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# FET theory for a simple switch

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#### tmd_63

##### Member level 2
Ok. Simple question.
What type of FET can be used to act as a switch.
It will have a battery supply on one side that will have between 15V and it's on load terminal voltage of a minimum of 11.5V.
The output side could draw up to 4A.
The control will be such that a 15V turns the switch off and a low of 10V or less will turn the FET on.

If you have a part number would be good.

simple answer .. pretty much any power MOSFET can be used to do the job. IRF540 ?

which signal to connect to which pin?
control to gate obviously, but does the battery connect to source or drain?

What you are describing is "high side switching", as opposed to "low side switching" where the switch is between the load and ground. High side switching is generally more difficult because the gate voltage is with respect to the souce pin, and the souce is not at ground. But your control voltage is with respect to ground. So some translation is required. One common solution is to use a P-channel enhancement-mode FET as the switch with the source connected to the battery. Then use a voltage translation circuit (like a PNP transistor) to invert the control signal and translate it so it is also with respect to the battery. Another solution is to use an N-channel enhancement mode FET as the switch and drive the gate with a specialized charge pump gate driver circuit that translates the control signal to be with repect to the FET source. For more information google "high side switching".

This is a difficult set-up to understand.
I need to be able to have a seamless changeover from mains supply to battery. Using a 12V battery but without the supply output going above 12V. The battery needs 13.8V to 14.4V to charge properly.
So I need to be able to disconnect the battery while it is being charged and to get the battery to take over the output when the mains is disconnected or drops out without the output dropping below 11.8V.
I did see a simplified method using a special FET such that the charging current uses the protection diode to allow current flow into the battery and the dropping charging voltage then causes the FET to change and allow battery power into the output circuit. But I cant get my head around how it does it.
Can anyone provide a circuit and explanation?