# extrinsic minority carriers

#### mentaaal

Joined Oct 17, 2005
451
I have one more question which is bothering me! In an instrinsic semiconductor, ni = pi and NiPi = Ni^{2} which is fine, makes sence and all but what i dont really understand is for an extrinsic semiconductor, this rule still applies, say for example you wanted to find the amound of minority carriers in an N type semiconductor which would obviously be holes, the formula says that the Na (number of holes) = Ni squared / Nd (number of donors)

What i am saying is why is it that the number of holes multiplied by the number of electrons is equal to the number of holes multiplied by the number of electrons in an intrinsic semiconductor? I mean why couldnt i just as easily say the number of holes PLUS the numer of electrons in an extrinsic semiconductor = to the number of holes PLUS the number of electrons in an intrinsic semiconductor.

I have already asked this question to my lecturer and I didnt realy understand anything he said about it. He seems to think that this should be obvious but i am sorry i am just not seeing it.

#### Papabravo

Joined Feb 24, 2006
12,405
In an intrinsic semiconductor the production of hole electron pairs is a function of temperature as has already been noted

Rich (BB code):
In an extrinsic semiconductor, n-type for example, n >> p
and Nd is the number of donors and Na is the number of acceptors

Nd + p = Na + n

since Na is nearly 0 in an n-type
we are left with

n ≈ Nd

pn = ni^2 / Nd
similarly for a p-type
Rich (BB code):
np*pp = ni^2, pp ≈ Na, np = ni^2 / Na

For the extrinsic semiconductor the concentration of holes or electrons is according to the number of donors or acceptors and is not as dependent on temperature as in the intrinsic semiconductor. You'll have to admit tha concentration is a far more desireable thing to try to control instead of temperature.

#### mentaaal

Joined Oct 17, 2005
451
In an intrinsic semiconductor the production of hole electron pairs is a function of temperature as has already been noted

Rich (BB code):
In an extrinsic semiconductor, n-type for example, n >> p
and Nd is the number of donors and Na is the number of acceptors

Nd + p = Na + n

since Na is nearly 0 in an n-type
we are left with

n ≈ Nd

pn = ni^2 / Nd
similarly for a p-type
Rich (BB code):
np*pp = ni^2, pp ≈ Na, np = ni^2 / Na

For the extrinsic semiconductor the concentration of holes or electrons is according to the number of donors or acceptors and is not as dependent on temperature as in the intrinsic semiconductor. You'll have to admit tha concentration is a far more desireable thing to try to control instead of temperature.
In the equation: Nd + p = Na + n, what does the p and n stand for? I know what Nd and Na stand for as you clearrly stated but dont follow for p and n in this context...

I agree what you are saying about how extrinsic semiconductors dont respond as much to temperature, this makes sence but still dont see why the extrinsic electrons and holes must adhere to the intrinsic multiplicative rule..

#### Papabravo

Joined Feb 24, 2006
12,405
n is the electron concentration and p is hole concentration. A doped semiconductor is like two things superimposed on each other. the doping doesn't make the intrinsic semiconductor go away.

If I could write conveniently with subscripts it might have been clearer.
n-sub-n would be the electron concentration in the doped n-type semiconductor. while n-sub-p would be the electron concentration in a p-type semiconductor. In your original post you used ni and pi which are rendered in books as n-sub-i and p-sub-i.

The reason they must follow the rule is because....ta-da, the material must be electrically NEUTRAL! No net charge -- anywhere.

Sorry for the confusion

#### mentaaal

Joined Oct 17, 2005
451
Ok i undestand your equations about how n is rougly equal to Nd but i dont understand why in an N type, pn = ni^2/Nd. why must pn equal to this??

Is there some sort of derivation that leads to this result? Everthing leading to this i understand, its just the dammed minority carriers equalling the above formula that is freaking me out!

#### Papabravo

Joined Feb 24, 2006
12,405
The derivation I'm looking at is from Millman, J., and Halkias, C., Electronic Devices and Circuits, McGraw Hill, 1967, p. 105.

I'm sure other authors have similar treatments. This must be one of those things you just have to look at repeatedly. Maybe someone else can offer a metaphor that is more understandable.

The equation in question for n-type semiconductors may be confusing because there is a difference between p*n with no subscripts and p-sub-n written pn.

Lets try again
Rich (BB code):
n*p = (n-sub-i)^2

Nd + p = Na + n   {no subscripts here}

n >> p  {n is very much greater than p}

Na = 0 for an n-type material

n ≈ Nd

p-sub-n = (n-sub-i)^2 / Nd
p-sub-n is the concentration of holes in the n-type material

The last equation is the same equation as the first equation except the symbols have been changed and subscripts added. The derivation is what it is and you either understand it or you don't.

#### mentaaal

Joined Oct 17, 2005
451
The derivation I'm looking at is from Millman, J., and Halkias, C., Electronic Devices and Circuits, McGraw Hill, 1967, p. 105.

I'm sure other authors have similar treatments. This must be one of those things you just have to look at repeatedly. Maybe someone else can offer a metaphor that is more understandable.

The equation in question for n-type semiconductors may be confusing because there is a difference between p*n with no subscripts and p-sub-n written pn.

Lets try again
Rich (BB code):
n*p = (n-sub-i)^2

Nd + p = Na + n   {no subscripts here}

n >> p  {n is very much greater than p}

Na = 0 for an n-type material

n ≈ Nd

p-sub-n = (n-sub-i)^2 / Nd
p-sub-n is the concentration of holes in the n-type material

The last equation is the same equation as the first equation except the symbols have been changed and subscripts added. The derivation is what it is and you either understand it or you don't.

I understand the equations you are using but i think the heart of my problem is this:

n*p = (n-sub-i)^2

I mean, why is this equation used at all? LIke i could just as eaily right
total carriers = ni + pi (for an intrinsic semiconductor) = 2ni

then Psubn could be something like 2ni - Nsubn

Obvoiusly this is completely wrong but its just to illustrate my confusion. I is it known that the minority carriers follow the multiplicative rule as opposed to my fabricated additive one?

#### Papabravo

Joined Feb 24, 2006
12,405
That original equation is called the mass-action law. It is bassed on approximating the behavior of both electrons and holes as if they were classical particles. The product of the concentrations (n*p) is independent of the Fermi level but does depend on the temperature and the energy gap. It is valid for both intrinsic and extrinsic semiconductors. Now here is the important point. Regardless of the individual magnitudes of n and p the product is always a constant at a fixed temperature.

It's derivation comes from an integral of the density of states times the Fermi function. The result is an exponential function. As you know when you multiply two exponential functions together the exponents are added. When you do this the values for the energy of the conduction band and the enrgy of the valence band combine in such a way that only the difference remains. This the the energy gap of 1.21 eV in Silicon or .785 eV in Germanium.

So (n-sub-i)^2 is just the constant that (n*p) must be equal to and it should be
Rich (BB code):
(n*p) = (n-sub-i)^2 = Nc*Nv*exp(-(Ec-Ev)/kT) = Nc*Nv*exp(-Eg/kT)
where Nc and Nv are constants related to the effective mass of the carriers in the valance band and the coonduction band, Ev and Ec are the energy levels of the valance and conduction bands and Eg is the energy gap between the conduction band and the valence band.

Last edited:

#### mentaaal

Joined Oct 17, 2005
451
You sir, are a legend!

Thats exactly what i wanted to know! Just the why of why it follows that rule in the first place. Excellent! Now i can get back to studying the rest of my physics for my exam next thurs!

Thanks for all the help!

#### Papabravo

Joined Feb 24, 2006
12,405
You are most welcome.