[SOLVED] Basic Mosfet Question

[SeriousTyro]

In class today we looked at two extremes of a NMOS.

Vgs=0 => Id = 0
Vgs= large => Vds= 0

I don't understand how Vds=0 when Vgs=large, assuming the circuit doesn't breakdown when Vgs is large?

Thanks,
Tyro

 

[D.A.(Tony)Stewart]

NMOS FET is a voltage control switch
When Vgs=0 the drain is pinched off so no current. high impedance and Drain may pull up to Vcc, say if a resistor is connected Drain to V+.

When Vgs is large the Drain switch to Source is low resistance so Vds=0, even if pull up. R from Drain to V+
Rds is often << 1 ohm. :roll: when on.

 

[SeriousTyro]

Ahh I see it now.
When Vgs= 0, the induced inversion layer will contain a high amount of electron concentration thus it can been seen as a wire, ie R is small. Thus Vd=Vs -> Vds=0.

However I don't agree when you mentioned it was pinched off. When Vgs = 0, there is no inversion layer thus there is no channel for electrons to flow.
Pinch off refers to saturation when the induced channel becomes shorter then the gate length where there is a larger concentration of electrons at the source.
Electrons still flow in pinch off because of the electric field due to Vds.

 

[js]

Use this animation:

**broken link removed**

 

[D.A.(Tony)Stewart]

almost
when Vgs=0 the n-type FET switch is open.

from wiki.." In the case of an n-type switch, the body is connected to the most negative supply (usually GND) and the gate is used as the switch control. Whenever the gate voltage exceeds the source voltage by at least a threshold voltage, the MOSFET conducts. The higher the voltage, the more the MOSFET can conduct. An N-MOS switch passes all voltages less than Vgate–Vtn. When the switch is conducting, it typically operates in the linear (or ohmic) mode of operation, since the source and drain voltages will typically be nearly equal."