The Mystery of Fermi Energy in Accumulation-Mode FETs

[Batmaniac]

Greetings,

I've been doing some literature research and can't seem to find any answers to my question.

In an accumulation-mode FET (i.e. only relevant capacitance is that of the gate dielectric), how does one translate the applied gate-voltage to the Fermi energy of the active region?

Say we can measure the threshold voltage (approximately the conduction band in an n-channel FET?), and know the geometric capacitance of our gate dielectric, and the work function of our gate metal, there should be some way to calculate the shift in fermi energy (i.e. where have I lifted the Fermi energy to relative to the conduction band) caused by the gate-effect, accounting for the "shielding" effect of the gate dielectric.

If anyone has any general thoughts/things to look-up/paper/references/anything, I'd be grateful!

Thanks!

 

[eljose79]

Hello,

Thank you for your question. The relationship between the applied gate voltage and the Fermi energy of the active region in an accumulation-mode FET can be calculated using the following equation:

EF = (Vg - Vt) + (Cg/Cox)*(φm - EF)

Where EF is the Fermi energy, Vg is the applied gate voltage, Vt is the threshold voltage, Cg is the gate capacitance, Cox is the oxide capacitance, and φm is the work function of the gate metal. This equation takes into account the "shielding" effect of the gate dielectric and allows you to calculate the shift in Fermi energy caused by the gate effect.

I would recommend looking into papers and references on the theory of FETs and semiconductor physics for a more in-depth understanding of this relationship. Additionally, you may find it helpful to consult with colleagues or experts in the field for further insights and discussions.

Best of luck in your research!