ku637
Advanced Member level 4
Hello,
I need to calculate the CANH to CANL capacitance in following circuit.
In the datasheet of the device the CAN input differential capacitance (Cdiff) is given as 10pF and additionally CANH and CANL (Ci) input capacitance is given as 20pF.
e.g https://www.onsemi.com/pub/Collateral/AMIS-30660-D.PDF
I need to calculate the total effective capacitance from CANH to CANL
So i understand the Cdiff=10pF will always be there between CANH and CANL.
Also, i understand that if effective pin capacitance is the matter of interest i need to add all the capacitance i.e if for e.g CANH pin then Cdiff+ Ci +C1 is the total capacitange.
But the question is lead to lead capacitance from CANH to CANL ,then
looking from transceiver side C1 and C2 appears in series (through GND) can we take Cdiff +C1||C2 as the effective lead to lead capacitance?
Or is it Cdiff + C1 + Ci will be the right answer?
Thanks for any help.
I need to calculate the CANH to CANL capacitance in following circuit.
In the datasheet of the device the CAN input differential capacitance (Cdiff) is given as 10pF and additionally CANH and CANL (Ci) input capacitance is given as 20pF.
e.g https://www.onsemi.com/pub/Collateral/AMIS-30660-D.PDF
I need to calculate the total effective capacitance from CANH to CANL
So i understand the Cdiff=10pF will always be there between CANH and CANL.
Also, i understand that if effective pin capacitance is the matter of interest i need to add all the capacitance i.e if for e.g CANH pin then Cdiff+ Ci +C1 is the total capacitange.
But the question is lead to lead capacitance from CANH to CANL ,then
looking from transceiver side C1 and C2 appears in series (through GND) can we take Cdiff +C1||C2 as the effective lead to lead capacitance?
Or is it Cdiff + C1 + Ci will be the right answer?
Thanks for any help.