Calculating Carrier Concentration in Semiconductors Using the Hall Effect

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SUMMARY

This discussion focuses on calculating carrier concentration in semiconductors using the Hall Effect. The formula for carrier concentration is given as n = wBJ / eV, where w is the sample width, B is the magnetic field, e is the charge, and V is the Hall voltage. The main challenge identified is determining the current density, J, which can be expressed as J = I/(wt), where t is the thickness of the sample. Participants suggest finding resistance, R, through alternative relationships to derive the current, I, necessary for calculating J.

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  • Understanding of the Hall Effect in semiconductors
  • Familiarity with electrical current and resistance concepts
  • Knowledge of basic semiconductor physics
  • Ability to manipulate equations involving voltage, current, and resistance
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Master J
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A question on semiconductors.

I need to determine the carrier concentration and type.

I have worked out the type but its the conc. that is getting me.

I have the conc., n, as:

n = wBJ / eV

w is the width of sample, B the mag. field, e is charge, and V the Hall voltage...these are all known EXCEPT J. Now the only other pieces of info I have are the sample dimensions and the battery voltage that supplies the current.

I can't for the life of me see where I find J. I just can't get it!

Any pointers in the right directions?
 
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one way to start is by noting that J = I/(wt) , where t is the thickness of the sample ..
 
Yes, but i do not know the current? It leaves me in the same boat...
 
okay .. what if you considered the relation v = IR >> I = v/R .. you can get it don't give up :)
 
Dont have R tho... :p

i only have the info as stated in the question. Damn I am confused!
 
COME ON! :o .. just break things up >> I can't help you with this one, I will just give hints .. you said you want R , so can you find another relation from where you can get R from what you are given in the question (of course other than R=V/I) >>> there is one please try to make some efforts I know sometimes it is not easy but atleast try harder ..
 

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