DS1307 backup power

Peter Verkaik

I have a DS1307 rtc circuit, however it is mechanically not possible
to add a lithium 3V coin cell. There is however a 9V 700mAh battery.

I was thinking of using a 78L02 regulator (output = 2.6V) but the
bias current is 3.6mA, which would drain the battery in·less than 200 hours.
So that is no option.

Perhaps a resistor divider using·2.7Mohm and 1.5Mohm resistors:
9V * 1.5 / (1.5 + 2.7) = 3.2V
Battery current = 9V / 4.2Mohm = 2.15 uA
Since the ds1307 backup input consumes 0.5 uA
the backup voltage will drop towards 3V.

Anybody have any experience using this setup (or other ideas)?
I would like to hear some figures on how long the 9V battery would last,
also consering the battery self-discharge.

regards peter
·

Tracy Allen

Hi Peter,

I see no reason why the voltage divider scheme wouldn't work. Also, LEDs even when biased at a couple of microamps forward current make pretty decent parallel voltage regulators.

I guess the 9v battery will power the rest of the circuit as well, which will turn off or go to sleep at times? Note that the DS1307 only achieves the 0.5uA current from the backup power when the main Vdd power turns off, and also be sure that the scl and sda lines are low (not pulled up to a higher I2C voltage).

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Tracy Allen
www.emesystems.com

Peter Verkaik

Hi Tracy,

Yes, the remaining circuitry·is powered·via a PNP transistor that can be turned off.
And I just found out there are nowadays 9V lithium batteries as well (1200 mAh), usually used
for smoke detectors.

regards peter