[SOLVED] Measuring current without using shunt resistance

[KlausST]

Hi,

You know a FET has a gate and a source. And as I told you before V_gs is the voltage between these two pins.
Then please connect a voltmeter to these two pins as FvM told you, and see what happens.

It seems you are not reading our posts...Please just try...

Klaus

 

[farrukhtalib]

I read all your posts and try to learn. Please have a look at the attached fig and correct it if you find any ambiguity

https://obrazki.elektroda.pl/4005425400_1444026902.jpg

 

[KlausST]

Hi,

As you can see the MOSFET is switched OFF now. So at least one of your questions is answered.
***

To make it more visible for you, I recommend to use the third voltmeter as true V_gs voltmeter.
(Now you measure the voltage between gate and GND. This is not of interest.)
--> Measure the voltage between gate and source instead and you will better understand the circuit/datasheet.

***
After you have done this, you can read V_gs to be very near to zero.
But as soon as you press the button (to switch ON the FET) you will see 38V at V_gs. This is too much.
Therfore you need a resistor in series with your pushbutton lines. Use 15k and see what happens.

***
post your new schematic (using the "insert image" button).

Klaus

 

[farrukhtalib]

So what I learned from you is
1 One can put source power rail to gate of PMOS even if is > then V gs
2 One should subtract negative voltage from source power rail voltage to get the voltage with in the safe boundaries of V gs specified in data sheet. which is in this case is around 15 Volts

https://obrazki.elektroda.pl/8570859600_1444075241.jpg

 

[KlausST]

Hi,

Try to see it from the FET's view.
It has three connections, G, D, S.
It does not know what GND is.
It can only see V_gs, V_ds, V_gd.
Now to the currents.
DC gate current is considered to be zero, therefore I_d = I_s (to be exact: -I_s)

Klaus