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Mesfet
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 « on: July 15, 2008, 11:55:23 11:55 »

Hello
I have just bought a sealed lead-acid battery (6V 4.5Ah , 24HR)
first I want to know what is the meaning of last number (24HR) does it relate to charging time and current charging limitation ?
after that , would you please tell me , will there be any danger or problem , if I use an normal regulator such as 7809 , a 5.6 ohm (2W) resistor and a diode to charge it ?(the charging current would be almost 410m)

In short, I  mean,  do I have to use special charger which produce constant current at accurate voltage or I can use the famous and simple circuit that I mentioned it.

Please excuse me if my English isn't well

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jzaghal
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 « Reply #1 on: July 15, 2008, 01:53:55 13:53 »

Hi,

To charge this type you will need to provide 7.4V limited to 450ma (Or less if you wish.), so that you charge it in 10Hrs.

Not Sure about the 24HR, but I don't think this is needed to provide charging information.

I am sure it must be 7.4V / 450ma.

I would normaly use an LM317T to provide this voltage.

Bye.
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Mesfet
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 « Reply #2 on: July 15, 2008, 02:51:40 14:51 »

Dear jzaghal
Thanks a lot for replaying to my question.
But there is remain a question , to the best of my knowledge ,  Voltages on head of batteries is constant.
for instance , in this simple circuit
12V----/\/\/\__-----\  (A)
__|__
6V     ---
|
/////
A point of (A) the voltage is always 6 V (or a bit more or less than 6 Volt)
Even we use current mirror source instead of this simple circuit.
In all other circuits that I have seen for charging this kind of batteries, there is a calculated voltage and a  resistor between this voltage and a battery, of course this resistor is necessary , there is no doubt, but  if we provide higher voltage than voltage that companies recommend (as you mentioned , for my batteries is 7.4) for example 8.3 , and use a bigger resistor for example 5.6 ohm, instead of 3.3 ohm (7.4v-6v)/3.3R~450m, there would not be different.
Do you agree with me?

I was wondering if others let me and others know their ideas.
Regards,
 « Last Edit: July 22, 2008, 04:18:13 16:18 by Mesfet » Logged
hemlig
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 « Reply #3 on: July 15, 2008, 03:19:03 15:19 »

@Mesfet

have a look at this link:

The answer on your question is that you might destroy the cells if you apply excessive voltage on them.
When lead batteries are used in a car the charger controls both current and voltage.
The battery also behave different in different temperatures.
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Mesfet
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 « Reply #4 on: July 15, 2008, 05:03:19 17:03 »

Thank you so much for your attention
I red that article two or three days a ago, but after I saw your advice , I red it again with more attention. Thank you so much for your advice.
But I think , It's impossible for battery cells to meet the voltage on other side of resistor. As you well know battery cells meet V - IR out of kirchoff's circuit laws and current will be Vsource-Vbattery)/R

I do not know what I cannot understand, in fact I am quite sure you are right out of lots of circuits that show your idea is completely true, furthermore , I cannot test my idea (charging this battery by high voltage and high value resistor in order to limit current out of dangers , I have exploded a rechargeable battery and I know how much it is frightening and dangerous

kind regards
 « Last Edit: July 16, 2008, 02:16:32 14:16 by Mesfet » Logged
Walkura
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 « Reply #5 on: July 16, 2008, 08:50:58 08:50 »

Have a look at this site .
http://www.batteryuniversity.com/partone-13.htm
Also have a look at this site where you have a design of a better charger
http://www.geocities.com/vk3em/sla-charger/sla-charger.html
Personaly i don't like the setup with resistors .
Its dissapative and i prefer to have matters controlled by either processor or dedicated chip .
Almost all manufacturers have special chips for chargers .

Good luck .
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looser
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 « Reply #6 on: July 19, 2008, 02:45:17 02:45 »

Hi mesfet
Walkura is right to say better if you control both current and voltage while charging.

But there is no risk of incresing the voltage for charge. As you increase the voltage the voltage drop on the battery will remain same but the current is to increase. This will shorten your charge time. Check the total charge the batterie plate has. Try to supply the same amount.
Ie: 6V * 4,5A= 27VAh
As conveniently you need to supply about 40% additional charge for losses.
27 * 40% =37,8 VAh

If you prefer 450ma the total charge time will be
37,8=7,2V * 0,45A * T
T   here will be about 11,66hr  ------- 12hours round up

Notice this amount is true if the battery is totally flat. ( the voltage at the terminals falls to say 5V )

Now whatever the power supply you have in your hand you can use. Try not to charge with one tenth of rated current ( here 450mA).
But Lead acid batteries are so durable to charge - discharge irregularities as well as deep discharge and excess charge.

If you prefer to charge it with one Amp ; welcome.
Your charge time will be short only by the calculation above.

Good charges.

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Walkura
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 « Reply #7 on: July 19, 2008, 08:43:41 08:43 »

Given that you keep a close eye on the voltage its no problem to charge with a higher voltage .
As long as you switch off in time there is no problem .
What i usualy do with microcontroller is limit the charging current till my battery reaches a preset value (11 volt for a 13.8 volt cell)
Above 11 volt i charge with what ever current the battery will accept
,when i reach gassing voltage i limit it again till my voltage is to full cell value .
Le me add to this that when you charge with a very high current you will have a shallow charge .
After some resting time your voltage dropped and will (can) take another portion of charge .
So in a way there is a limit to your charging current or at least limit to the being usefull.
AT the following link you will find a book about battery's and charging .
They give a quite good explanation about battery's and how to treat them .