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?
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
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
Oops - forgot to mark the zener diode for you on the diagram. The end of the diode that has a band on it gets connected to the base of the transistor and the 1.2K resistor.
thanks! oh, i was just too lazy to draw the LEDs correctly but i meant for them to be in parallel. so how does the circuit prevent the batteries from overcharging? they're only 60mAh each.
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
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.
im trying to follow the calculations and how the current flows to each component, but i cant. its too confusing for me. i keep thinking that the resistor that limits the current to the battery to 6mA also limits the current to the LEDs to 6mA since they are in series with the resistor, while plugged in. actually, did you mean for the adapter to be only used for charging? that would make sense then. i still dont get how the 317 limits takes the voltage down to 4.3v. i thought it only limits the current through the adj pin. and how does this stop the batteries from being overcharged? i dont know. its too much for me. maybe ill just leave out the battery part altogether. thanks for helping out. i really appreciate it. i wish i hadnt thought about using batteries in the first place. would have saved me 20 hours of worthless internet browsing.
it claims that it will automatically shut off charging when the battery equalizes with the charge voltage with the use of the zenner diode shutting off the base. i dont know. im not following it completely. this might work if i also put a switch between the battery, wall, and LEDs so that only either the battery or the wall provides power to the LEDs.
i dont know. it still doesnt make much sense to me how it works, and how i can use it correctly. oh well.
Yup - I missed that spec. The value of R1 would need to be changed to 210 ohms, but that creates another problem: there won't be enough current to light the LEDs while the battery is charging. So a modified circuit is needed:
Replace the LM317 with an LM7809, add
2 diodes, 2 caps and change the resistor values (except R2)and the zener as shown below:
No wonder I have been convinced what crap rechargeble batteries are all this time. that circuit cannot charge NiCd or Lead/acid battery to any reasonable capacity, maybe 2% Most recently I started with Ni-Mh batteries when I got a digital camera which came with charger and 2 batteries. after the first 2 or 3 times recharges I started noticing that I can't make more than 5-7 pictures before it shuts down because the batteries are weak. I don't think I ever had a charger that was any good. I think I measured the current while using the ZOOM it was sucking nearly an amp if the camera draws that much current then they should at least make room for 4 batteries but I haven't seen a camera that takes 4 batteries yet.
You are righ that only 6mA are available with that circuit but if the batteries still have charge , They'll supply most power less the 6mA.
while the 317 can be used to supply constant current I don't know any way of terminating charging besides using microprocessor. After finally reading something and seeing the 1.42V figure I have decided to try and let the voltage go that high with my 4.8V stack. I set the voltage source to 1.43 * 4 and used resistor that would allow around 7 mA and it took about 14 hours for the current fall down to lot less than 1mA. Then I discharged it (starting with 7 or 8 mA) and got about 9 hours before the voltage dropped to about 1.1v per cell. The charging doesn't realy stop but is passing wery little current through the battery, small enough not to cause any damage to it. As the battery charges the voltage rises, it takes relatively short time to reach into the 1.30 range and then about 14 hours to rise the additional .1V, when finally the battery is
1.42 and supply 1.43 you only have .01/ 20ohm = 500nA
Same way I terminated discharging: compare the voltage to a reference voltage and take the appropriate action (shut off the power supply, disconnect the battery, interrupt the circuit, whatever.)
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