Doped semiconductor-to find the resistance?

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SUMMARY

The discussion focuses on calculating the resistance of an extrinsic semiconductor with a doping concentration that varies according to Nd = N0 Exp(-x/L). The resistance formula is derived as R = L/(AμeN0Exp(-x/L)), but the user questions its accuracy due to the spatial variation of doping concentration. The correct approach involves integrating the differential resistance, dR = dx/(AμN0e^(-x/L)), to account for the changing doping concentration along the length of the semiconductor.

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  • Understanding of semiconductor physics, specifically extrinsic semiconductors
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Homework Statement


An extrinsic semiconductor sample of cross section A and length L is doped in such a way that the doping concentration varies as Nd=N0 Exp(-x/L)
Assume that mobility μ of the majority carriers remain constant.What is the resistance of the sample?

Homework Equations


σ=neμ
ρ=1/σ
R=Lρ/A=L/(Aμen)

The Attempt at a Solution


In the eqn n is the free electron concentration.
I thought this will be eqal to doping concentration.
then,R = L/(AμeNoExp(-x/L))
But this is not the right answer.

Can this resistance be varrying with x since doping concentration depend on x?

Thanks.
 
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humanist rho said:

Homework Statement


An extrinsic semiconductor sample of cross section A and length L is doped in such a way that the doping concentration varies as Nd=N0 Exp(-x/L)
Assume that mobility μ of the majority carriers remain constant.What is the resistance of the sample?

Homework Equations


σ=neμ
ρ=1/σ
R=Lρ/A=L/(Aμen)

The Attempt at a Solution


In the eqn n is the free electron concentration.
I thought this will be equal to doping concentration.
then,R = L/(AμeNoExp(-x/L))
But this is not the right answer.

Can this resistance be varying with x since doping concentration depend on x?

Thanks.
The resistance, dR, for a very short length, dx, of semi-conductor is \displaystyle dR=\frac{dx}{A\mu N_0 e^{-x/L}}

Integrate that .
 
That was really helpful.Thank you.
 

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