nitro joe

My Feedback: (2)

I have a 55MA 4.8V wallwart charger.

I need to reduce this to 30 MA

Anybody know the formula to size a dropping resistor?

This will be for a 4-cell 300 MAH NiMH pack.



Electronics 101 Anybody?


Joe

Rodney

We need more info in order to answer your question. What is the internal impedance of the charger? What is the no-load voltage of the charger?

RJConnet

Don't bother, the 55 mA charge cuurrent will not hurt your pack a bit. Just reduce your charge time by 30 or 40 percent to keep it in the ball-park.

nitro joe

My Feedback: (2)

OK.

The charger is the std. unit that came w/ my new Hitec Laser 4.

Mod. CG-25A

Input 117vac 50/60 HZ. 4VA (Rated)

Output( RX side)4.8 vdc 55MA (rated)

No load / open circuit 7.3-7.5 vdc avg. tends to "float" (RX output)(measured)

Secondary (charge side) 2.3M ohms. (measured)

Primary ( A/C side ) .328K ohms. (measured)

Readings taken W/ my old Radio Shack VOM, cant vouch for it's accuracy. ( the Fluke's at work)

I really don't want to go 25MA over the C/10 of 30MA if a simple dropping resister across the charge leads will do the trick.( Plus I'm cheap.)

Oh, no bother I like fooling around with this stuff, as I'm sure most of you do as well.

NV Flyer

My Feedback: (1)

Could you put a variable resistor in there? And adjust it while measuring current with a multimeter.
Then either leave the pot in or replace it with a fixed resistor of the correct value.
Shouldn't be to much trouble.

nitro joe

My Feedback: (2)

NV

Yeah, that would be even better.

Still would need a clue about pot type and a rough value.

Maybe there's a cyber schematic/parts list out there for a little circuit?

Links anyone? I'm ready to solder up a little something

Joe

mulligan

You would need a pot. (or resistor) with a 1/2W rating. I, too, suggest the use of a pot. The pot. max. resistance should be a little above the internal resistance of the battery pack, since you are looking to cut current to about half.

- George

RJConnet

If you really insist, remember the resistor will go in series with the positive lead not across the two leads. Just get a 22 ohm resistor and cut one of the leads of your walwart and insert the resistor. Then measure the voltage dropped across the resistor with the power on and the battery pack connected. You will probably see about 0.5 to 0.7 Vdc, divide that voltage by 22(ohms) and you will have the current (Adc). Example, .66Vdc divided by 22 equals 0.03 or 30 mA. Just remember, the voltage across the resistor divided by the resistance (in ohms), equals the current. If you need, go up or down in resistance to get exactly the current you want.

traskos

I think you want to leave the voltage stay since it is for a 4.8 volt pack already. You want to lower the current which means the resistor wants to be in parallel. That way the voltage remains the same and you split the current between battery and resistor. The resistor should take 25 ma. at 4.8 volts. I can`t remember ohms law to come up with the answer, but I`m sure it will be forthcoming.

traskos

By putting the resistor in series you will drop the voltage also. Part of the total output voltage will be across the resistor and the rest across the battery.

JimTrainor

Ohms law is V=IR or R = V/I

So if the charger regulates the current (i.e. varies the voltage) to hold 55 mA and is in the neighborhood of 4.8 volts, then a ~200 Ohm shunt would be needed to shunt 25 mA.

But it's not just resistance in parallel here The circuit would look like a resistance and battery on one side, and a plain old resistance on the shunt side.

The batteries resistance is quite low but it's own potential acts to reduce the current. To complicate matters the cells' potential will increase as it is charged. The charger will match this plus a small delta as required to push 55 mA.

I have a feeling there is a "gotcha" in this. In absence of further thought, I'd guess that if you charged a dead battery most current would initially flow through the battery, then as the battery charge state rose the current would balance out on each branch, then you might even reach a point where the charger's voltage equals the battery's potential and will go no higher. This would happen when the charger's voltage is sufficient to drive 55 mA thought the shunt... at which point the battery would stop charging. So you might need a larger shunt resistance then anticipated to avoid that.

If you add a resistor in series you won't have the voltage required to charge the battery.

It would seem that reducing the charge time is simpler.

