Hello,
I'd like to verify my understanding of the LM74801's reverse current protection mechanism as it relates to my design.
I have read through the LM7480-Q1 data sheet and the Body Diodes Basics App note.
For my circuit, I am using an NCEP023N10LL FET for the reverse polarity protection FET. The Rds(on) range is between 1.8 and 2.3mOhms. According to the LM7480-Q1 datasheet, the Vac_rev is between -6.4min, -4.5typ, -1.3max. For the LM74801 to open the FET from a reverse current condition with the hysteretic ON/OFF control mechanism there must be at least -1.3mV and at most -6.4mV measured across the drain and source of the FET.
With the low Rds(on) of the NCE023N10LL, that would mean my circuit could see at minimum -0.56A and at most -3.56A being drawn out during a reverse current condition, is that true? Or am I not completely understanding this operation?
Calcs:
minimum = Vac_rev(max) / Rds(on)(max) = -1.3mV / 2.3mV = -0.56A
maximum = Vac_rev(min) / Rds(on)(typ) = -6.4mV / 1.8mV = -3.56A
*no minimum Rds(on) given for the FET
Thank you,
