Hello,
Would the attached schematic be correct to bypass the CAP+ (8V) rail when Vin (12V) is disconnected ? First time using a P-Channel MOSFET so i am in doubt if this would be correct.

 

I guess adding a pulldown resistor at the gate should be ideal here because without it the gate would be floating when Vin is disconnected.

 

hi anish,
Is this what you are trying to do.?
E
Better image.!

 

You want the regulator powered by CAP+ when there is no VIN present?

How about a pair of Schottky diodes?

 

Is Vin >= Vcap? That is, is the intent to keep Vcap charged by Vin until Vin is removed? (NB: The mosfet conducts in both directions when on and by the body diode when off.)

If so, why do you need any mosfet switch?

 

I guess adding a pulldown resistor at the gate should be ideal here

You need the pulldown resistor if using the P-MOSFET.

 

hi,
As suggested I would use a couple of schottky diodes.
E

 

I would only use one diode and am not even sure that is needed. In fact, if my assumption about the plan is correct, Vcap is simply parallel to C17. There could still be a reason for C17 (like lower esr).

 

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.

I guess the mosfet could be replaced by a Schottky diode as suggested. I am guessing this is what @Sensacell suggested
@jpanhalt the 10uf is a decoupling cap suggested in the datasheet of AP5100.
Can't seem to understand how one diode can do the job.

 

hi anish,
Is this what you are trying to do.?
E
Better image.!

Yes about rite.

 

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.?
E

 

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.?
E

Disconnected from the pin. Vin = 0V. Either way Vin should not reach or power the CAP+ rail.

 

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
E

 

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
E

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,
Two diodes should cover both a 12V disconnection and a loss of the power source for the 12V PSU.

E

 

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.

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?

Is CAP a capacitor or some other type supply, e.g, battery? My comments are from the assumption it is a capacitor.

 

Scenario 1
When Vin = 0 but still with cable connected and CAP = 8V. The Vout should source power from the CAP.

Scenario 2
When Vin = 12V and CAP = 8V. Based on the schematic in Post# 9, Vout sourcing power from either would be fine.
CAP is a capacitor bank.

Is it true that in scenario 2 the combined voltage at Vcc would be 20V.

 

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?

This is to prevent the 8V from powering the 12V rail when the cable is left connected tho not powered.

 

If your capacitor will tolerate 8V but not 12V minus diode drop, you are cutting the margin of safety pretty thin. Why do you want 2 supplies, one to charge the capacitor and the other to run the circuit?

 

Is it true that in scenario 2 the combined voltage at Vcc would be 20V.

hi,
No.
E