ehsjr said:
Hello,
Could you help me with a personal project that I am working on?
I need help creating a simple rechargable battery circuit.
I am making an acrylic flower with LEDs that make the head glow white.
This is a birthday present for a friend.
My original plan was to just have a wall adapter to power the LEDs, but
now I think it would be cool if I can use rechargable batteries, so
when the flower is unplugged, it will still light up for a short time.
I see a lot of charger controllers for batteries, but they're too
complex for me to understand and I don't know anything about charging
batteries. I'm just an architecture student, so my knowledge of
electronics is very basic. Im afraid to use them because I don't know
what to connect the pins to or how it works.
I also don't have much space to put the components in, so a simple
circuit would be better.
I was wondering will it work if I use a 6v wall adapter to directly
charge a depleted 7.2v battery? Suppose the batteries are drained to
2v. Will the 6v adapter bring the batteries up to 6v? Or will it not
work at all? Will the charging stop automatically or keep going and
overcharge the batteries?
Here is a diagram drawing of what I want to do.
http://www.thomashuang.net/project.gif
Thank you for any advice you can give me.
Thomas
Groan - as drawn, nothing will work right.
You show 3 LEDs at 3.3 volts each in series,
so 9.9 volts would be needed, not 6 volts.
You should never let a 6 cell (7.2V) NiCd pack
discharge to 2 volts - that can ruin the cells.
Your switch circuit will prevent battery power
from reachingh the LEDs - and as mentioned before
6 volts is too low in any event.
Get a 12V DC adapter and do this:
----- 2N3906
+12V ---Vin|LM317|Vout---+ +---+---- ---+----+----+
----- | | | e\ /c | | |
Adj [R1] | [R2] - | | |
| | | | | [R3] [R4] [R5]
+----------+---+ +-----+ | | |
| | |+ |+ |+
[NiCd] [Zd] [LED][LED][LED]
| | | | |
Gnd ---------------------+-------+----------+----+----+
R1 = 10 ohms, R2 = 1.2K, R3,R4,R5 = 220 ohms
Zd = 5.6 volt zener diode.
When the 12 volt supply is available, the LM317 will
limit the current to ~ 125 mA, which will protect the
NiCd pack (assuming the cells are rated at 1200 mAh
or more) from overcharge. That 125 mA will provide
current for the LEDs and current to charge the NiCds.
The 220 ohm resistors will limit current through each
LED to ~ 23 mA when the batteries are fully charged
down to ~ 13 mA when they are almost fully discharged.
The 5.6 volt zener diode will hold the base of the 2N3906
at 5.6 volts, so the transistor will conduct as long as
the pack is above ~ 6.2 volts. When the pack discharges
to ~ 6.2 volts, the transistor will turn off and prevent
the pack from discharging too far.
Here's a "picture" of an LM317
-----
| o |
|_____|
/______/|
| ||
|______|/
| | |
| | |
Adj Vin Vout
The + side of the LEDs is marked on the diagram,
and the emitter (e) and collector (c) of the
2N3906 transistor are also marked on the diagram.
Ed
The over discharge protection is nice but perhaps unnecessary, I am not sure how much
leds let trough bellow it's forward voltage but it must be very / extremely low and discharging NiCd down to 1.1V per cell is acceptable according to what I can remember reading.
If indeed the LEDS are 3.3V not 1.7 , putting the batteries in parallel (or just using 1 )
and driving the leds in parallel through about 18 or 20 ohm resistors will burn up lot less power (in the resistors.)
1200Ma? I am looking at similar 4.8V pack and it is only rated at 60mAh.
Oh yes I just looked at the circuit again and it says 60mAh, however if you put then in series
you don't get twice the capacity just twice the voltage.
Come to thing of it, at this low rating , 3 leds in parallel is stressing the 1 or 2
batteries too much, but if it is only really 1 time use .....
Maybe use only 2 LEDs in series. you will have to find out the led forward voltage.
As far as charging goes you should probably aim for 1.42 - 1.43 constant voltage per cell
(for simplicity). I am not sure how much I destroyed my 4.8V pack prior to taking measurement and coming to this conclusion of 1.43 V but I hope not too much.
Older batteries tend to have higher internal resistance and reach higher voltages faster
( DURING CHARGING ) than brand new ones, when that happens charging stops or slows down to a crawl. With my setup the current after long charging period was down to about 100 -200nA
which I considered perfect trickle charging situation.
I am not a fan of battery stacks but trickle charging is supposed to be able to charge embedded weak batteries.
Anyway..
So if you use the 2 batteries in parallel ( and all LEDs in parallel) ,
you will need 4.3V supply.
6V wall wart maybe sufficient but 7V to 12 would be better.
Now use the LM317 to build regulated 4.3V power supply.
Since the regulator maintains 1.2V between ADJ and Vout. using the prescribed 240ohm:
_____________
| | 1.2V / 240 = 5mA
| LM317 | 5 - 1.2 = 3.8V 4.3V - 1.2 = 3.1V
| | 3.8 / .005 = 760ohm (R2) 3.1 / .005 = 620ohm (R2)
-------------
" " "
| | | Vout
| | L---------------
Adj | | |
| Vin |
-------------- | (5V)
| | R1 | 4.3V out R3 R4a
| | |---------| | | |--------| |--------|
| L-----o--| 240ohm |----o------>| ----| 30 ohm |----o---| 43 ohm |--o
| | |---------| | |--------| | |--------| |
| ----- 1N4001 | |
| |R2 | | V LED 1a
O + | | 620 ohm BAT. 3.6V ===== ===
from | | --- |
O - wallwart| | | |
| ----- | |
| | | |
--------------o-o---------------------------------------------o---------------|
|
GND ===
I think to calculate the charging resistor from the 1/10 capacity charging current (6mA),
I would use 1.25V ( at which voltage the battery only holds fraction of it's capacity )
and start with 6mA from here.
So:
1.43V - 1.25V / .006 = 30 ohm
Now we need to account for the leds current( 2 in parallel? ) at the same time hmmm.
At this higher voltage 4.3V lets allow 23mA ( assuming the LEDs can handle it ).
So:
4.3 - 3.3 = 1
1 / .023 = 43ohm (resistors with each of the 2 leds )
When batterie supplies the current and is at it's nominal voltage of 3.6 - 3.75V the current will be .3 / 43 = 6.9mA
For 20mA at 3*1.25V ==> (3.75 - 3.3) / .02 = 22.5
at 3.6V 22ohm only gives 13.6mA which maybe way to dim
and at 4.3V there is 1 / 22 = 45mA flowing through them LEDs.
Can they handle that?
!!!!looks like we have a major problem.
Maybe burning the 4V on resistors is a better approach after all, that way the current difference ( with larget resistors ) between charging voltage (8.58 - 8.6 ) and the
operating voltage (7.2) is not as large percentage wise.
In this case you'll most likely want to use the transistor/resistor/zener protection so
that the batteries don't go bellow 6V. At first glance there is something fishy about that sub-circuit even if the 3906 is a PNP transistor, so I can't tell you wheather to use 6.2 or 6.8V zener.
Alternatively the LEDS could be disconnected when charging.
If you do that then you could use what I drew.
As an after thought I inserted a diode after the regulator ( because regulators don't like to see voltage on their output and not on input.) That brings the voltage requirement to 5V,
so instead of LM317 and R1 and R2 you could use (LM)7805.
|---------|
| L7805 |
| |
| |
|---------|
I I I
Vin GND Vout (5V)
I'd have to think about how to use additional transistor to turn off the leds when voltage is above or better yet current flows through R3.
Right now I have to run.
Ciao for now,
Sam.
