Thank you so much for your reply ।।।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.
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.where is that negative Vgs comes into the picture.
Thank you so much for your kind reply and time.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.
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.What about Drain/Source voltage
|Thread starter||Similar threads||Forum||Replies||Date|
|D||Basic OpAmp question||PCB Layout , EDA & Simulations||11|
|S||A very basic question about transformers and schematics.||Power Electronics||9|
|Y||basic avoing grating lobe formula logic||Wireless & RF Design||0|
|[VIDEO] Basic RLC Circuits (1944, Army)||Off-Topic||0|
|W||Basic impedance calculation with LM741 op-amp buffer circuit||Homework Help||8|