Simple Battery Recharging Question : Electric Bicycle

For an electric bicycle project, I plan to use a battery pack (four 1
volt 7.2 Ah in series-paralell for 24 volt 14.4 ah)

Instead of routing the charging process through the motor controller,
will have the battery pack detachable for convenient charging (I live i
an apartment)so I can charge it seperately/directly in my room withou
having to haul the entire bicycle up the building every time I need t
charge.

Will I need a special 24 volt charger? Here in Thailand, they only have 1
volt car battery chargers. They are huge and not practicle for m
application. However, I found a 220--->24 volt AC-DC adapter (Thailan
uses 220 AC) The small size of this adapter seems practicle. There are tw
led lights on it which I assume deal with the charging signal. It is rate
at 1.8 amps.

forgive me if my question is real basic.. So, is this 220-24 volt 1.8 am
adapter a charger in itself. Meaning, can I simply detach the positive an
negative terminals of the battery pack from the load, hall the pack to m
room, and connect this adapter to the positive/negative terminals of th
battery pack (the same output leads that go to the load) to the wall fo
charging? Or will I need to add some other circuit/components to th
battery pack?

Can someone clarify the input/output routing betwee
batteries-loads-chargers. Are the positive and negative output of th
battery also used for input when charging, or is a seperate circui
altogether that does this.

thanks in advance

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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.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
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You will need to add other components. Basically, a battery

So, using these three components along with the 24 volt, 1.8 amp powe
supply (adapter) would regulate .833 amps per hour??? (17 hours = 14.16
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 adapto
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 negativ
feed directly to the negative battery terminal? What does the Adj termina
of the regulator represent?

I'm not so acquainted at reading/understanding schematics but gettin
better. I might be able decipher better with a graphic schematic, thoug
this courier version is probably already basic at best.

Thanks

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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

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

 

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

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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.

 

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