[SOLVED] fermi level in doped semiconductors

Sreenath N

i was going through the basics of energy bands in metals and
semiconductors.i have one doubt.
how do interpret the following,
in p type semiconductor the fermi level is close to valance band and
in n type semi conductor the fermi level is close to conduction band.
how can we explain this?

Rgds,
Sreenath.

William R. Frensley

Sreenath N wrote:
> i was going through the basics of energy bands in metals and
> semiconductors.i have one doubt.
> how do interpret the following,
> in p type semiconductor the fermi level is close to valance band and
> in n type semi conductor the fermi level is close to conduction band.
> how can we explain this?
>
The Fermi level moves to wherever it needs to be to assure that the overall
system is charge-neutral. In an n-type semiconductor, we introduce fixed
positive charges (donors), which must be balanced by mobile negative charges
(electrons). The excess electrons must reside in the conduction bands,
because the valence bands are full. To have excess electrons in the
conduction band, the Fermi level (electrochemical potential for electrons)
must lie near the conduction band. A similar argument can be made for
p-type doping. If you have Java installed on your computer, run the program
that is invoked by entering this in your web browser:
http://www.utdallas.edu/~frensley/Un.../fermil01.jnlp
You can adjust the doping levels and see how the Fermi level changes.

- Bill Frensley

Peritas

Sreenath N wrote:
> i was going through the basics of energy bands in metals and
> semiconductors.i have one doubt.
> how do interpret the following,
> in p type semiconductor the fermi level is close to valance band and
> in n type semi conductor the fermi level is close to conduction band.
> how can we explain this?

Take silicon for example. Boron acts as an acceptor, where phosphorous
is a donor in silicon. So p-type silicon would be doped with boron (or
some other suitable acceptor), and n-type silicon would be doped with
phosphorous.

Now look at the relative position of these elements on the periodic
table. Boron has one less valence electron than silicon, and
phosphorous has one extra. So, when you put phosphorous in a silicon
lattice it will give up (donate) it's extra electron. Boron, however,
has room to accept an electron.

Given that the valence band is fully occupied in the absence of
external excitation, it would stand to reason that the donated electron
goes to the conduction band. So if there are a bunch of donated
electrons in the conduction band, then naturally the fermi level would
be nearer to the conduction band. ...and vice versa for boron.

I've glossed over many details, but I hope this approach helps.

Regards

Igor Khavkine

Sreenath N wrote:
> i was going through the basics of energy bands in metals and
> semiconductors.i have one doubt.
> how do interpret the following,
> in p type semiconductor the fermi level is close to valance band and
> in n type semi conductor the fermi level is close to conduction band.
> how can we explain this?

Different Fermi levels are in one-to-one correspondance with ground
states of differents total amounts of charge. In doped semi-conductors,
the donor and acceptor impurities also contribute to the total charge,
although the electrons sittting (or absent) on the impurities in the
ground state don't count toward the total number of electrons in the
valence band.

In n-type semiconductors, impurities carry excess charge that can move
into the conduction band at higher temperatures. This means that it
requires less energy to excite single electrons from the ground state
into the conduction band. Theoretically, the minimum energy of single
excitations is measured from the Fermi level to the bottom of the
conduction band, in which case it is consistent that the Fermi level is
closer to the conduction band when the sample is n-doped.

Similarly, when the sample is p-doped, it becomes easier for some
electrons to escape from the valance and become bound in the accepting
impurities. In other words, the energy needed to put holes in the
valance band is now lower, which effectively brings the Fermi level
closer to that band.

Hope this helps.

Igor