Making an NiMH battery charger - need help from electronics gurus

[Loonix 0]

Well, I know that some quite smart people frequent these forums, so since I can't find my answers anywhere else, here goes...

An NiMH cell is 1.2 V, and according to wikipedia (http://en.wikipedia.org/wiki/Nickel_metal_hydride_battery), the charging voltage is 1.4-1.6 V. I need to charge a batterypack made up of 3 cells in series, and two such triplets in paralell (so, 3x2 batteries in total). I'm not sure about the mAh of the batteries, but let's say 1000. That gives my batterypack 3.6V, 2000mAh. To slowcharge (my preferred method, since it is the simplest), I charge at C/10 = 200mAh, but what voltage? Would that be 3.6+margin (0.2-0.4), or (1.4*3)-(1.6*3) ? Is what I'm trying to do even possible? See any other problems with this?


I decided to make myself an underwater flashlight, for freediving, so now you know what I need this for...

• Quote

[quade 4]

I'm sure billvon will be with you shortly.

quade -
The World's Most Boring Skydiver

[1969912 0]

Tons of stuff on the web. Most seem to use a Maxim 712/713 IC:

http://www.qsl.net/eb4eqa/batt_charger/batt_charger.htm


Maxim 712/713:


http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1666


Not sure about charging cells in series. The datasheet should address it.

"Once we got to the point where twenty/something's needed a place on the corner that changed the oil in their cars we were doomed . . ."
-NickDG

[billvon 3,120]

>To slowcharge (my preferred method, since it is the simplest), I charge
>at C/10 = 200mAh, but what voltage?

Whatever voltage is needed to get 200mah. I'd recommend going to C/20 to be safe. Expect to need 1.6 volts per cell (4.8 volts total) to charge over temperature.

Note that you have to regulate CURRENT, not voltage, to accomplish this.

[-Scatter- 0]

The simplest trickle charger will be a voltage source and a power resistor. I agree C/20 end of charge is better than C/10.

For a nominal 3.6v pack, that'll be an end of charge voltage of about 4.5v. You need to have C/20 current going into the pack at 4.5v. For total pack capacity of 2000mAh you'll want a charge current of 100mA. That gives you one equation:

(Vin-4.5V)/R = 100mA

Now, for a "fully discharged" battery, say 2.4v, we need to set a charge current. Say we'd like to charge at C/10 when we're fully discharged. That gives us

(Vin-2.4V)/R = 200mA

Solving both equations simultaneously will give the desired open circuit voltage for the charger and the correct resistance. In this case, Vin = 6.6v and R = 21 ohms.

To be safe the resistor needs to be sized to not overheat at max power dissipation, which would be a short instead of a battery. 6.6^2/21 = 2.07 watts, so a 5 watt resistor would do the trick. 21 ohms isn't a standard value, but a 5 watt, 22 ohm would be just about perfect.

Alternatively, if you have some particular power supply or wall wart you want to use, you can use the same procedure: set the final charge current and the open circuit voltage, and then calculate the resistance and initial charge current, Just remember to take the resistance of the secondary into account if you're using a wall wart.