# Calculating resistance for battery charger - easy for some lol! (help please)

Discussion in 'General Electronics Chat' started by turbine, Jun 28, 2011.

1. ### turbine Thread Starter New Member

Jun 28, 2011
8
0
Hi Guys,
I am making a quick car battery charger from a wall wart to get the car started (have a habit of leaving the park lights on). I am using a wall wart from an old portable TV rated 15v@ 800mA and my target is 13.7 volts @ 700mA
I quickly realized that the 15 volt wall wart puts out ~ 21 volts unloaded and then I remembered that I couldn't quite remember how to apply Ohms law in this case.

I need a bit of help.

I want to calculate the correct "dropping" resistor to place inline with the positive of the wall wart to limit the current (so as not to burn out the wall wart) and to bring the voltage down to the desired 13.7v.

When I calculate for the correct resistance using Ohm's law, how should I consider the voltage part of the Ohms law equation?

Do I use the target (13.7v) or the Voltage that my dropping resistor will "see" ; the difference between the 21V and the ~ 12.5 volts of the car battery? Ie a potential of 8.5v ?

OR

Should I calculate based on the desired voltage 13.7 ?

Help calculate that resistor and its required rating in watts. Please!

2. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,953
1,826
Oops... this is an automotive modification, and violates the TOS.

psst... charging a car battery with 800 mA ain't gonna be fast! Sears sells a wondeerful jumper cable package with a portable battery inside, plus a charger! You can leave that plugged in and fully charged until needed (I just charge mine every few months for a day or so), and when you leave your lights on you give yourself a jump and get onto your trip.

3. ### #12 Expert

Nov 30, 2010
18,094
9,679
You have 2 variables. The wall wart voltage per current and the battery voltage per charge condition. There ain't no resistor that will fix that and give you optimum charging rate.

However...you can put an 18 ohm 25 watt resistor in series with the wall wart and it will be protected from anything but a direct short. 22 ohms 20 watts if you want it protected from a direct short. Radio Shack has 10 ohm 10 watt resistor in a 2 pack. Put them in series for 20 ohms. Still, these won't protect the battery from over charging.

The right way to do a bulletproof charger is to use voltage regulator chips to make both a float voltage charger and a current limit protector for the wall wart.

Last edited: Jun 28, 2011
4. ### strantor AAC Fanatic!

Oct 3, 2010
5,010
3,029
I think you may be seriously underestimating the amperage requirement. you won't get a charge with that wall wart that could be described anything close to "quick". I suspect the battery may even continue to discharge while you have this wall wart hooked up to it.

anything amp rating starting with "milli" should be thrown right out.

If you want a "jump start", see this product:
http://www.grainger.com/Grainger/WESTWARD-Jump-Starter-1YMN2?Pid=search
It puts out 1000A peak to jump start the car.

This one could be referred to as a "trickle charger" - the kind you let set overnight or long periods of time:
http://www.grainger.com/Grainger/WESTWARD-Battery-Charger-1JYU1?Pid=search
it has settings for 2A, 40A, and 60A

5. ### turbine Thread Starter New Member

Jun 28, 2011
8
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Oh sorry, I read the restrictions, didn't realize it was any auto question Ermm perhaps we could just pretend it's a general 12v battery charging question? If not I'll withdraw it if possible.

I realize it will take some hours to charge @700mA so no jump starts, that's fine.

It's really the Ohms law stuff that make me curious. I stated to think it through and then realized I had a few questions, thus the post. This is where I got stuck:

When calculating for the correct resistor (using ohms law) with a wall wart rated @ 15v and putting out ~20v unloaded, should I calculate for the voltage the resistor "sees", that is to say the difference in potential between the ~21v and the 12.5 volts of the flat battery (so about 6.5 volts) or the 13.7v I am aiming for?

That the bit I am not sure about. One way I can see it is that 6.5 volts is the correction measurement (potential difference) and the other is the desired 13.7v

What I am going for here is nothing fancy at all. Just a quick and "dirty" solution with the stuff I have on hand. I really have to or make or buy a proper charger. The plan with this is just have something I can put on for a few hours and get the car cranking over, then drive off.

I just aim to restrict the voltage and current so the wall wart and battery aren't damaged.

Last edited: Jun 28, 2011
6. ### #12 Expert

Nov 30, 2010
18,094
9,679
As soon as you connect, the 21 volts won't be there!

I repeat: You can not calculate a resistor to satisfy 2 moving voltages at the same time.
You can connect the circuit and try to adjust a restance for the conditions present on that day.

7. ### turbine Thread Starter New Member

Jun 28, 2011
8
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So it's going to be a very ballpark figure, I hear you.
I know the 21v will be gone with a load on. I assume the wall wart will put out something close to it's rated 15v if I can restrict the current to within it's stated range and the car battery will move a little as it charges.

If Ohms law states that R = E / I

Voltage/Current= Resistance

Which is the correct way to think of the Voltage in this case?
The desired 13.7?
The voltage potential difference between the charger and the battery voltage?

8. ### #12 Expert

Nov 30, 2010
18,094
9,679
I'm thinking 15 volts fully loaded wall wart into 12.5 battery voltage at .8 amps is 3.125 ohms.

Nearest value is 3.3 ohms 4 watts.
Then, the next time you REALLY empty the battery, the resistor smokes and the wall wart fries.