Perhaps it's like the Borg say: "resistance is futile"

RJConnet

C'mon guys, just put the 22 ohm resistor in series with one of the walwart leads, plug-in the battery and turn everything on and you will have pretty close to 30 mA of current. Don't worry about the voltage across the resistor since it is only in the order of 0.6 volts and the walwart is putting out a peak voltage of probably around 12 volts no load. You can't read it with your ordinary VOM analog or digital because it is pulsating dc out of a half-wave rectifier in the walwart. You would need an oscilloscope or a peak reading meter to measure it. Don't try to shunt the load to adjust the current through the battery or you will run into a real bag of worms. IMO

adam_one

Hi !

I think you guys have forgotten something…
If the charger is designed to charge at a constant current of, say, 55mA, then it will try to compensate for changes in the load's impedance by adjusting it's output voltage accordingly.
That means, if you put a resistor in series with the battery, the charger circuit will rise the output voltage, in order to keep the same charging current.
Prove me wrong…

(electronics is easy)

greenboot

My Feedback: (1)

How about switching the charger on and off instead of lowering the current. I'm not sure, but maybe that would prevent overheating and "effectively" charge at the desired rate. I'm thinking of those little Christmas tree light blinkers you insert at the wall plug. Maybe they don't make them anymore. We had them when I was a kid.

Another thought, would some kind of "blinker" allow me to leave my batteries plugged into the wall wart all the time?

Tom

PS: the above information is from someone that really doesn't know.

Rodney

No need to worry about most wall warts compensating for added resistance, they do not. most are simply transformers and diodes that depend on the winding resistance to limit the current to about 50 to 70 ma, they do not have constant current circuitry built in. Just adding a resitor will do fine in limiting the current to a lower value.

JPMacG

My Feedback: (2)

You can make a current limited charger easily from your wall wart using an LM317 regulator IC and a resistor. Choose the resistor to give 30 mA or whatever you need. See this link for details: http://www.uoguelph.ca/~antoon/gadgets/sccs.htm

In order for this to work your wall wart will need to have enough voltage allow a drop across the regulator and still charge your pack at the desired current.

The diodes and LED arn't required as your wall wart already has them. I use this scheme to charge the 290 mAh pack for my GWS Cub using a wall wart from an old answering machine.

theox

Mr. nitro joe and Mr. greenboot,

Check out 'Keep your batteries in the ready state
without damaging overcharge' over at

http://www.rcbatteryclinic.com/

Brilliant.

nitro joe

My Feedback: (2)

JPMacG

Cool, I just printed that one out.

I stumbled across a version of that last night.

Do you electronic guys think this is a better way to go, instead of the simple series or parallel resistor?

There's a RadioShack just up the road.

www.ncws.com/rcrock/charger.htm

JimTrainor

I think the LM317 solution mentioned by JPMacG is kind of neat and simple. And I see your link is similar.

The data book is on National's web site:

http://www.national.com/pf/LM/LM317.html

Look in the "Application Notes" table at item "LB-35: Linear Brief 35 Adjustable 3-Terminal Regulator for Low-Cost Battery Charging Systems". That has a couple of charging circuits. Look at figure 4. It's the same thing as your link.

I never gave my charger much thought until reading this thread. Pretty simple. I think I'll try the LM317 approach myself rather than spending $$$ on chargers that do little more.

JPMacG

My Feedback: (2)

I should have mentioned before that a filter capacitor may be required as well. The output from some wall warts is just rectified ac. A capacitor will be required to smooth it out so that the current regulator performs as expected. I use a 100 uF 25 V electrolytic capacitor for my 30 mA charger. Radio Shack carries these for a dollar or two. If you are unsure about the output of your wall wart then use a capacitor - it cant hurt. Be sure to observe the polarity marked on the capacitor otherwise it might burst.

Alternatives to Radio Shack are Mouser (www.mouser.com) and DigiKey (www.digikey.com). Their prices and selection are much better than RS, but you have to pay shipping.

The advantage of regulated chargers is that with them the charging current is constant over the duration of the charge. My Futaba wall wart is labeled 60 mA, but I measured over 150 mA into a discharged battery, gradually tapering off to about 65 mA as the battery charged.

nitro joe

My Feedback: (2)

Well....

That settles that, as far as I'm concerned!

