Need to reduce current of charger

[NewbieX]

I just got a teeny 300ma nimh battery for a discuss launch glider.

I have a hitec cg335 charger as well as a number of wallwarts in the 55ma to 70ma range for 4.8v receiver packs. The lowest the Hitec will go for a standard charge is 500ma, which translates to 50ma once it peaks and goes into trickle mode. I think the hitec would be fine to quick charge, but I'm also concerned about trickle charging the 300ma batt.

If I just put a resistor in line, that's only going to reduce voltage, correct? Is there any way I can cheaply reduce the current of one of my walwarts to about 30ma, which would allow me to trickle charge the little battery safely and effectively.

What is another option, short of spending $120 bucks on some bells and whistles charger? I'm mostly flying slope gliders now, so the need for a high end charger isn't all that critical.

Thanks,

Graham

[Dan Baldwin]

You can't put a resistor in series with the CG335. It needs to be connected directly to the battery to sense the peak. If you were to put a resistor in series, the charger would just put out more voltage (as high as it can) to hold the current you have set. This charger will work fine as long as you don't leave the battery connected too long after the charge ends.

You can, however, put a resistor in series with the wall wart to reduce charge current to about 30 MA. I'm not sure what value you will need, but 20 ohms might be a good starting point.

Dan

[MatC]

Series won't work, you're right.

The simple way is to put a resistor (and diode, see below) in parallel with your cells. Assuming you are charging 4.8v cells, and the supply gives 500mA, and you want 30mA, the resistor should be 10.2 ohms (4.8/(0.500-0.030)). Power would be 1.7W.

However the problem is that when the charger goes into trickle mode, the charger will supply 50mA, and that resistor is still there, so it will discharge your batteries very quickly.

So you need a diode in series with the battery, and the resistor in parallel with both. Voltage drop across a silicon (ie most of them) diode is 0.6v, so the resistor should now be 11.5 ohm, and power is now 2.5W. Go for at least 5watt resistor, and take care they will get hot. You are bleeding off around 95% of curernt, so that res. value is quite critical. (eg 11 ohms gives 10mA charge, 12 ohms give 50 mA.) Use precision wire wound. If you mix a 10, 1 and 0.5 ohm in series, they each need power of about 5W, 0.5W and 0.25W respectively (that includes a factor of 2 for safety).

You can make a 1W resistor out of four .25W resistors by arranging two in parallel, in series with another two in parallel. This gives a resistance for the network that is the same as each individual resistor, but you'll probably have to buy the 5W specially.

Make sure your diode can handle 500 mA.

All info at your own risk, good luck, sorry if too much info :)

[MatC]

Dan: not sure I agree there. You don't know the wallwart voltage, only the current it's supplying. Wallwarts are notoriously bad at regulation (they are made to a price, and a very low one at that). If it is approx 5v and supplying 50mA to 4.8v with a 4 ohm resistor and you add a 20 ohm, it will now supply about 10 mA.

However if it is 15v and supplying 50mA to 4.8v and you add that 20 ohm resistor, it will give about 45mA. Situation even worse if the voltage is higher, although it probably isn't.

The point about peak charge not being detected is true.

If you want _really_ cheap and cheerful, get a 12v battery and use a resistor of 150 ohms (power of around 1W). That will happily supply 48 mA but it will not detect full charge so don't leave it on too long :)

All info at your own risk, ensure an adult is present, and I really do mean it about the resistors getting hot: add heatsinks to them.

[Mr.RC-CAM]

I have a hitec cg335 charger as well as a number of wallwarts in the 55ma to 70ma range for 4.8v receiver packs.Been there, done that. I used a wallwart with series resistor with great success. The resistor value will depend on the number of cells and which output you use from the R/C wallwart charger. With a low mAH 4-cell pack, I use the Rx output, with a 1/2 watt 25 to 100 ohm resistor, in series with one of the charger leads. This trick requires an ammeter to correctly chose the resistor value. Do not skip the ammeter verification.

