Realtime Clock Power Questions

SkydivinGirl

Retro Girl
Joined
Dec 16, 2008
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Hello fellow geeks!

As you may have seen in my Commodore PC10-III thread, I installed a 6V battery pack to bypass some leaked acid damage for the power line on the realtime clock. Wiring the battery pack's positive line to the power pin on the chip worked great, but I think there are other things pulling the voltage down and running the batteries down.

According to the RTC chip specs, the chip can go into low power mode needing only 2V to operate. Standard operating mode is 5V with a maximum of 6V. After only a short amount of time (a week to a week and a half), my 6V power pack was only showing 2V at the clock with the system turned off.

So, how do I go about fixing this problem? I'm thinking that I can cut the trace that connects the clock's power pin to the main board. After that, I can put in a 3V battery with a diode for standby mode but also run 5V from another point on the board to the clock's power pin for when the computer is powered. I think this will work, but I have a few quick questions before doing this.

1. Should I put another diode between the 5V jumper wire and the clock's power pin? The reason I think I might want to do that is to keep the battery from sending power back along the jumper to other parts of the motherboard when the computer is turned off.

2. When the computer is turned on, it will effectively have two power sources (3V from the battery and 5V from the jumper). What actually happens in this situation?

3. Am I going about the completely wrong? If so, what should I do?

Thanks for your help!

Heather
 
Would a jumper or switch somewhere on the circuit be an easier route? That way you can chose to have the power flowing through it or not, though that said I havn't ever come accross a PC (albeit a retro one) that's got this feature so it's hard for me to recommend. Just a suggestion anyway.
 
2. When the computer is turned on, it will effectively have two power sources (3V from the battery and 5V from the jumper). What actually happens in this situation?

i would imagine,depending on the circuit, that it charges the battery while powering the clock ic.

in a similar way to most other clock circuits,like the amiga for instance.
 
Would a jumper or switch somewhere on the circuit be an easier route?
The original circuit does not have a switch and I don't want to have to manually do anything. :)

i would imagine,depending on the circuit, that it charges the battery while powering the clock ic.
With the diode in line, the batteries will not be charged. I suppose a better question is this: Would the voltage potential from the 5V line keep the 3V line from the battery from discharging while the 5V line is active?

*EDIT* I just answered my last question. I'll go with my 5V from the board + diode solution unless someone else has some other options that may be better.

Thanks!

Heather
 
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