Charging a Backup Battery

Brambo · #1 07-14-2007, 01:38 PM

I'm thinking about installing a second battery to act as power source for GRT screens during startup. The battery requires a max charging amp of 2.1. So, doing the math if I assume 14volts from the alt, and put a 10ohm ressistor in the line from alt to battery that should give 1.4 amps charging. Problem is, Radio Shack has 10ohm resistors that come in 1/4, 1/2, and 1 watts. What is the differance?

Also, I thought I would put a 3amp diode right after the resistor.

Bill Rambo
RV-7A

Ironflight · #2 07-14-2007, 03:46 PM

At the serious risk of doing math in public...p=vi, so with your 10 amp resistor you're going to be running about 20 watts - those resistors are going to be a bit small, and you're going to let all the magic smoke out.

For my setup, I am using a 4.5 Ah sealed battery that I trickle charge with about 1 watt of current (I forget the resistor size, but I think I have it in the "File" section of the GRT Yahoo group). The batter doesn't care what current it is charged at - I just have to make sure IO have enough of a voltage difference to drive current into it. Your battery might well be different - I know battery experts, and they are a lot smarter than I am!

Paul

Rick_A · #3 07-14-2007, 03:57 PM

LSE has diagram for a backup battery for a Dual Lightspeed setup. It should also meet the charging requirements for the GRT b/u bat.

They don't even bother with a resistor. So is it really needed?

http://www.lsecorp.com/Manuals/PS_Diagram.htm

Walt · #4 07-14-2007, 06:02 PM

Providing a constant current source is not that simple. The battery also has an internal resistance and off hand I'm not sure what that is. For exampe, if you have 1 amp flowing through a 10 ohm resistor (and assuming 4 ohm battery) you are going to have 10 volts dropped across the resistor and only 4 V across the battery. The resistor will also be disipating 10 watts of power (I think the largest resistor they sell is 10 watts and at max power that sucker will burn you). If you are going to use a resistor you need to know the battery resistance to make the correct calculation and then get the appropriate size resistor to handle the power (also if it calls for a 10 watt dissipation I would go with at least a 20W resistor to maintain a reasonable margin).
Whew... that took more brain power than it should have

Walt · #5 07-14-2007, 07:25 PM

If it were me I'd use the same schematic as in the LSE link above and just use a 2 or 3 amp fuse or CB to limit the current in the battery. If the battery is kept pretty close to full charge the charging current is really based on the voltage difference between the batteries/alternator output which is normally going to be very low. The only time you would have a higher current is if the battery was discharged quite a bit in which case you would blow the fuse. If you put a resistor in the circuit I don't believe you will ever get a fully charged battery. If you do need to charge a fully discharged battery you can do so from a current regulated battery charger

Below is a desription of a contant current curcuit you can use if you feel you must (but overkill IMO):

The circuit (lower right) illustrates using a LM317 variable voltage regulator as a constant current source. The voltage between the adjustment terminal and the output terminal is always 1.25 volts, so by connecting the adjustment terminal to the load and placing a resistor (R) between the load and the output terminal, a constant current of 1.25/R is established. Thus we need a 12 ohm resistor (R) to get 100mA of charge current and a 1.2 ohm, 2 watt resistor for 1 amp of current. A diode is used in series with the input to prevent the batteries from applying a reverse voltage to the regulator if the power is turned off while the batteries are still connected. It's probably a good idea to remove the batteries before turning off the power.

If you absolutely must use a resistor use the circuit in the upper left and you can use this this formula to figure it (if you use 14 and 10 you will need a 2 ohm resistor for a 2 amp limit which give you 4 watts so a 10 watt resistor would do):

A simple method of charging a battery from a higher voltage battery is shown in the circuit below to the left. Only one resistor is needed to set the desired charging current and is calculated by dividing the difference in battery voltages by the charge current. So, for example if 4 high capacity (4000 mA hour) ni-cads are to be charged at 300 mA from a 12 volt battery, the resistor needed would be 12-(4*1.25)/0.3 = 23.3 ohms, or 22 ohms which is the nearest standard value. The power rating for the resistor is figured from the square of the current times the resistance or (0.3)^2 * 22 = 2 watts which is a standard value but close to the limit, so a 5 watt or greater value is recommended

Brambo · #6 07-14-2007, 09:53 PM

Thanks for the reply. I believe I'll will just go with a 3 amp fuse. That seems to be the simple way.

Bill Rambo

Radomir · #7 07-14-2007, 11:56 PM

Don't forget the diode...

Ironflight · #8 07-15-2007, 06:21 AM

Quote:

Originally Posted by Radomir

Don't forget the diode...

And one of the reasons for the resistor is so that you don't have too much current for the diode ... make sure to check the rating on that! (Or you'll end up creating a NED - a Noise Emitting Diode - the emit a large noise - once!)

Paul

frankh · #9 07-15-2007, 06:56 AM

I'm a bit confused here as it seems we are talking anout different things..

1) you want to add a second 12V battery..or...
2) you want to add a lower voltage battery say a 6v?

If then you don't need any circuits or resitors you just simply hook the second battery in parralel with the first. Adding a diode is also required to prevent backflow when your cranking battery voltage is pulled down by the starter.

You will however get a forward voltage drop from the diode...I.e if your charging at 14V from the alt then you'll probably end up with 13V at the second batt. Ad a fuse to protect the diode.

If 2, then you'll need a way to step down the volts from the circuit shown.

Hope I'm reading this right...its not 6am yet...

Frank

Walt · #10 07-15-2007, 07:17 AM

Quote:

Originally Posted by frankh

I'm a bit confused here as it seems we are talking anout different things..Frank

1) you want to add a second 12V battery..or...
2) you want to add a lower voltage battery say a 6v?

If then you don't need any circuits or resitors you just simply hook the second battery in parralel with the first. Adding a diode is also required to prevent backflow when your cranking battery voltage is pulled down by the starter.

You will however get a forward voltage drop from the diode...I.e if your charging at 14V from the alt then you'll probably end up with 13V at the second batt. Ad a fuse to protect the diode.

Forward VD on a standard diode is 0.7 while the schotky is closer to 0.2. The schotky shown is more than capable of handling the current, fuse protects battery and diode.

Quote:

Originally Posted by frankh

If 2, then you'll need a way to step down the volts from the circuit shown.Frank

I showed the second resistor method only to illustrate the calculation necessary to limit current to some value. The numbers I used in my example (14/10) represent the alternator ouput vs a mildly discharged back up battery.

Quote:

Originally Posted by frankh

Hope I'm reading this right...its not 6am yet...

Yes you are reading it correctly!