Could anybody guide me basics calculation for N and P channel depletion/enhancement mode MOSFET.

It's difficult to do calculation for P channels.

Please help me out .

 

Calculation of what exactly?

 

They are pretty much identical. If you are fine with n channels think about switching ground and Vcc for the P channel circuit and then replace the p channel with an N channel (keeping gate source and drain connections). Once you do a few times it will feel like second nature.

 

Actually I wanted to know.

1. How to decide Vgs for N / P channel and at that time what should be the drain voltage

2. What is the perfect condition a mosfet to get turn ON.

Let's say a P channel Mosfet FDS8858CZ has -2.1V Vth .

Now how to bias MOSFET perfectly if I want to switch VCC to load. Let's say 100 mA load connected at low side


How do we calculate gate voltage Vgs and what should be the drain voltage... !!!

Please any body will do calculation for me .

 

Comment still awited ।।।।

 

If you are switching a load then basically you don't calculate Vgs; you just switch the FET hard on. If the p-FET is specified as a "logic-level" type then Vgs = about -5V should do that. If it's not a logic-level type then Vgs = about -10V is the usual value.

 

If you are switching a load then basically you don't calculate Vgs; you just switch the FET hard on. If the p-FET is specified as a "logic-level" type then Vgs = about -5V should do that. If it's not a logic-level type then Vgs = about -10V is the usual value.

Thank you so much for your reply ।।।
Still I do not understand the fundamentals.

I know if switching through p channel VGS is negative but far away from real fact .

Let's take a examples! !!

I have 3.3V logic signal from microcontroller and that is connected to gate terminal of MOSFET and a 3.3V vcc is connected to source of FET.

Would P CH mosfet turn ON.
Microcontroller gives only 0V and 3.3V so where is that negative Vgs comes into the picture....!!!

 

where is that negative Vgs comes into the picture.

Vgs means "V(gate) minus V(source)". If the p-FET source is at 3.3V (Vcc) and the gate is at 0V then Vgs = V(gate) - V(source) = -3.3V.
To ensure the FET is switched fully on you would choose one with a rated Vgs(threshold) of around -1V.

 

Vgs means "V(gate) minus V(source)". If the p-FET source is at 3.3V (Vcc) and the gate is at 0V then Vgs = V(gate) - V(source) = -3.3V.
To ensure the FET is switched fully on you would choose one with a rated Vgs(threshold) of around -1V.

Thank you so much for your kind reply and time.

Now I think I have understood the logic to drive N and P channel MOSFET.

Vgs is greater (Vgs-Vt) is this relationship is trure for both N and P Channel ।।

What about Drain/Source voltage.

Is there any condition for drain/Source voltage for N/P channel plays any role to turn ON the MOS.

A bit of further more explanations will clear my doubt .

Thanks again. !!!

 

What about Drain/Source voltage

When the FET is turned fully on the voltage drop from drain to source will be small, determined by the product of current and the drain-to-source resistance Rds(on), which is a figure you'll find in the FET's datasheet. The lower the value of Rds(on) the better.

 

@Alec_t you said that vgs is simply the difference of the voltage at the gate and the source is this true for both p and n channel mosfet?

for my first example im going to use a p channel mosfet
assume that:
Vs= 15
Vg = 10
so my Vgs is simply = -5


and for my second example im going to use a n channel mosfet
assume that:
Vs= 0 (since source voltage is always at the ground level)
Vg = 10
so my Vgs is simply = 10


is this correct??? pls correct me if i have some error....TIA

 

You are correct.

 

Welcome to AAC!

vgs is simply the difference of the voltage at the gate and the source

By definition, Vgs is Vg - Vs.

 

Vs= 0 (since source voltage is always at the ground level)

That is definitely not true, there is no universal rule that says that an N-fet has to have its source connected to ground. Otherwise you have it correct.