How Do You Know When to Use P Holes or N Electrons in Solving Conductivity

Sceptile

Heres the question:
A sample of pure silicon, which has four valence electrons is doped with gallium, which has iii valence electrons, to give a concentration of 10^23 m^-3.

a. Is the textile n-type or p-type?

b. If the electron mobility is 0.14 m^2/Vs and the hole mobility is 0.05 k^ii/Vs, what is the electrical conductivity?

I know this is a p-blazon, but desire to make sure. And the equation for conductivity is electrical conductivity = (number denisty)*(electron charge)*(mobility), but I don't know how to incorporate electron and hole mobility into this.

An explanation and working out on how you solve this question specially part b) would actually be helpful. Thanks.

thebigstar25

okay proficient, the sample is p-type, and the reason is that Ga is from group iii in the periodic table and since Si is from group iv, then Ga is considered as acceptor ..

as for office(b):
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the full general formula for the conductivity is,

conductivity = q(n*Mn + p*Mp) ..

where q = 1.6x10^-19, n : concentration of electrons, Mn : electron mobility, p : concentration of holes, Mp : hole mobility ..

Tin can you lot become from in that location? (i am not allowed to solve it for you), please try more and if you lot needed more assist inquire again..

Sceptile

Ok thanks so concentration of electrons(n) is the number of electrons per unit volume and the concentration of holes(p) is the number of holes per unit volume. What I don't get is how would you find these quantities, are they specific to the type of material for due east.g. in this case it is silicon doped with gallium. I establish this formula for "n" which was n=N/V where Due north is the number of objects in volume V, that didn't aid much though :Southward.

I just demand to know how or where practise you observe the values for 'n' and 'p' in this question.

Thank you once more for the assist

thebigstar25

hint:

Try to make use of the neautrality equation:
n+Na = p+Nd

Where Na: concentration of acceptors
Nd: concentration of donors

Besides, you may demand this equation:

ni^2 = n*p , where ni: the intrinsic concentration ..

Have another look at your question, and try to decide what values you have and what you need to calculate ..

thebigstar25

Heres the question:
A sample of pure silicon, which has four valence electrons is doped with gallium, which has iii valence electrons, to give a concentration of 10^23 m^-3.
a. Is the material north-blazon or p-type?

b. If the electron mobility is 0.14 yard^two/Vs and the hole mobility is 0.05 thousand^2/Vs, what is the conductivity?

we agreed that role(a) was fine ..

I volition endeavour to explain office(b) in a clearer way ..

lets try to analyze this trouble in a logical style, at the beginning we had a sample of silicon with an n (concentration of electrons) and p (concentration of donors) and that sample has no net charge ..

so, we doped Ga atoms , and since Ga is from grouping 3 and Si from group 4, then Ga atoms are considered as acceptors to Si (I assume you sympathise this part) .. since we doped atoms (not ions) and then the Si sample volition nonetheless be neutral (that is why i suggested to use the neutrality equation) ..

now lets get dorsum to what role(b) is asking, yous are asked to find the conductivity, and as I mentioned in a previous post:

conductivity = q(n*Mn + p*Mp) .. >> this is the general formula

you are given Mn (electron mobility), Mp (hole mobility), and q is just 1.6x10^-19c, then in guild to go your answer you lot all the same demand to know what are due north and p ..

ii unknowns need two equations (which I suggested for you to use)

one equation tin be used (the neutrality equation) :

due north+Na = p+Nd , your sample is only doped with acceptor so Nd=0

and you know that ni^two = n*p, (I recollect ni should be given somewhere in your book equally abiding, I believe that ni = 1.5x10^10 cm-3 << check this value againa)

I tried to explain it in a elementary manner equally much as I could, it is straightforward now solve the two equations to get your answer!!! I tin not do this for you!

Good Luck ..

bionicworld

I accept exactly the aforementioned question and I tried to solve it using simultaneous equations you suggested, but I ended up with a stupid quadratic and two impossible answers - that the value of northward is 0 or negative....

Also, we oasis't been taught about of the formulas you're using and then I'm pretty certain we shouldn't take to use them. We only know the first one...

Furthermore, the semi-usher is a p-type, extrinsic semi-conductor, so why are nosotros using a formula that we were told is for intrinsic semi-conductors...? The provision of hole and electron mobility could be to stop people from guessing the first answer from the info in the second part - ie, if y'all only had hole mobility, you'd know to use that formula - p*q*Mp, right?

Lastly, WTH does the "concentration of x^23" refer to? Electron concentration, gallium concentration or what? Because I'll bet it's useful in some manner and you haven't even suggested a apply for it... Unfortunately the stupid question isn't clear nigh what it is either... if it is the concentration of gallium, how can we effigy out n or p from it?

thebigstar25

hmm, I thought I made this question clearer (simply it seems Im terrible in explaining :due south) ..

