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Voltage vs ma rating for reusing chargers - 3/6/2003 9:13:15 AM   
gkjake


 

Posts: 13
Joined: 3/5/2003
From: Fort Leavenworth, KS,
Status: offline 		All,
Have read Red's (he hee read Reds -- sounds funny) battery clinic and how to reuse and test at home chargers already on hand etc. I was wondering if the output voltage made much difference when choosing one to start with. I read somewhere in this forum that you must have at more output voltage than your combined cell voltage (over 7.2V for a 6 cell NiCd). How important is this and will it come to light anyway once you check the current of the circuit with your Ammeter and get the real deal? I took Juice in college but all I can remember is twinkle twinkle little star V=IR and that you add voltage for a series and add current when hooked in parallel. I'm sure there should be some easy mathmatical formula I could do to calculate what my charge rate in ma would be for a 7.2V NiCd with 6 cells @ 600ma when hooked to a --for example -- an output of 15V and 200 ma. Some one who remembers this stuff help me out.
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Voltage vs ma rating for reusing chargers - 3/6/2003 8:23:09 PM   
Jazzy



Posts: 1106
Joined: 1/8/2002
From: Erie, PA, USA
Status: offline 		Well,
It's not so much Ohm's law but how far you push the wall wart beyond it's design parameters.
Generally speaking, say your wall wart is rated at 6V, 175mA output. Charging a single cell or a number of cells whose voltage is much less than the wart's rating will yield a higher mA output.
If charging beyond the wart's voltage rating, for instance a six cell pack, (7.2V nominal), the mA output of the wart will decrease.

I mention the wart's design parameters because I have fried one or two warts by exceeding them.

The math may say one thing but the actual results may yield another. Your best bet is to use your meter.

Using your example and the equation: W=E*I, it can be shown what your wart's output would be at 7.2V.

W=15V*.200A
W=3 , 3 watts
To find the current, I, at a lower voltage one must rearrange the equation to read: I=W/E.
I=3W/7.2V
I=.417A or 417mA

From what I remember, that is how one would do it. If I had my new scanner operational I'd post a sheet of Ohm's law and derivations of it.

Hope this helps,
Jeff

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Thanks - 4/25/2003 9:39:34 AM   
gkjake


 

Posts: 13
Joined: 3/5/2003
From: Fort Leavenworth, KS,
Status: offline 		Jeff-- don't know if I mentioned it before but I really appreciate the help--thanks much

Greg

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Voltage vs ma rating for reusing chargers - 4/25/2003 10:12:07 AM   
mglavin



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Joined: 11/29/2001
From: Elverta, CA, USA
Status: offline 		Interesting, but. If the charger is rated for 200mA output @ 15V and the cell count is five or 7.2V static. Presuming the combined cells would consume 7.2V under charge is unrealistic, IMO. The math or formula is accurate, but the V rate at which the battery pack will be charged is unknown. Most likely is it will be much higher than 7.2V as the charger is capable and rated to deliver 15V. It knows not what's at the other end. A simple exercise with a DVM will reveal the actual V under load.

I am not convinced that the charger will charge at a higher mA output simply because the V is well below the output rating. There is no load other than the cells IR...

As Jazz mentioned math and theory are often worlds apart from reality.

< Message edited by mglavin -- Apr 25 2003 5:26AM >

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yup--test it - 4/25/2003 10:16:07 AM   
gkjake


 

Posts: 13
Joined: 3/5/2003
From: Fort Leavenworth, KS,
Status: offline 		Yes-- I agree with you-- the only sure way I have found is to test it with a voltmeter in series with the positive lead going from charger to battery to see how many ma it is putting out. If anyone knows how to calculate this mathmatically I would appreciate it.

Thanks

Greg

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Voltage vs ma rating for reusing chargers - 4/25/2003 12:10:12 PM   
Steve Lewin


 

Posts: 904
Joined: 3/18/2002
From: Reading, UNITED KINGDOM
Status: offline 		[QUOTE]Originally posted by mglavin
Interesting, but. If the charger is rated for 200mA output @ 15V and the cell count is five or 7.2V static. Presuming the combined cells would consume 7.2V under charge is unrealistic, IMO. The math or formula is accurate, but the V rate at which the battery pack will be charged is unknown. Most likely is it will be much higher than 7.2V as the charger is capable and rated to deliver 15V. It knows not what's at the other end. A simple exercise with a DVM will reveal the actual V under load.

I am not convinced that the charger will charge at a higher mA output simply because the V is well below the output rating. There is no load other than the cells IR...

