I am trying to program an ISL12022M RTC and am having trouble interpreting the register map (self taught with little experience). The documentation says that the RTC registers (SC,MN,HR,DT,MO,YR,DW) are BCD representations. In order to allow write capabilitiy into the RTC registers the WRTC bit(bit 6 of address 08h is set to '1'.The map looks like this:
The FAQ example from the Intersil site tells me that to set the WRTC bit I need to send DEh (slave address) 08h (register address) and 41 (Enable WRTC bit, other bits remain in default). Why not hex? Why 41 and not 40? And what does SC22 in SC bit 6, SC21 in bit 5, etc. mean?
I've read the documentation until I can't see anymore and I've searched until I am just getting more confused. Any help is appreciated.
Well, it looks like these values in the map are nibbles. The range for the first register is 0 - 59. When represented in BCD, 4 bits are needed for the digit in the ones place and three bits are needed for the 10's place. So, bits 0 - 3 belong to the first nibble; bit 0 = SC(register name)1(first nibble)0(first bit). Bits 4, 5 and 6 belong to the second nibble. Bit 4 = SC(register name)2(second nibble)0(first bit). Bit 7 is not needed.
The example sheet from Intersil has a typo; the WRTC value needs to be 40h or 41h.