(>> taking a deep breath) I will attempt again .. first, I will answer your questions:

-------------------------------------------------------------------------
(I am not sure what equations they have been teaching y'all, and so the only fashion to answer your questions is to utilize what I know)
-------------------------------------------------------------------------

the semi-usher is a p-type, extrinsic semi-conductor, and so why are we using a formula that we were told is for intrinsic semi-conductors...?

I believe that you are refering to the equation ni^2 = north*p .. well to be more accurate it is ni^two = no*po (where no and po: are the concentrations of electrons and holes, respectively, provided thermal equilibrium is maintained) this equation is valid for intrinsic or doped cloth ..(why is that true?)

since no = Nc*exp(-(Ec-Ef)/kT), po = Nv*exp(-(Ef-Ev)kT)
>> no*po = Nc*Nv*exp(-(Ec-Ev)/kT) = Nc*Nv*exp(-Eg/kT)

when nosotros accept intrinsic material, ni = Nc*exp(-(Ec-Ei)/kT), pi = Nv*exp(-(Ei-Ev)/kT)
>> ni*pi = Nc*Nv*exp(-(Ec-Ev)/kT) = Nc*Nv*exp(-Eg/kT)
since with intrinsic cloth we have ni=pi, then we can say ni^2 = Nc*Nv*exp(-Eg/kT) = no*po ..
(where ni and pi are the intrinsic electron and pigsty concentrations) ..

I hope that answers your question .. :)

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The provision of hole and electron mobility could exist to terminate people from guessing the first reply from the info in the second part - ie, if you lot only had hole mobility, you'd know to use that formula - p*q*Mp, right?

I woud say that it is not necessary true all the time, as I accept mentioned the general formula for conductivity:

conductivity = q(northward*Mn + p*Mp)

y'all can use conductivity = q*p*Mp if you know for certain that the value of q*n*Mn tin can be neglected when compared with q*p*Mp, and you can not exist sure unless yous found out what are n and p ..

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WTH does the "concentration of 10^23" refer to?

indeed it is not clear what vlaue this 10^23 represents, merely I would suggest to cosider it as the Ga atoms (concentration of acceptors,Na) ..

---------------------------------------------------------------------------

I promise I answered all your questions regarding this problem, if in that location are equations you are non familiar with please mention what exactly are they, as well information technology would be better if y'all write all the equations related to the problem so I would be able to assistance..

newbiephysics

thebigstar25, i think you are forgetting that we are dealing with an extrinsic semiconductor, the formula y'all accept presented is used in computing conductivity of an intrinsic semiconductor.

this problem can just be solved where the supposition is made that all the electrons are accustomed so the corporeality of holes per grand^iii is equivalent to the concentration given then thus n=p

using the extrinsic p-type semiconductor formula for electrical conductivity we get

conductivity(sigma) = p*|e|*hole mobility

thebigstar25

well, I have in my book case of doped Si and they were using the relation ni^2=n*p ..

Anyway, you can brand an approximation that Na = p if Na is much greater than ni ..

And as I mentioned in a previous mail service, the electrical conductivity in general = qnMn+qpMp (intrinsic or extrinsic) you can not ignore 1 term unless you know for sure that it would be too pocket-size compared to the other one, sometimes yous would have a sample doped (say with acceptor) but the doped concentration is not large enough, then in the case you can not say that the conductivity is only qpMp, you got my point there?

bionicworld

bigstar25 - Thank you heaps for your help and although you lot may exist right about the use of that formula for extrin. and intrin. we (beingness get-go years) were but told to use it for intrin. NOT extrin. so I and the other students I checked with just used the electrical conductivity formula with hole mobility just. Too, our newbie first yr books haven't told us about ni^2 = np thing yet :)

newbiephysics - you're right - that concentration was for the acceptor and turns out nosotros were supposed to assume that the concentration of Ga = p since each atom of Ga has 1 'hole' spot available.

thebigstar25

hmm, well, I did mentioned in a previous mail service that the number 10^23 was near likely for the Ga atoms (concentration of acceptors) , also I said in case this Na much larger than ni (which information technology is in this instance =1.5x10^16 yard-iii) and then Na=p ...

For the conductivity in this question the dominant term is as you said q*p*Mp .. And I nevertheless insist that these kind of approximations tin can not be practical all the time (just proceed that on mind in case yous will take an advanced class) ..

And I appologize in case I dislocated you, just I believe I didnt mention wrong information..

Hande_O

Hmm I was but post-obit the whole thread and wanted to mention that thebigstar25 is right.

p*n=ni^2 formula is valid for both intrinsic and extrinsic conductors.

The problem is when we try to solve the doped semiconductor electrical conductivity trouble, we always presume that Nd+ (ionized portion of dopants) >>Na_ or Na_>>Nd+. Hence say if the semiconductor is p type, p>>n, and then nosotros cancel n on the RHS of Nd+ + p=Na_ + due north and Nd+ on the LHS and use the new equality p=Na_. For general cases where p and n values are comparable, we need to solve p or northward values numerically. Therefore nosotros only consider either p>>n or n>>p cases to solve for the conductivity by hand.

If y'all wonder, my reference is Kittel's Solid State Physics edition three.

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