Need help with charger for GP645 spotlight

[tony22]

I have an old GP645 powered spotlight that I got from my FIL. It used to work great but does not work right now. I figured that since it's so old I probably just need to get a new battery. I took it apart and checked out the contacts and continuity after removing the battery. Everything checks out fine. I checked out the charger and noticed it was putting out 12 VDC. The GP645 is a 6 volt battery. Not being sure how these unregulated chargers work I figured maybe once the battery is hooked up the voltage would drop to the correct level. I hooked the charger up after putting the old battery back in and read the voltage across the battery terminals - it was still aproximately 12 VDC. Isn't the charger supposed to charge at the same voltage as the battery? Do I have a bad charger?

 

[xul]

I hooked the charger up after putting the old battery back in and read the voltage across the battery terminals - it was still aproximately 12 VDC.

Battery is open-circuited?

Also, you can make yourself a shunt from several of these

Hosfelt
5 Watt .47 Ohm Wirewound
$0.19
p/n 5W-.47

in series or parallel or both and measure the current into the battery.

 

[tony22]

Battery is open-circuited?

:ohgeez:Should have checked that. Hang on...

It reads 3.45 MOhms. I guess that's open-circuited enough. :duh2:

But am I correct in assuming the charger will load itself down to 6 VDC when it's feeding a good battery?

 

[xul]

http://www.batterysales.com/downloads/GP645.pdf

Squeeze all the info you can from the graphs above to make up your battery equivalent circuit while being charged*, but
assuming 5.5 v for an open circuited discharged battery and higher than 0.02 ohms internal resistance Rint when being charged rather than what Rint is when being discharged, and
Vc = charger voltage
Ic = charge current
then
5.5 + Ic(Rint) = Vc,
so at Ic = 0.5A the charger voltage = 5.5 + 0.5(>.02) = >5.51 v
assuming the charger is current limited.

If the charger were an ideal voltage source, it and the battery would be at 12 v and the charging current would be enormous.

An ideal voltage source delivers its set voltage regardless of the current drawn, and its dual, an ideal current source, delivers its set current regardless of the voltage it takes to achieve it.
So you must never short an ideal voltage source or open-circuit an ideal current source.

*I guess you could simulate a battery being charged by using a huge Zener diode and a series resistor.

 

[tony22]

Thanks xul. This helps. If I look at the Battery Voltage and Charge Time graph, it shows charge voltages on the order of 13-14 volts. So I guess it shouldn't harm the battery, but I still wonder about the charger since it is listed as a 6 volt device. If it's rated to deliver 300 mA at 6 volts then...

5.5 + .3(Rint) = 6

which would yield a charging restistance of about 3.6 Ohms. Does that sound reasonable?

 

[xul]

it shows charge voltages on the order of 13-14 volts.

I think that's a misprint.

Be back in a few weeks.