How to reduce Off-time delay in switch over condition using P-channel MOSFET

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nansri84

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Hi,

We are using a P-channel MOSFET (FDC642P) as a switch from main DC supply to battery supply. There is a time delay about 1sec, in which he in-built S-RAM looses its content. Any suggestion to reduce the time delay?? Your help is highly appreciated.
Circuit attached.






 

What is Vbat? And can I assume that 10K goes to ground? And you realize that if Vbat is less than about 3.6 volts, it's going to be sucking current through the parasitic diode of the MOSFET, right? And the load is connected to the drain?
 

barry said:
What is Vbat? And can I assume that 10K goes to ground? And you realize that if Vbat is less than about 3.6 volts, it's going to be sucking current through the parasitic diode of the MOSFET, right? And the load is connected to the drain?
Click to expand...

Thanks Barry.

Main DC supply passes thru the circuit thru 10K.
Vbat is the battery supply to which the circuit to be switch over when the main DC supply gets cut-off.
The Load is connected to Drain.
 

I agree with barry that the battery is potentially charged from 4.2V node. Depending on the battery type this might conflict with safety regulations.

Presumed that you don't need to cut-off the battery supply, you'll usually flip drain and source terminals.

But for a full assessment, battery type, main voltage supply range and minimal RAM supply must be known. As a generally comment, for battery and RTC backup only, I would try to come along with a passive diode switch.
 

The battery type is 3.7V/890mAh Li-on battery. The main supply is max 36V and the operating voltage is 4.2V.
 

nansri84 said:
The battery type is 3.7V/890mAh Li-on battery. The main supply is max 36V and the operating voltage is 4.2V.
Click to expand...
Now I'm totally confused. First of all, that is not the schematic symbol for a P-channel MOSFET; that's the schematic symbol for a box with 6 terminals.

Are you saying your 36V supply is connected to the bottom of that 10K? That makes no sense. Where does the 4.2V come from?
 

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