Simple Battery Recharging Question : Electric Bicycle

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

If you connect all four batteries in parallel, you can use a 12 volt automotive charger to charge the combination. You should be able to find a small charger rated at 10 amps or less (the sort of thing you might use on your car overnight if you leave the lights on).

No - a 24 volt regulated power supply will not charge a 24 volt battery. You will need something producing about 28.8 volts to charge a nominal 24 volt battery.

A battery charger is normally connected directly to the battery terminals.

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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.

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...?

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? What does the Adj terminal of the regulator represent?

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.

Thanks This message was sent using the sci.electronics.basics web interface on

See the reply below

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

Yes

What does the Adj terminal

See below for how it works.

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.

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

14.161

out?

though

on

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

The information (in Thai) confirms so as well as the sticker says 'for 24 volt lead acid batteries'

(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

If it says "for 24 volt lead acid battery" on the label, I would guess it's a charger for a 24 volt lead acid battery. With 1.6 amp output it won't hurt your batteries, so just go ahead and hook it up to them and monitor the battery voltage. A 24 volt battery that has sat for a while since its last charge will have a voltage of no more than about

26 volts (if it's an AGM) or about 25.5 (if it's a flooded battery). That voltage should go up as soon as you connect it to the charger/power supply. Keep tabs on it. Look for the voltage to rise to 29 volts or so (on a flooded battery, somewhat higher on an AGM), at which point keep a careful eye on it. One of the led's should come on at some point. If the voltage goes up much higher and the charger/power supply gives no signal such as an led lighting, then you will have to monitor battery voltage yourself each time it charges. There is also the possibility (actually the likelihood, since it is obviously a switcher) that the power supply has voltage regulation. If it is regulated at about 29 volts you are in luck, you can leave the battery on overnight or however long you find it usually takes to replenish them, without keeping an eye on them or worrying too much about overcharging.

Don't use that site to get to the newsgroups. You don't have a newsreader that will show the ASCII original, so you should go on Google Groups, where there is an option to show the original message as it is stored in the newsgroups, and you can read the message as it is shown in the original ASCII. After you get to a message in this thread through Google Groups just click on options and "show original." When you reply, do the same -- click on options and "reply" instead of just using the reply button.

Ugh.

I cannot confirm it - I don't know. Ed