Hello everyone,
I'm developing my own board using the nRF5340. I am quite lost because of the capacitance matching that I need to perform with the 32M Crystal and the 32k Crystal, and I would like to do it with external capacitors.
Following both the PS pdf (https://infocenter.nordicsemi.com/pdf/nRF5340_PS_v1.2.pdf) and this additional resource that I found online https://infocenter.nordicsemi.com/pdf/nwp_015.pdf?cp=12_12 I am trying to figure out what capacitance should I use.
Note that the load capacitance of the 32M crystal in the nrf5340DK is the same of our crystal (8 pF)
Assuming we work with internal capacitors disabled, then Cpin = 5.5 pF
We know from the link above that
C' = Ct + Cpcb + Cpin = 2 * C
Ct + Cpcb + 5.5 pF = 2*8 pF --> Ct + Cpcb = 11.5 pF
Is this calculation correct? Also, IF we enable and keep default settings (from the blinky example) about the internal capacitors, what is this default value? I couldn't find anything with respect to the 32M Crystal.
Instead, I know that for the 32k crystal (load capacitance = 8 pF) the internal set capacitance Ct is by default at 7 pF.
Assuming we work with internal capacitors enabled and to default, then Cpin = 7 pF (typical from the datasheet)
We know from the link above that
C' = Ct + Cpcb + Cpin = 2 * C
Ct + Cpcb + 7 pF = 2*8 pF --> Ct + Cpcb = 9 pF --> Cpcb = 2 pF
Are these calculations correct? Is it reasonable to assume to have a Cpcb so small? I know that it depends by many thing, but I 'd like to have a reasonable dimensioning for the capacitors that I need to place.