Instructables

[help] How to make AA Ni-Mh charger?

Hello!
I have one DC Adapter (output 4.5v 500mA) , battery holder , some diod....
I wanna to make a charger which can charge 2AA Ni-Mh at same time.Please help me !!!
Thank!!
sorry for my bad English


b-stro4 years ago
 Basically, You want to run current backwards through the batteries at a level that will not overcharge them. Assuming you battery holder puts them in series, you'll need around 2.8 volts to charge the batteries. (Although the voltage of each cell is 1.2, a slightly higher voltage is required to effectively charge the batteries). The most important issue to consider is overcharging. If you overcharge NiMh batteries for too long, they will overheat, and degrade. The easiest way to avoid this is to trickle charge the batteries. You can do this by charging them at less than 10% of their total milliamp-hour capacity. This value is usually listed on the batteries or their packaging. Modern NiMh batteries contain a catalyst that will prevent degradation due to overcharging, as long as it is no higher than 10%. So if you have 2700 milliamp-hour batteries, your current should be no higher than 270 milliamps. Remember: if your batteries are in series, their currents are not cumulative (their voltages are). You can regulate the current by adding resistors (calculate using V=IR), or by using a current-regulator like the LM317T. Even with this safety feature, it is best not to leave the batteries in the charger too long. To calculate the charging time, divide the total milliamp hour capacity of your battery by the current of your circuit, and multiply that value by 1.5 (due to the less than perfect charge efficiency) this will give you your approximate charge time in hours. Diodes can be added as an extra safety precaution, to prevent the batteries from discharging through the circuit, but shouldn't be necessary with your setup. before calculating any values for your circuit, make sure you measure the actual voltage output of your DC adapter - they often deviate from the listed value. Your final circuit would be something like this:
DC adapter positive lead----Positive end of battery holder--batteries--- negative end of battery holder--- resistor/current regulator---DC adaptor negative lead
login721 (author) 4 years ago
Thank, you helped me alot.

login721 (author) 4 years ago
I forget to post .I used diod 1n4004:
  • Peak Reverse Voltage: 400V (Max)
  • DC Blocking Voltage: 400VDC (Max)
  • Forward Rectified Current: 1A
  • Silicon Rectifier
  • Diffused junction
  • Maximum Average Forward Voltage Drop: VF= 0.8 Volts
  • Non-repetitive Peak Surge Current: IFSM= 30 amps for 1 cycle
  • Maximum Reverse Current @ 25 degrees C: IR= 10uA
but i think the voltage drop is more than 0.8 (i think higher than 1.5v or the heat when i solder make it worth???)
.

 

b-stro login7214 years ago
you may have damaged your diode by overheating it in the soldering process, or the packaging may just be misleading. 
 There are a few things you can do here:

1. wire the diode to a source with a known voltage (a battery maybe), and measure the voltage across the diode to find the actual voltage drop.

keep in mind you only need 2.8 volts across your batteries to charge them, so you may not need to worry about the voltage drop.

2. If  the voltage drop is still a concern, you could obtain a second diode of the same specifications. This would reduce your voltage drop, and would not harm the diodes. This is because your circuit will be operating well below the maximum voltage and current limits of your diode.

3. Omit the diode from your circuit. As I understand it, as long as your DC adapter is plugged in, it will supply power to the batteries. The purpose of a diode in a circuit like this is to prevent the batteries from discharging through the circuit when they are not being supplied current. As long as you remove the batteries when the charger is unplugged,  you would technically not need a diode. If you were using another power supply, like a solar cell, a diode would be necessary, to prevent the batteries from discharging through the cell when no light is available to supply current.

4. Find a better diode. Just make sure it can safely operate within the conditions of your circuit.

good luck!
login721 (author) 4 years ago
 !OMG, the voltage is drop too much.I checked the resistance of diode, the multimeter said : 40ohms.
I dont know what to do next...Please help

login721 (author) 4 years ago
Thank all you very much , :D.
login721 (author) 4 years ago
Finally i done!
This is wire map...please check it for me .
[IMG]http://i247.photobucket.com/albums/gg132/chipmapple/final.png[/IMG]

Thank!!
b-stro login7214 years ago
 Looks good! If you're still wondering about the voltage, It would be 2.8 volts for 2 batteries because they are in series (about 1.4 per single battery). One other thing to remember is the voltage drop on your diode, you don't have the specs listed so I can't tell. It should be listed as forward voltage on the packaging or data sheet. just subtract its forward voltage from your supply's voltage to calculate your new voltage (if you haven't done so already).
login721 (author) 4 years ago
I forget to post image
http://i247.photobucket.com/albums/gg132/chipmapple/HHH.jpg
login721 (author) 4 years ago
Thank for your help :D .
I'm having a trouble : i wired 2 battery as the image , but i dont know about DC's Voltage . 2.8v for 2 battery or 2.8v for each battery ????

Thank for advance !
*sorry for my bad english
lemonie4 years ago
If you really want to build something, have a look at these projects:
Intelligent NiCd/NiMH Battery Charger
simple LM317 device
USB powered

"some diode" is not really enough

L

orksecurity4 years ago
I'd recommend buying one. The chemistry for NiMH batteries is scary.
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