greenwanderer108 said:
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