As Jazz mentioned math and theory are often worlds apart from reality. [/QUOTE]
I'm sorry but you've completely missed the point. If you connect a 6V battery to a charger rated for a maximum of 15V the output of the charger is dragged down to 6V. There's no way you can pull a 6V battery up to 15V. Batteries are voltage sources, they decide what voltage they take. You can't apply Ohm's Law to them as though they were resistors. In your terms the charger certainly does know what is at the other end and the output voltage is forced down by it. Your simple exercise with the DMM will clearly show this.

Because of this and the way the wall wart chargers are constructed at the lower voltage the output CURRENT of the charger will increase. A simple measurement with an ammeter on any wall wart will confirm this too.

Steve

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Voltage vs ma rating for reusing chargers - 4/25/2003 7:38:00 PM   
Rodney



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Joined: 12/8/2001
From: FL
Status: offline 		One of the limiting factor on current is the internal impedance of the charger. This is primarily limited by the size of the wire used on the transformer in the charger. If you notice, the wall warts usually get quite warm after a few minutes, this is due to the self heating by the current thru the internal impedance and will increase as you lower the number of cells you are trying to charge.

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Voltage vs ma rating for reusing chargers - 4/25/2003 7:54:14 PM   
gkjake


 

Posts: 13
Joined: 3/5/2003
From: Fort Leavenworth, KS,
Status: offline 		Steve,

I think we are saying the same thing and I agree with you on the voltage. My point is you can't be positive exactly what voltage and current will go to the batteries throughout the charging process. This varies with internal resistance, battery type and # of cells etc. The best way is to hook your voltmeter/ammeter in series and test. I think we are saying the same thing but the best method is to test it.

Greg

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Voltage vs ma rating for reusing chargers - 4/25/2003 8:05:23 PM   
Steve Lewin


 

Posts: 904
Joined: 3/18/2002
From: Reading, UNITED KINGDOM
Status: offline 		We're almost saying the same thing. However the current from a wall wart charger is mainly determined by the internal resistance of the CHARGER not any battery connected to it, as Rodney says.

However if you take a wall wart apart you'll find that it's not the windings of the transformer which are the major part of the internal resistance it's the resistor stuck in series with the transformer. It's placed in there specifically to define the output current, a sort of poor mans constant current device.

But you're absolutely correct about one thing. If you want to KNOW what's happening measure it, don't just guess

Steve

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Voltage vs ma rating for reusing chargers - 4/25/2003 9:02:54 PM   
gkjake


 

Posts: 13
Joined: 3/5/2003
From: Fort Leavenworth, KS,
Status: offline 		Steve,

Thanks, you definitely know more about it than I do. Hey by the way--just got a JR XF421 5 channel radio off ebay and it came with a charger-- one lead for the radio itself and the other for a the battery for the receiver/servos etc. I have a BEC so obviously don't need that one and want to convert it into a trickle charger (its rated at 5.8 V 50 mah) but it has three leads - positive, negative and a sensor lead --guess that makes it the light on the charger light up. Can I just splice the connector lead to the positive lead and then use it to trickle charge my batteries? Obviously I will check this using the methods we have talked about but it is the sensor lead that is confusing me.

Thanks
Greg

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Voltage vs ma rating for reusing chargers - 4/25/2003 9:36:41 PM   
hebertjj


 

Posts: 323
Joined: 1/18/2002
From: Knoxville, TN, USA
Status: offline 		Jake,

As someone noted, most of these wall warts are just a transformer, a diode to convert to sort of DC, and a resistor to limit the current. If it's a 12 volt ( for instance ) and the internal resistance of the transformer, the resistor, and the battery is 140 ohms, the diode gives a .5 volt drop, and the battery is discharged to 4.5 volts; that means you got 7 volts (12 - .5 - 4.5 )dropping across 140 ohms which will give you 50ma. This goes down as the battery voltage goes up, but that's an estimate to start with.

Since you don't really know what the internal resistance of the wart is, (as others have said) or it may even be a more complicated constant current supply, plug in an amp meter and measure the current when plugged in. If the current is less than twice the rated current of the wart, you're probably OK.

Joel

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Voltage vs ma rating for reusing chargers - 4/26/2003 4:46:24 AM   
Jazzy



Posts: 1106
Joined: 1/8/2002
From: Erie, PA, USA
Status: offline 		Ok, Steve supplied additional info from my initial computation. Thanks Steve.
For those in question, these wall warts are mostly constant power chargers. Therefore, as the cell number/voltage decreases from the rated amount the current will increase. Inversely, as the cell number/voltage increases from the rated amount the current will decrease.
Still, due to the limits of the wall warts it is a good idea to check the output with a meter and watch it for overheating. (The latter is just a safety precaution.)

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