1
\$\begingroup\$

In a MOSFET if VGS is kept constant then VDS is increased till entering a saturation mode, a pinch-off accrue near the drain. This indicates that the E field between the gate and the bulk is not evenly distributed due the increasing positive potential at and near the drain region. This phenomena will also affect the resistivity of the channel. Does this means that the resistivity near the drain have the highest value? If so the dissipated power is mostly generated near the drain?

\$\endgroup\$
1
  • \$\begingroup\$ Yes, I believe it does. \$\endgroup\$ – Andy aka Apr 3 at 9:52
0
\$\begingroup\$

Interesting question.

I simulated for a year electronic semiconductor devices.


Channel resistivity is a 2D function: ρ(x,y).

ρ(x,y) changes as VDS change.


To answer your question:

Does this mean that the resistivity near the drain have the highest value?

Yes, it does.

Is the dissipated power mostly generated near the drain?

Yes, it is.


The external circuit will see the overall channel resistance R which is the 2D integral of ρ(x,y).

You increase the channel width W, you reduce R.

The electric field vector E(x,y) and the current density J(x,y) are 2D functions as well.

\$\endgroup\$
5
  • \$\begingroup\$ In a saturated FET, the depletion region between the drain and the channel (the pinch-off region) is not considered as a resistive region -- it is non-conductive -- but carriers are flowing in it (similar to injected carriers across the base-collector junction of a BJT). This essentially is why the drain current DOESN'T increase with drain voltage -- that current is determined by what is generated by the source and channel at that end. You are correct that most power is at the drain end. FET channel lengths are very small, and this heat spreads throughout the device quickly. \$\endgroup\$ – jp314 Apr 3 at 18:29
  • \$\begingroup\$ Thank you very much for your answer. That's what I thought. But what confused me was an answer to a similar question stating the opposite (lower pinch resistivity - electronics.stackexchange.com/questions/77198/…). How someone can confirm this theory (by simulation or by measurement)? Is there any publication or a reference to explains it? Am I missing some quantum phenomenon here? A reference states that the velocity of electrons increases as they pass through the pinch-off region to the drain. \$\endgroup\$ – Michael Apr 5 at 4:33
  • \$\begingroup\$ If we further increase VDS, the Pinch-off becomes more severe, but the current remains constant, and according to the following relation Pd should increases (Pd = (VDS ^ 2) / Ron), till the channel fail (at the site with the highest resistivity). \$\endgroup\$ – Michael Apr 5 at 4:33
  • \$\begingroup\$ Michael: all devices simulators solve the so called "Drift-diffusion equations" starting from the boundary and initial conditions and from semiconductor data. In this context E(x,y) and J(x,y) are the unknowns. Once the simulator has finished (usually after 12 hours or more) you can evaluate current I at a certain section, plot or evaluate resistivity etc. If you change VDS, you have to run another simulation. The pinch-off condition or inversion condition can be seen after the simulation has ended. \$\endgroup\$ – Enrico Migliore Apr 5 at 7:26
  • \$\begingroup\$ Back in 1995, devices simulators like IBM's DAMOCLES, did not use quantum equations. They solved the drift-diffusion equations. \$\endgroup\$ – Enrico Migliore Apr 5 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.