Why can current IDS go through the cutoff area when VDS>VGS-VTH? (MOS saturation)

[yann_sun]

There is no continuous channel, why can current IDS go through the cutoff area when VDS>VGS-VTH?

 

[Weiming]

mos, saturation

When VDS increased larger than VDSsat, its channel is pinched off, and there is the region of depletion between drain-channel. However there is also electric field(E) so the charge can pass the depletion region pushed by E.[/quote]

 

[shaikhsarfraz]

nmos saturation pinch off

Hi,
As you have said Vds>Vgs-Vt, so there is no channel, but the field is so strong that the drain region extends to the source, and under the influence of drain voltage the carriers cros the channel and are collected at the drain.

Thanks
Shaikh Sarfraz

 

[Hommmer]

pmos saturation

on a more physicall approach you can say that the electrons from the current flow, actually "jump" from the doped region to the channel, 'pushed' by the electric field (Vds) applied between the n+ doped regions.

 

[vijay_nag]

MOS saturation

the electrons when reach the pinch-off region experience a strong positive charge at the drain and are pulled across it.

 

[invent]

Re: MOS saturation

Hi,

There is one book I read: "Analog Integrated Circuit Design - Johns & Martin (1997)"

which says that there is actually still a small channel exits. The electrons traveling through is therefore velocity saturated.

But i still think it is because of the large potential difference between the pinch-off point with the drain. (voltage at pinch-off point is VGS - VT)