It's the LM317 circuit for me.

He11, for the cost of a couple grease burger Happy meals I'll have nice little 30MA charger.

My hats off to all you guys, I'm really glad I found this board.

And a special :thumbup: for JPMacG and Jim Trainor for their electronic mini seminar.

Can you tell I'm Excited?

Building this,as apposed to buying ready made... is like building a
kit, as apposed to buying a RTF plane to me. to those who can relate.

Joe

Ladyflyer

My Feedback: (2)

Quote : Adam _one

"Hi !

I think you guys have forgotten something…
If the charger is designed to charge at a constant current of, say, 55mA, then it will try to compensate for changes in the load's impedance by adjusting it's output voltage accordingly.
That means, if you put a resistor in series with the battery, the charger circuit will rise the output voltage, in order to keep the same charging current.
Prove me wrong…

(electronics is easy)"....... For some


In this case the "constant current" is pretty well maxed out . The little wall wart charger will not allow for a current compensating rise in voltage on the load side of the resistor.

The simple wall wart charger is pretty much limited by saturation. As such a series resistance will indeed drop the charge current.
Actually a series resistance will drop the CURRENT in a series circuit (Which is what the charge circuit in question is ). The VOLTAGE at the battery will be lower . The voltage at the SUPPLY or INPUT side of the resistor will be higher and the load side (battery) will be lower. That is why they refer to a VOLTAGE DROP across a resistance.

There are some old radios floating around that charged NiCads directly from rectified AC mains,with only a SERIES dropping resistance to LIMIT the charging CURRENT . Not a very good method to be sure . The safety factor is dubious and the current regulation is poor .

The current regulated LM 317 IC scheme is a very good one . With that scheme the output will go up (limited by the input voltage and regulator characteristics) Current regulator will adjust the voltage to maintain the desired current during the charge.

NiCads should be charged at a constant current . The impedance of the wall wart factory charger limits the charge current by design. A series resistor would also drop the current but the regulation would become poorer as the charge current is decreased. Probably not enough to be of concern with the small decrease you are seeking.

There is one caveat ,depending on the circuit of your charger an LED charge indicator may be very dim . It also may not light at all.

Also bear in mind some inexpensive VOMs add a bit of series resistance to the circuit while you are measuring . When the meter taken from the circuit the current can rise above the value intended. The difference may be a bit more than you are aware of.
Although the resolution is less ,higher current ranges in many meters use a lower series resistance and have less affect on the circuit.



Still, IF you want to keep it SIMPLE as so many seek to do ,the resistor method will work for you . Not ideal but elegantly simple and effective.
In some areas of the world they seek to keep it simple . Simpler yet a shorter charge duration as already suggested works too.

RJConnet

Thank you Lady Flyer,

Fifty years working in the electronics field have made me a staunch devotee of the KISS theory. Keep It Simple Stupid.

Bob C.

adam_one

Ladyflyer,

Thanks a lot for the redundant explanation (as it was already been given by Rodney - Titusville, Florida, 02-26-2003 3:56 PM)
Rodney wrote:
No need to worry about most wall warts compensating for added resistance, they do not. most are simply transformers and diodes that depend on the winding resistance to limit the current to about 50 to 70 ma, they do not have constant current circuitry built in.
However, if you read carefully my text you'll notice that I've written the following:
If the charger is designed to charge at a constant current…

So the word If gives a conditional term to my statement, right?

RJConnet:
I do care about my NiCads that's why I do prefer to charge them with a constant current source of about 1/10 of their capacity and I supposed that the most of us do the same.
So, please explain why I am stupid?

raymcm

How do you know the state of your pack? Do you fully discharge? If not how long do you charge at C/10 for ???

To step back to some of the previous replies, constant current cannot be achieved by a resistor as the pack voltage rises during charging and the voltage drop across the resistor will change and so will the current ???

Using the 55mA will cause absolutly no damage to the pack, just charge for less time.

If you really want to care for your packs then buy even the cheapest of delta peak chargers and charge at 0.5 - 1C and you will not overcharge them and it doesn't matter if they are half or fully discharged before you charge either....

Before anyone comments, the NiCd memory effect is a myth lingering from the very early days of NiCd and mainly caused by continuos charging at C/10

Raymond