Disconnected from the pin. Vin = 0V. Either way Vin should not reach or power the CAP+ rail.hi a,
Ref one diode.
When you say Vin looses power .
Does this mean it is disconnected from that pin or is the 12V line still connected back into the non active 12V psu.?
I would design it with worst case scenario, that with Vin not powering yet connected. In this case would it be ok to have the diode in series ? I guess that would not let the smoke out from anything around.hi,
If it was not disconnected, then the 9V would be connected to the non powered output of the 12V psu, which is not desirable for the 12V psu.
If fully disconnected only one diode is required, from the 9V supply
If Vin = 12V and Vcap = 8V, the diode in series with Vin is not needed, unless you want it to drop Vin. In that case, why use a Schottky?Thank you all for the reply.
The idea is only to power the regulator when Vin is OFF. Charging the CAP is not in the scope. CAP+ acts as a secondary power when Vin looses power. Ideally Vin rail is +12V and CAP+ is 8V so charging the CAPs from the +12V rail would be a no no.
Can't seem to understand how one diode can do the job.
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by Jake Hertz