Homework Help: Doped Si carrier concentration

1. Apr 17, 2010

Master J

Hope you can give me a hand here guys.
I'm finding the whole carrier concentration thing a bit confusing!!!

Si doped with say 10^17 P (/cm3). At 300K, calculate n, p, E_F (Fermi energy).

Now I know the relations such as n=(N_c)exp[(E_c-E_F)/kT], N_c given

I just cant figure out where to start. If we assume full ionization, then should n just be the doner concentration?? But then how do I find p?

I know, from charge neutrality, that n + Nd = p + Na (assuming full ionization).

2. Apr 17, 2010

thebigstar25

"Si doped with say 10^17 P (/cm3)" , it is doped with what exactly?

3. Apr 17, 2010

Master J

Maybe it wasnt clear sorry. The P meant Phosphorous, ie. electron doner.

4. Apr 17, 2010

thebigstar25

i thought p=per .. now it is clear ..

the first step when you assumed full ionization was a good one .. and the reason of that is , because the ionization energy of Phosphorous in Si is 45 mev, which is so low and for that reason we can say that we have full ionization situation ..

the statment "If we assume full ionization, then should n just be the doner concentration" is not correct , the full ionization just leads you to the fact that Nd+ = Nd (you get my point?) ..

another good step you did, is using the neutrality equation n + Nd = p + Na ..

1) since Si is not doped with acceptor what does that tell you?
2)you can assume that n (approximately) = Nd in a case where Nd is much greater than ni (the intrinsic concentration of Si, i think this one is either given or you have the values Nc, Nv, and Eg) ..

3) following 1 & 2 you should get n, then get p ..

4)then Ef should be obtained easily ..

5. Apr 17, 2010

Master J

If its not doped with acceptors, then Na = 0

So we have: n + Nd = p

Yes I have Nc, Nv and Eg.

p is just the intrinsic value, since no acceptors were added??

If we assume full ionization, then is n just the intrinsic electron conc. PLUS the doner density(one electron per doner)??

6. Apr 17, 2010

thebigstar25

ok good Na = 0 since Si is not doped with any acceptor ..

is the Fermi level the same for both cases when Si is doped (with acceptor or donor or both) and when it is not doped at all ? >> ( answer this question while having a look at the expression used to find p) ..

7. Apr 17, 2010

Master J

Assuming my answer to find p is right?

ni^2 = Nc.Nv.exp[-Eg/kT] .... gives me intrinsic conc. squared....hence p, which is just the intrinsic conc?

Well, the Fermi level moves towards either the valence or conduction band, depending upon which is the amjority carrier, according to:

E_F = (Ec +Ev)/2 -(1/2).ln(p/n) + (3/4).ln(Mp/Mn)

...

8. Apr 17, 2010

Master J

Ok so here's an attempt!!!! :-)

(ni)^2 = Nc.Nv.exp(-Eg/kT) = 2.66*10^19

n = ni + Nd = 10^17 ( so yea, ni doesn't really matter in comparison).

Therefore,

p = (ni)^2 / n = (ni)^2 / Nd = 266

Now 266 holes / cm^3 seems REALLY small!!! Is that right???

9. Apr 17, 2010

thebigstar25

the neutrality equation is supposed to be n+Na=p+Nd (correct me if im wrong) ...

When i suggested that in case Nd much greater than ni then Nd = n .. I didnt mean to push you to say n=ni+Nd?.. If that confused you .. Just plug p = ni^2/n in the neutrality equation and find n .. When you find n , then you can easily find p ..

>> i believe that your value for n and p is correct .. my comment was all about why u wrote n = ni + Nd ... You can repeat solving the question taking in consideration my last note and you will see that you will get the same answer..

10. Apr 18, 2010

thebigstar25

im so sorry i didnt pay attention to this post .. your answer was not right!!!

you can not assume that p is just intrinsic value !! .. you mentioned that the Fermi level would not be the same in case you dont have doping with a case where you have doping ..

I think that p = Nv exp[(E_F-E_v)/kT], so how it would be the same? if you have a second view please provide us with it ..

11. Apr 27, 2010

Master J

But isnt the intrinsic conc. squared always equal to np??

And the intrinsic conc. squared does not depend on E_F:

(n_i)^2 = Nv.Nc.exp[-Eg/kT]

So with that I can get p = (n_i)^2 / n .....where n is approx. = N_d

12. Apr 27, 2010

thebigstar25

yes ni^2 is always equal n*p .. What i meant is that in the case of doping p wont be the same as p when you have no doping, you get my point?, so thats why you have to calculate it with the relation ni^2=np .. Im sorry in case if i misunderstood what you said ..