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So how negative can I drive the Vgs of an N-channel Mosfet before damage occurs? Datasheets specify BVDSS and BVGSS but these are "forward" breakdown voltages -- ie the absolute maximum Vds before avalanche breakdown of the drain body pn junction, and the maximum gate drive voltage before the electric field breaks down the gate oxide. I can't find the "negative" breakdown voltages in the device specifications.

Does the susceptibility of the gate oxide to an excessive e-field then suggest that the maximum "negative" Vgs is the same as the BVGSS specified but in the opposite polarity?

EDIT: Here's a portion of the p-channel device I was looking at that prompted the question: enter image description here

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All datasheet specified Vgs_max for example.

enter image description here

In this case "negative" (minus) 20V is the maximum value.

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  • \$\begingroup\$ yep, it is obvious from that data sheet, thank you. The datasheet of the device I was looking at (a P-channel, CSD25213W10) does not specify it... Unfortunately, I can't add the pic here... It specifies it only as BVGSS Gate to Source Voltage; VDS = 0V, IG = –250μA –6.0 V (min) I asked the question in terms of an N-channel to avoid any confusion, knowing that i would understand how any given answer would apply to the p-channel device. So, although the datasheet for the CSD25213W10 does not specify the maximum reverse Vgs, it is then understood to be +6V? see updated question question.... \$\endgroup\$
    – jrive
    Commented Apr 6, 2020 at 19:16
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    \$\begingroup\$ Yes, it will be +6V because if the MOSFET substrate was not short together with a source (due to physical construction) it will be a fully symmetrical device. \$\endgroup\$
    – G36
    Commented Apr 6, 2020 at 20:21

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