Check the attached jpeg file (schematic n description) regarding circuit of IRF6216pbf MOSFET.
IRF6216pbf is a HEXFET mosfet from International Rectifiers.
I am using it to switch on/off the PIN diode based RF channel.
TO switch OFF the channel, PIN diode should be biased with -75V and to switch it ON, PIN diode should be biased with +1.84V.
AS shown in figure, I am getting +1.84V at the Drain terminals only when I left the Vgs (gate terminal) open .
Kindly advise me on this circuit as well as on my requirement.
Observation: 100 ohm POT(R84) in the source terminal needs to be removed & instead of +1.8V ,minimum +5V to be given at Source terminal.
Now , if gate is at +5V, then -75V will be available at drain (or R86) & if gate is at 0V, then +5V (source terminal voltage) will be available at drain (or R86).
But I am unable to understand that How presence of a POT (R84) is affecting it?
The VGS(th) (Gate Threshold Voltage) specs are -5.0v max with the following conditions VDS = VGS, ID = -250μA
Useable Vgs start from about -7v but the needed Vgs will depend on your load current requirement.
You should start from Vgs=-7v so 1.84v + 7v = 8.84v (it's better if you go higher, up to about 15v for lower Rds-ON)
To turn the mosfet off you can just ground the gate (Vgs=+1.84)