Does the MOSFET in the Schematic will work

[Bhanu_Muppalla]

I using the MOSFET IRFB3004PBF switching. 56.2V is a source voltage coming from the charger and 48 is the nominal voltage of the batter.
switch S1 is for switching Vgs.
Vgs=5 v.
Rgs=10k ohms (connected to the gate terminal and battery ground).
battery is 13s3p Nominal=48v Peak=54.6v

Will MOSFET works if the switch is on? Please suggest edits to make the circuit if it doesn't.
Thanks in advance.

 

[KlausST]

Hi,

It will not work.

switch S1 is for switching Vgs.
Vgs=5 v.

Vgs = voltage between Gate and Source.

Your 5V is between GND and gate (when switch is closed)
but your Source potietial is not the same as GND potential.
Just use a voltmeter and connect it to Gate and Source and see what it shows.

VGS will never be 5V.

Klaus

 

[cupoftea]

yes
IRF3004

https://www.infineon.com/dgdl/Infineon-IRFS3004-DataSheet-v01_01-EN.pdf?fileId=5546d462533600a4015356363a642149

...You may not do that.....you will blow the fet...you cant just short a battery to a higher voltage.

You would need a resistor in there to drop voltage and limit current......you can use an isolated psu module and "ground" it to the battery pos voltage...then you can switch the fet on, with a bit more circuitry.

To drive an NFET, you need VGS to be at least 5V.....pref 10V...."VGS" means VG - VS

 

[betwixt]

I agree with Klaus on this one.
With the switch open you exceed the maximum rating of VGS which is +/-20V, you have 48V.
When the switch is closed you still exceed it with 43V.

It would work if the 5V source had it's negative side on the source pin, R1 was between gate and source and you used a MOSFET with maximum voltage rating higher than 40V, it should be at the very least rated to 56.2V and preferably somewhat higher to allow a safety margin.

Brian.