1
\$\begingroup\$

Many books (Chenming-Hu Ch.6 Page 15, Neamen Page 413) use the term 'channel voltage' \$V(y)\$ to mean 'the potential in the inverted channel at a point \$y\$ distance away from the source along the channel length'.

First part: What do they mean by 'potential in the channel'? Is it the same as the potential at a point inside a resistor (because the channel in a MOSFET is similar to a resistor)?

If so, is this potential constant along the direction perpendicular to the current flow, as in the resistor?

If it is constant, then how is 'potential in the channel' any different from the surface potential (i.e., the potential at the semiconductor-oxide interface \$\psi_{s}(x)\$, also a function of \$x\$?

Second part: When a drain voltage \$V_{ds}\$ is applied (the source and the body are grounded and a channel has been formed), the band diagram of the MOSFET along the channel is expected to show Fermi level splitting because the system is out of equilibrium.

Relevant image :MOSFET when channel is formed and drain voltage is applied

What I don't understand is the band diagram in the \$x\$-direction at a point \$y\$ distance away from the source and in the channel...

Band diagram at y in the channel

The x-axis runs from left to right in the above image.

The image shows that the Fermi levels split inside the body of the semiconductor and the amount of the splitting is \$eV_{ds}\$ at the drain, and by extension, \$eV(y)\$ inside the channel. Why does this happen? Is there any connection between the 'potential in the channel' interpretation of \$V(y)\$ and this splitting?

Legend for the band diagrams :

  • Black dotted lines - intrinsic level
  • Red dotted line - Quasi Fermi level for holes
  • Red solid line - Quasi Fermi level for electrons
  • Black solid lines - Bottom of conduction and top of valence bands
\$\endgroup\$
  • \$\begingroup\$ Does this question belong here or over at Physics SE? \$\endgroup\$ – transistor Nov 17 '15 at 19:04
  • \$\begingroup\$ In theory there should be enough competent people here to answer it. \$\endgroup\$ – Fizz Nov 17 '15 at 19:05

Your Answer

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

Browse other questions tagged or ask your own question.