Re: Simple Battery Recharging Question : Electric Bicycle
- From: "greenwanderer108" <greenwanderer108@xxxxxxxxx>
- Date: Fri, 14 Oct 2005 21:08:49 -0500
>greenwanderer108 wrote:
>>>You will need to add other components. Basically, a battery
>>>charger consists of a DC source and a charge regulator. The
>>>regulator helps insure that you do not overcharge the batteries.
>>>In the case shown below, it also serves to protect the AC-DC
>>>adapter, as the discharged battery might draw more current
>>
>>>from the adapter than it can safely provide.
>>
>>>You *might* find that your 24 volt AC-DC adapter provides
>>>sufficient voltage under load in a 17 hour charge configuration.
>>>(That means you would supply the battery with .833 amps for 17
>>>hours). 3 components, in addition to your AC-DC adapter are required:
>>>an LM317 3 terminal voltage regulator mounted on a good heatsink,
>>>a 1 ohm, 2 watt (or higher) resistor, and a 1N5401 diode:
>>>(view in Courier font)
>>>
>>> -------
>>>+24V ---Vin| LM317 |Vout---+
>>> ------- |
>>> Adj [1R] 2 watt or higher
>>> | |
>>> +-----------+--->|--- To 24v battery + terminal
>>> 1N5401
>>>
>>>Gnd -------------------------------- To battery - terminal
>>>
>>>There are other, better ways to charge a battery, but the circuit
>>>shown is probably the best you can do simply with your adapter.
>>>
>>>Ed
>>>
>>
>>
>>
>>>You will need to add other components. Basically, a battery
>>>charger consists of a DC source and a charge regulator. The
>>>regulator helps insure that you do not overcharge the batteries.
>>>In the case shown below, it also serves to protect the AC-DC
>>>adapter, as the discharged battery might draw more current
>>
>>>from the adapter than it can safely provide.
>>
>>>You *might* find that your 24 volt AC-DC adapter provides
>>>sufficient voltage under load in a 17 hour charge configuration.
>>>(That means you would supply the battery with .833 amps for 17
>>>hours). 3 components, in addition to your AC-DC adapter are required:
>>>an LM317 3 terminal voltage regulator mounted on a good heatsink,
>>>a 1 ohm, 2 watt (or higher) resistor, and a 1N5401 diode:
>>>(view in Courier font)
>>>
>>> -------
>>>+24V ---Vin| LM317 |Vout---+
>>> ------- |
>>> Adj [1R] 2 watt or higher
>>> | |
>>> +-----------+--->|--- To 24v battery + terminal
>>> 1N5401
>>>
>>>Gnd -------------------------------- To battery - terminal
>>>
>>>There are other, better ways to charge a battery, but the circuit
>>>shown is probably the best you can do simply with your adapter.
>>>
>>>Ed
>>>
>>
>>
>> So, using these three components along with the 24 volt, 1.8 amp power
>> supply (adapter) would regulate .833 amps per hour??? (17 hours =
14.161
>> amps) A little confused as the power adapter is rated at 1.8 amps.
>
>See the reply below
>>
>> And how would I know that the battery was charged from this regulator.
>> Would it send a signal to the led(s) that are already on the adaptor
>> or...?
>
>You know it is charged based on how long it has been connected to the
>charger. 17 hours is more than sufficient to charge it. It will not
>send any signal - it just depends on you connecting the battery to it
>for charging. See below for more detail
>
>
>> For the +24 volt input into the LM317 regulator, how does that work
out?
>> Is that simply the positive fead going into the LM317 and the negative
>> feed directly to the negative battery terminal?
>Yes
>
>What does the Adj terminal
>> of the regulator represent?
>>
>See below for how it works.
>> I'm not so acquainted at reading/understanding schematics but getting
>> better. I might be able decipher better with a graphic schematic,
though
>> this courier version is probably already basic at best.
>
>I know what you mean. It would be nice if we could draw graphics
>directly here instead of having to use ACSII art. I can draw
>it for you as a jpg and send it via email if you want. But a lot
>of stuff gets posted here as ASCII art and you could miss out
>on it until you learn to read ASCII art schematics. Reading
>ASCII schematics does take some getting used to.
>
>>
>> Thanks
>>
>> This message was sent using the sci.electronics.basics web interface
on
>> www.Electronics-Related.com
>
>
>
>The circuit uses the LM317 as a current regulator.
>Here's how it works: The circuitry inside the LM317 holds the voltage
>difference between the Adj pin and the Vout pin to 1.25 volts. That
>means 1.25 volts is across the 1 ohm resistor. (I originally computed
>it based on a 1.5 ohm resistor and later changed to 1 ohms, but forgot
>to change the .833 to 1.25 amps in the post. More on that later) Using
>the .833 amps figure based on a 1.5 ohm resistor: Ohms law says that the
>current throught the resistor will be .833 amps:
>E(volts)=I(current)*R(resistance);I=E/R; I=1.25/1.5 = .833
>It does not matter that the adapter is capable of higher amperage -
>the current is regulated by the circuit such that less than the
>full current capability of the adapter is used.
>
>The adapter *might* provide sufficient voltage under load.
>If it is a regulated adapter, it won't. The circuit depends
>upon the adapter providing sufficient voltage, under load, to provide
>a high enough voltage at the output of the circuit to push the
>.833 amps (or 1.2 amps as drawn) into the battery. The LM317 needs
>an "overhead" of about 3 volts, there's a 1.25 volt drop across
>the resistor and about .7 volts across the 1N5401 diode, so the
>input voltage at Vin needs to be ~ 28 volts or higher.
>
>Going back to the resistor - 1.5 ohms vs 1 ohm - 1 ohm is a better
>choice as it is commonly available. I should have changed the .833 to
>1.25 amps to eliminate the confusion. You could make a 1.5 ohm
>resistance using 3 1 ohm resistors (which I'll draw) but it is
>just simpler to use the 1 ohm resistor by itself.
>
>1.5 ohm resistance between point A and B diagram:
>
>A---[1R]---+---[1R]---+---B
> | |
> +---[1R]---+
>
>
>Charge time is an approximation. The rule of thumb is to charge the
>battery by putting in 120 percent of the charge you took out. You can
>do that quickly or slowly. If you do it quickly, you need to include
>precision control of the charging duration to prevent damaging the
>battery - not something included in this circuit. When you do it slow
>enough, which this circuit does, you have a wide latitude of duration of
>the charge that will not damage the battery. Anywhere from 12 to 24
>hours is fine with a slow charge rate (C/10 - C/20) even if the battery
>is only partly discharged. And in use it will never be 100% discharged.
>
>Regarding sending a signal/knowing when the charge is complete:
>A circuit to do that, and more, can be drawn - but it puts you in
>the realm of a better charger - one that is not made in an attempt
>to use your existing adapter. It also requires more than 3 components,
>so it is not as simple as the one already posted.
>
>Ed
>
Thanks for all the info. The only problem I had is with the SCII art was
realigning them properly in word, etc.
Anyhow, I found a picture of that adapter I mentoned. Aparently it's a
charging circuit in itself for 24 volt batteries. Here is the picture
http://www.farsai.net/q06.jpg
The information (in Thai) confirms so as well as the sticker says 'for 24
volt lead acid batteries'
http://www.farsai.net/ (it's displayed 3/4 way down the page code G48-3
Can you confirm that it is in fact a charger, not simply a power supply.
That's what I assumed the two lights were for? Thanks
This message was sent using the sci.electronics.basics web interface on
www.Electronics-Related.com
.
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