i am rithy now i study at university
I have just bought a sealed lead-acid battery (6V 4.5Ah , 24HR) unfortunately I don't have any experience about this kind of battery. 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
Thanks in advance
The term 24HR normally means that the battery has a capacity of 4.5Ah when using a 24 hour discharge current (4.5Ah / 24h = 18.75mA).
6V, 4.5Ah would represent 27Wh (P = A * I) of energy at such a load.
The capacity of the battery is given for a specific discharge current. With a higher discharge current, you will get less energy from the battery. With a lower discharge current, you will get more energy from the battery. And the usable capacity will vary significantly with the used discharge current!
A lead-acid battery is _not_ charged by a constant current, in contrast to NiCd or NiMH batteries.
A lead-acid battery is charged with a constant voltage. In your case, a voltage not above 7.5V (2.5V / cell). It can also be a good idea to limit the max charge current, to avoid overheating the battery. The optimum is a charger that switches between current-controlled and voltage-controlled charging. In most situations, the inner resistance in the battery will limit the charge current, but depending on charge state and temperature, you may need the external current limitation. It will also help if you get a shorted cell, since the other cells would get a much too high charging voltage in such a situation.
You can not use a 7809 to charge your battery, since the voltage is much too high.
The datasheet for the battery will contain a lot of useful information.
It will tell what voltage you should charge with if the battery is kept constantly charged (such as in emergency lights or an UPS). And it will show a different charge voltage if the battery is in cyclic use.
The datasheet will also show how the capacity of the battery is affected by temperature and discharge current. And it will describe the internal resistance in the battery and the largest allowed charge and discharge currents.
Thank you so much for your attention I red the 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)
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
Yes, during a large part of the charge cycle, your serial resistor will keep down the charge voltage.
But when charging batteries, it isn't just a question of limiting the current and integrating amount of energy you have put into the battery.
Some battery technologies (often called chemistries instead of technologies) have specific requirements for max current allowed. Some have hard limits for max battery temperature. Some have hard limits for max cell voltage.
If you use a 9V charger with a series resistor, the charge voltage will be below 2.5V/cell as long as your lead-acid battery has a charge current high enough that you get your 1,5V voltage drop over the series resistor. But as soon as the charge current drops, the voltage drop over the resistor will decrease and your charge voltage will continue to climb above 7.5V.
But under some parts of the charge cycle, your series resistor will limit the charge current too much.
The chemistry of a lead-acid battery do not like a high voltage, even if the charge current may be low.
That is also why the datasheet for a lead-acid battery specify a higher allowed charge voltage for cyclic use since the high charge voltage will then be applied to a battery that isn't fully charged. When continuously charging a battery, the datasheet will specify a lower charge voltage since continuous charging means that you will apply the the voltage even after the battery is 100% full.
If you charge continuously, then you should probably select somewhere around 2.2 .. 2.3V / cell. In your case no more than 6.9V.
For cyclic use, you should probably have 2.4 .. 2.5V / cell - in your case max 7.5V.
And in both cases, the charger should have current-limiting. Max charge current will probably be somewhere around the Ah figure * 0.3 [A] or in your case about 1.35A.
The above figures are reasonable, but if you visit the web site for the battery manufacturer, you may be able to find more specific parameters.
The charge voltage will affect the lifetime - number of years and/or number of cycles - you will get out of your battery. For optimum performance, you should also have temperature compensation, where you decrease the charge voltage with a couple of mV/cell for each degree Celsius above room temperature. You may increase the charge voltage if the temperature is below 20C. Usually, the compensation factor is around 3mV/C for each cell for continuus charge and 4mV/C for cyclic use, but the battery manufacturer may have more specific figures available.
In no particular order:
http://www.keil.com/forum/docs/thread12851.asp
http://www.keil.com/forum/docs/thread12832.asp