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S Aug 7 at 17:02 history edited Tim Williams CC BY-SA 4.0
add a brief note; also specific part numbers, and justification
S Aug 7 at 17:02 history suggested jwo CC BY-SA 4.0
add a brief note
Aug 7 at 16:57 review Suggested edits
S Aug 7 at 17:02
Aug 6 at 6:16 vote accept RUMBUFDSI
Aug 5 at 19:07 comment added Tim Williams @Jens Depends on magnitude of foldback. ;) Given the original circuit was an [attempted] CCS, I expect that power dissipation is acceptable; and if not, I have the note mentioning it. Given they keyword, one can search on design of foldback limiting, or ask a new one; it's off-topic of this question, I would say.
Aug 5 at 17:15 comment added Jens Will it enable the load current at all at startup?
Aug 5 at 15:37 history edited Tim Williams CC BY-SA 4.0
added 567 characters in body
Aug 5 at 15:33 comment added Tim Williams @RohatKılıç Actually, I will add an edit, for a reason no one mentioned yet..!
Aug 5 at 15:30 comment added Tim Williams The the question says 12V and made no concerns about reversal, cross-wiring, etc., so are not a concern. For example, it might be supplied from a (hard wired) battery or SMPS, or be part of a larger module where the 12V is by design, or already protected against such, etc. It's a building block, not a complete standalone module. That's how I understood the question.
Aug 5 at 15:03 comment added Rohat Kılıç @TimWilliams yes, it's PMOS. The body diode's cathode at S, anode at D. In case of a reverse-polarity input it'll be forward-biased (again, ignored the diode D in your schematic). Though the OP said a reverse-polarity protection is implemented in an early stage but what if there wasn't any.
Aug 5 at 13:09 comment added Tim Williams @RohatKılıç Check again -- it's PMOS, source to V2. The arrow shows the substrate diode pointing up.
Aug 5 at 12:57 comment added RUMBUFDSI This answer is extremely helpful, thank you very much!
Aug 5 at 12:52 comment added Rohat Kılıç +1 for the answer and the excessive work on the existing image. One other thing worth to note is the reverse polarity: VT2's body diode will be forward-biased (ignoring the D diode in your circuit) so the incandescent bulb will still light up but there won't be any current limiting or S/C protection.
Aug 5 at 12:44 history answered Tim Williams CC BY-SA 4.0