I read in a datasheet of mosfet that its Rds(on) 1.35ohm @ VGS=10V, IDS = 2.7A.

When in ON state, is IDS = VDS / RDS(on) ?

Thank you.

Thanks.

Correct me please if i'm wrong, in saturation state, Rds is very large, then how come IDS is maximal (for constant VGS) in sturation state?

In saturation state RDS is very small.

But in stauration state, Rds = VA / IDS, isnt it?
And VA (Early Effect Voltage) is very large, therefore Rds is very large also.

Where am i wrong?

The numbers from a data sheet do not mean the FET will all wise work like that.

In your example from the first post: 10 volts is applies from G-S. A current source (2.7A) is applies to the D. The FET looks like a 1.35 ohm resistor. The voltage D-S is 3.6 volts, under these conditions.

Appling 10 volts to the gate will not cause 2.7 amps. It will cause the FET to turn on and look much like a 1.35 volt resistor. The current is a function of what is on the D.

Early Effect Voltage???? Are you talking about the first nano-seconds of turn on? Saturation is talking about being on (DC) not AC.

I know the early effect voltage (VA) is normally applied with BJT's. I don't know if it is applicable with FET's. It is actually the variation of the width of the base as the reverse voltage between collector base varies. Typically, the gate of the FET's as virtually isolated, high impedance input.

Originally Posted by Roff

A search finds several references to the Early effect in MOSFETs. It's a different effect, but with similar results.

I see, I try to dig on that for more details.

similar to the way the VA describes the linear dependency of IC on VCE in saturation region in BJT, it also describes the linear dependency of IDS on VDS in saturation region in MOSFET.