oPossum 1,083 Posted July 17, 2013 Share Posted July 17, 2013 If you have a module like this... and it is unreliable or not working at all, here is a possible fix. - Remove battery - Remove D1, R6 and R4 - Solder jumper wire in place of R6 - Inspect crystal soldering - fix if necessary - Replace battery The module should now work properly. It requires 5V power, but will work with 3.3V logic. The DS1307 can be replaced with the newer pin compatible DS1338 for operation from 1.8V to 5V, or the TI BQ32000 for 3.3V (only) operation. --- The battery backup circuit on these modules is of poor design. D1 and R5 provide a charging current to the CR2032 coin cell. Lithium primary cells such as the CRxxxx series should not be (re)charged. Removing D1 fixes this. R4 & R6 create a voltage divider between the battery and the DS1307. This raises the source impedance (spec sheet warns against this) and increases the current draw from the battery. The voltage divider was necessary because the battery voltage would float at about 4.2V volts due to the D1/R5 circuit! That would keep the DS1307 disabled if present on the Vbat pin. Replacing R6 with a jumper allows the full battery voltage to Vbat. spirilis and bluehash 2 Quote Link to post Share on other sites
spirilis 1,265 Posted July 17, 2013 Share Posted July 17, 2013 Lol! That is pretty amazing. Suddenly my epic fails in PCB design seem small in comparison... Sent from my Galaxy Note II with Tapatalk altineller 1 Quote Link to post Share on other sites
bluehash 1,581 Posted July 17, 2013 Share Posted July 17, 2013 Thanks opossum. Quote Link to post Share on other sites
cde 334 Posted July 17, 2013 Share Posted July 17, 2013 @@oPossum don't these have i2c pull-ups tied to 5v (Actually the ds1307/module's VCC which should be 5v)? Wouldn't these need to be removed as well? Quote Link to post Share on other sites
oPossum 1,083 Posted July 18, 2013 Author Share Posted July 18, 2013 There are 3k3 pullup resistor to 5V. So there will be about 300 microamps current flow thru each of two of the ESD protection diodes in the MSP430. This will usually not cause problems as long as there is something on the MCU's supply rail to sink this current. If the MSP430 is running at 1MHz or more and not sleeping, then it will use that current. Adding 6k8 pulldown resistors to SDA and SCL would reduce the voltage to about 3.3V and prevent current flow thru the ESD diodes. Quote Link to post Share on other sites
cde 334 Posted July 18, 2013 Share Posted July 18, 2013 But then wouldn't that voltage divider reek havoc on the i2c communications, increasing the pull-up time, making higher speed i2c unstable? And the 300 microamps, if the msp is asleep, would cause issues? Quote Link to post Share on other sites
oPossum 1,083 Posted July 18, 2013 Author Share Posted July 18, 2013 The 3k3 + 6k8 resistors with 5V are equivalent to 2k22 with 3.33V (th Quote Link to post Share on other sites
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