1
\$\begingroup\$

The conductivity of a semiconductor is given as (if extrinsic):

\$\sigma = qnu_e + qpu_h\$

where \$\sigma\$ is the conductivity, \$u_e\$ and \$u_h\$ are the electron and hole mobilities.

But does this equation always hold true? Is there any circumstances when conductivity decreases if a dopant is added (whether p-type of n-type?). What happens if a dopant is added which has a large binding energy to the lattice. Surely that'll decrease the conductivity!?

I am only a second year Engineering so I have not gone into that much detail into solid state physics/electronics.

\$\endgroup\$
4
\$\begingroup\$

Is there any circumstances when conductivity decreases if a dopant is added (whether p-type of n-type?).

Yes. Consider compensation doping.

Suppose silicon is initially doped with donors, making it n-type.

If you then begin to dope with acceptors, this will not immediately create a p-type material. Instead, the acceptors will provide locations for free electrons (provided by the initial donor dopant) to be captured, reducing the electron concentration, and thus the conductivity of the material.

If you can match the acceptor concentration very closely to the donor concentration, you can produce material that is very near intrinsic in its behavior. (Matching the concentrations is difficult in practice because the different dopants have different diffusion coefficients, etc.)

But does this equation always hold true? [\$\sigma = qn\mu_e + qp\mu_h\$]

This is still true. This just says the conductivity is proportional to the carrier concentration. It doesn't say anything about what dopants were added to achieve those carrier concentrations.

What happens if a dopant is added which has a large binding energy to the lattice. Surely that'll decrease the conductivity!?

Here, I'm not clear what you're asking about.

The binding energy of the dopant to the lattice doesn't have much to do with its contribution of carriers.

If you're talking about a dopant that requires a large energy to ionize, it has little effect because that means the captured state energy levels are "deeper" in the band gap, so less likely to be occupied, than for a dopant requiring little energy to ionize.

\$\endgroup\$
  • \$\begingroup\$ > If you then begin to dope with acceptors, this will not immediately create a p-type material [..] Wow I didn't think of that, cheers. \$\endgroup\$ – Lewis May 20 '17 at 21:40
  • 1
    \$\begingroup\$ FWIW, compensation is how many devices are formed. You have one dopant with low(ish) concentration but allowed to diffuse deeply into the material. Then an opposite-type dopant with stronger concentration but not so deeply diffused. Et voila, a pn-junction is born. \$\endgroup\$ – The Photon May 20 '17 at 21:44
  • \$\begingroup\$ Ohh I did not know that. I just thought the n-type regions and p-type regions were just created epitaxially or in bulk in the factory, didn't realise it's more complex than that. \$\endgroup\$ – Lewis May 20 '17 at 22:05
  • \$\begingroup\$ With ion implantation of dopants, you can dial up the acceleration voltage, or dial down. Then, from what I've heard the fab jocks state "thermal annealing will drive in the dopants". Perhaps the combined minds here can address this "drive in"? \$\endgroup\$ – analogsystemsrf May 21 '17 at 16:41
2
\$\begingroup\$

Doping does not always increase the conductivity. Suppose that we have an intrinsic sample with n = p but the hole drift mobility is smaller. If we dope the material very slightly with p-type then p > n. However, this would decrease the conductivity because it would create more holes with lower mobility at the expense of electrons with higher mobility. Obviously with further doping p increases sufficiently to result in the conductivity increasing with the extent of doping.

\$\endgroup\$
  • \$\begingroup\$ +1 Because you are correct, but I just wanted to point out that the dopant concentration you would need to be able to see this effect experimentally is quite low, not much more than the intrinsic concentration. (Unless you had some materials with vastly different electron and hole mobilities) \$\endgroup\$ – Matt Oct 5 '17 at 2:07

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.