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Thanks Andy...
Old Feb 17, 2003, 08:06 PM
Pierre Audette is offline
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Does the Maxim 1811 chip work for lipoly?
http://pdfserv.maxim-ic.com/arpdf/MAX1811.pdf
Old Feb 17, 2003, 09:52 PM
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Jeff,

I would think two diodes in positive line to the regulator would reduce the output voltage from the regulator. Each diode will drop .6 volt reducing the input to the regulator to 10.8 volts with a 12 volt input. Since the regulator drops about 3 volts I believe you would have to change the values of R3 and R2 to maintain the 8.4 volts.

Also as long as the diodes are conducting there will be 1.2 volts across both regardless of the current to the regulator so the bulb will always see 1.2 volts. I'm not sure what the resistance of two forward biased diodes is, but the current through the bulb will depend on it's resistance and the parallel resistance of the diodes. It could be calculated easy enough using ohm's law.

Grant
Old Feb 18, 2003, 02:25 PM
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.................
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Quote:
Originally posted by MX
I built one too and it works very well. I made mine with switches
to select 1 or 2 cells and 560mAh or 1020mAh capacity.

MX
what charge rate are you using for the 1020 cells? i hear they can be charged at 1C are you following the .8C method or the 1C method?, and what value of Resistor are you using to get that rate? i have parts on the way to make 4 of these and 2 of his gel cell chargers, what a great site!

Dylan
Old Feb 18, 2003, 03:53 PM
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Quote:
Originally posted by Dylwad
what charge rate are you using for the 1020 cells? i hear they can be charged at 1C are you following the .8C method or the 1C method?, and what value of Resistor are you using to get that rate? i have parts on the way to make 4 of these and 2 of his gel cell chargers, what a great site!

Dylan
I'm charging at 0.7C as recommended on the website for my
560 and 1020 packs, and about 0.9C for my 880 qualcomms
charged at the 1020 setting.

I don't know the exact resistor values, because to get the low
resistance I stuck about 8 10-ohm resistors in parallel. I just
added and subtracted resistors until my current measured where
I wanted it.

The website shows how to calculate the resistance needed.
Here's what I calculate:

1020 0.7C = 0.91 ohms 1C = 0.637 ohms
560 0.7C = 1.67 ohms 1C = 1.16 ohms


MX
Old Feb 18, 2003, 04:16 PM
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Simpler Charger


Jeff, Grant

Diodes do not obey Ohm's law and measuring their resistance won't tell you much. The problem with putting a charge monitor circuit across a diode is that even when the current is a fraction of a milliamp it still drops about 0.6V .

If the 1.5V bulb works then this might be the best solution, have you actually tried it Jeff. Incidentally the diode is in the circuit to prevent discharge of the battery into the L200, it is not a functioning part of the regulator circuit, it could be left out if you always have an input to the charger when a battery is connected. In any event putting a bulb in parallel with it will nullify its usefulness as a blocker.

Adding a second diode in series with the original one will not effect the voltage regulation as it is inside the feedback loop. All that an extra diode will mean is that the overhead of input to output voltage will go up by 0.6V.

Let's try and keep it simple so more people are encouraged to make their own gear. If anyone out there knows an ultra-simple way of getting an indication when the current drops below about 20mA, please post it.

A word of caution: The L200 regulator is capable of 2Amps output current, the diode quoted is a 1N4001 which is only rated at 1Amp so if you are planning on using this circuiit at above 1Amp, you will need to select a higher rated diode like 1N5401.

Andy
Old Feb 18, 2003, 04:18 PM
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@MX
Thats a nice looking box there.
Old Feb 18, 2003, 04:19 PM
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Even simpler charger


How about this then, The Maxim MAX1811 IC is a dedicated Lithium charger IC which has a status LED built in 100/500mA selector, it has recovery trickle charge for over-discharged cells and the best bit is that it only needs 4 or 5 components including the IC and it doesn't require an external shunt resistor.
Now for the disadvantages, it needs a fairly closely controlled input 4.35 to 6.5V, it is dedicated to one cell charging and it only comes in a surface mount package.

See http://pdfserv.maxim-ic.com/arpdf/MAX1811.pdf for data sheet.

I'm not sure if any of the normal distributers carry this IC but Maxim are pretty good with giving samples especially if you have a bona fide reason for wanting them.

I have some samples coming, lets see how it well works...

Andy
Old Feb 18, 2003, 07:17 PM
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@Andy

I seen that chip posted somewhere else and figured it would be expensive.

Keep us updated on your progress I will try to get some samples as well.

I guess you could use a 7805 regulator and a 9Volt for supply.
that would be 3 more components though. How about 3 AA for a feild charger?


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