9. ### turbine Thread Starter New Member

Jun 28, 2011
8
0
So you're going with the potential difference 15v - 12.5v = 2.5v.

and then

2.5v/0.8 A = 3.125 ohms

Then to get the power rating in watts P = E x I (voltage x current)

so... 2.5 x 0.8 = 2 watts and then double it to be safe yes?

Nov 30, 2010
18,094
9,679
Right.....

11. ### turbine Thread Starter New Member

Jun 28, 2011
8
0
Thanks buddy, I was all set to calculate it at 13.7 volts and then when I was having lunch I had a (dim) memory that "potential difference" might be a factor here.
Good thing I stopped by or I would have calculated it all wrong.

Thanks.

12. ### turbine Thread Starter New Member

Jun 28, 2011
8
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Hang on... going back the other way to get the voltage difference I get

E = I x R (voltage = current x resistance)

0.8 x 3.3 = 2.64v

Which subtracted from the 15v available is too much takes us back down to 12.5V , I want 13.7 a difference of 1.3v so maybe

R = E / I 1.3v/0.8A = 1.625 ohms?

And then going for power P = E x I

1.3v x 0.8A = 1 watt (double it for safety)

Phew hope thats right.

13. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,953
1,826
While all your calculations are correct you are picking to deliver the charging current into a fully charged battery. So when the battery is discharged you will be pumping even more current into it and burning off even more power in that resistor.

Battery chargers are quite complicated little beasties. If you do this don't be surprised if you see some smoke from somewhere.

14. ### turbine Thread Starter New Member

Jun 28, 2011
8
0
The resistor is an odd value, so I'll solder up a little bit of Vero with something between 5-10 1/2 watt resistors (depending on the values I have on hand) to achieve my 1.65 ohms @ somewhere between 2.5 and 5 watts.
If I have 0.5 v to "spare" I'll have a 1n4001 diode as the exit point to the positive. They are rated @ 1 amp so it can function both as a diode and a sacrificial fuse/current limiter.

This is all very "getto" but kinda fun.
May well end up just going to Kmart and buying one too if the fun ends up not being so amusing

15. ### #12 Expert

Nov 30, 2010
18,094
9,679
Actually, a real fuse would be a good idea.

Quoting, "While all your calculations are correct", you are picking points in time that are temporary. It doesn't matter if you get something perfect, it will change in a hour!

Really, I'm getting frustrated with you because you don't seem to understand that "There ain't no resistor that will fix that and give you optimum charging rate".

Play with it all you want, but your stated goal is impossible using nothing but a resistance. All I can say now is, "have fun" as that seems to be the real goal.
I don't object to having fun. I object to trying to calculate where the horse is without looking out the window.

16. ### turbine Thread Starter New Member

Jun 28, 2011
8
0
Thanks guys worked perfectly - wife is happy/car is running.

@12

What the?
My goals were pretty clear. If you noticed words such as "ballpark" "getto" etc you would see that a quick and "dirty" solution using only resitance WAS the goal.

I can sit down and design something around a fixed voltage and variable current source later. This for today was " quick and dirty".
The "moving" points aren't that volatile. A semi flat (not dead flat) car battery will only vary over a few volts and with the correct resistance a 15volt wall wart will be reasonable stable whilst kept under load. So checking and adjusting in field is the go..
Remember the idea here is to simply "top up" the battery for a few hours only and then disconnect and harm neither the wall wart or battery in the process...

My only question to you was how to read the voltage: total or potential between the charger output and battery so as to calculate using ohms law.

I got there is the end under my own steam. Your math was wrong in all your answers so I really don't know what you mean by being "frustrated" but I think you should take a less gruff approach with new members.

Cheers guys.

17. ### davebee Well-Known Member

Oct 22, 2008
539
47
Here's another way to look at it.

The wallwart must have a large internal resistance already: 21 - 15 = 6 volts @ 800 ma R = 7.5 Ohms

So if you connect it to the car battery, you'd want more like: 21 - 13.7 = 7.3 volts, R = 7.3/0.8 = 9.125 total Ohms

You already have 7.5 Ohms of wallwart internal resistance, so you only need to add 1.6 Ohms.

I'd suggest that you get a 1 or 2 ohm resistor of several watts power rating, connect the wallwart through the resistor to the battery, then measure what happens.

Measure the current, the voltage, feel both the wallwart and the resistor to make sure they don't get too hot.

I suspect the others are right - this will charge so slowly that it would take several days to build a charge enough to start your car, but if you have the time, it should work. 800 ma isn't that small a current.

18. ### #12 Expert

Nov 30, 2010
18,094
9,679
I didn't think I was being really awful. I guess it came across worse than I intended.

The frustration is about telling you what value of resistor will work to "top up the battery" when it isn't very bad off, and knowing that "so as not to burn out the wall wart" has not been achieved. With resistances as low as 2 or 3 ohms, "only varying over a few volts" has a lot of effect, like nearly an amp. When you have less than an amp in the first place, that's a problem!

I will go study Ohm's Law again, as it seems I didn't do well with that, either. Being wrong in all my answers shows a serious defect on my part.

Good luck with this. I'd say I see a very hot resistor in your future, but I'm not doing very well with Watt's Law either.