Keep in mind that common wall wart R/C Chargers are not constant current designs, so the charge current you choose for a nearly dead pack will be less at full charge. I prefer a current that is halfway between the two extremes; a brainless 14 hour charge time can be used if you do this.

Some folks have used their high current peak fast chargers to recharge low capacity packs. As mentioned, it requires placing resistors in parallel with the charge leads to shunt the charge current. When setup correctly, a 1C charge rate can be dialed in. It is important that the pack is immediately removed at the end of charge, or the shunt resistors will blindly begin a discharge effort. Or, you can introduce an isolation diode as mentioned by MatC, which prevents the unwanted discharge issue.

[NewbieX]

Thanks for the replies. As you've probably guessed, I'm an English Major and I'd rather be spouting poetry than figuring out resistor values.

I'm not overly concerned about the CG-335/CG340, because I can use that at 500ma to do a quick charge ~1/2 hour on the 300mah batt. I'm primarily concerned about the initial trickle charge and the occasional trickle for battery maintenance. Seems the walwart and resistor combo is my best bet.

So let's assume I'm using a Hitec walwart radio charger with the rx 4.8 V output and it runs at 55ma. I think I also have a Futaba walwart that does 70ma at 4.8v.

Should I just go pick up a few different 1/2 watt resistors 25-100 ohm, hook em up and test the current to see which one gets me closest to 30ma? I have a Craftsman multi-meter that can read 0-200 ma in 1 ma increments.

Looks like you guys just saved me a hundred bucks plus, beers all around. I was going to use this $8 battery as an excuse to buy a new charger that would do micro currents, but I'm glad there's a cheaper option.

All hail the DIY electronics forum.

Graham

[Mr.RC-CAM]

Should I just go pick up a few different 1/2 watt resistors 25-100 ohm, hook em up and test the current to see which one gets me closest to 30ma?That's what I recommend. Just buy a 5-Pack of 10 ohms and a 5-Pack of 22ohms (1/2W) from Radio Shack. You can simply stack these in series to create the resistor values you might need. There is no need to have a precise 30mA result. In this case, "close" is close enough. Aim for a current rate that is near 30mA when the 300mAH NiMH pack is about half charged.

Be sure to use the ammeter feature correctly. With current measurements, the multimeter goes in SERIES with *one* of the leads. Connecting it in parallel (across both leads) will probably result in blown meter fuse (or meter damage).

[Dan Baldwin]

That's what I recommend. Just buy a 5-Pack of 10 ohms and a 5-Pack of 22ohms (1/2W) from Radio Shack. You can simply stack these in series to create the resistor values you might need. There is no need to have a precise 30mA result. In this case, "close" is close enough. Aim for a current rate that is near 30mA when the 300mAH NiMH pack is about half charged.

Be sure to use the ammeter feature correctly. With current measurements, the multimeter goes in SERIES with *one* of the leads. Connecting it in parallel (across both leads) will probably result in blown meter fuse (or meter damage).

What he said


Looks like you guys just saved me a hundred bucks plus, beers all around.

Graham

Heineken :D


Dan

[ScopeTrig]

This is all very interesting, what if you wanted to charge 1, 2, or 3-cells with the CG335. Are there any possible solutions for this?

[MatC]

> 1, 2, or 3-cells with the CG335. Are there any possible solutions for this?

Assuming 500mA from the '335, and desired 30mA, using a shunt resistor, and diode as mentioned, and assuming the '335 can charge 3 cells and won't think there is a fault due to under voltage, and assuming it's not a blue moon or a Tuesday:

1 cell with diode = 1.8v, shunt current = 470 mA, resistor = 3.8 ohms, get a 2W one.

2 cells with diode = 3.0v, shunt current = 470 mA, res = 6.4 ohms, 3W.

3 cells with diode = 4.2v, shunt current = 470 mA, res = 9 ohms, 4W.

Again if you mix resistor values in series, the smaller values can have proportionally less power rating.

As per Mr RC-Cam's suggestion: check with your multimeter, and attach heatsinks to the res.

Stella please :D