Finding the conduction current density

In summary, at 300K, adding 10¹⁴ donor atoms/cm³ to a germanium sample will result in a resistivity of 60 ohm-cm.
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Homework Statement
A sample of germanium is doped to the extent of 10^14 donor atoms/cm^3 and 7*10^13 acceptor atoms/cm^3. At the temperature of the sample the resistivity of pure (intrinsic) germanium is 60 Ohm-cm. If the applied electric field is 2V/cm, find the total conduction current density.
Relevant Equations
##J = (n\mu_n + p\mu_p)qE ##
The current density is given by the formula
##J_e = (n\mu_n + p\mu_p)qE = \sigma E; \sigma \text{=conductivity}## ->eq1
The resistivity of intrinsic germanium is 60 Ohm-cm, the equation 1 becomes
##J_i=n_i(\mu_n + \mu_p)qE## ->eq2
##J_i=60 \text{ ohm-cm} ##
Applying the standard equations of charge neutrality
##N_D + p = N_A + n ## ->eq3
##N_D - \text{ number of Donor atoms} = 10^{14}##
##N_A - \text{ number of Acceptor atoms} = 7*10^{13} ##
##p - \text{ number of holes} ##
##n - \text{ number of electrons} ##
Using Mass Action law
##np = n_i^2## ->eq4
From eq3, 4 i can calculate ##n## and ##p##, but not sure how to proceed further? I feel three unknowns to be solved ##\mu_n, \mu_p,J_e## only 2 equations eq1, eq2 are available. How to solve?
 
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PhysicsTest said:
I feel three unknowns to be solved ##\mu_n, \mu_p,J_e## only 2 equations eq1, eq2 are available. How to solve?
I suspect the question is incomplete and you should also be given the values of electron and hole mobilities: 3800 cm²V⁻¹s⁻¹ and 1800cm²⁻¹s⁻¹ respectively.

I’ve seen this question before and it included the mobility values. See Q8 here: https://www.meritnation.com/ask-ans...s-materials-devices-and-simple-circu/10985139
 
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I am not sure if the question in the link is correct, because if you see
1609942429869.png

The above are the standard values at 300K as per the book. But in the actual question it does not mention the exact temperature, it only says at the temperature of the sample.
 
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PhysicsTest said:
I am not sure if the question in the link is correct, because if you see
View attachment 275754
The above are the standard values at 300K as per the book. But in the actual question it does not mention the exact temperature, it only says at the temperature of the sample.
I can’t suggest anything else.

However, I will note this. You have both donor and acceptor atoms added. It is not clear to me what will happen. I suspect they will partially ‘cancel out’, the net effect being the same as simply adding (10¹⁴ - 7x10¹³ =) 3x10¹³ donor atoms/cm³.

EDIT: The above is supported by the statement from https://inst.eecs.berkeley.edu/~ee105/fa05/handouts/discussions/Discussion1.pdf
“If we assume complete ionization and if Nd-Na >> ni, the majority carrier electron concentration is, to a very good approximation, just the difference between the donor and acceptor concentrations”
 
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FAQ: Finding the conduction current density

What is conduction current density?

Conduction current density is a measure of the flow of electric charge through a material. It is typically represented by the symbol J and is measured in units of amperes per square meter.

How is conduction current density calculated?

Conduction current density can be calculated by dividing the current (I) by the cross-sectional area (A) of the material. This can be represented by the equation J = I/A.

What factors affect conduction current density?

Conduction current density is affected by the material's conductivity, the cross-sectional area, and the applied electric field. It is also influenced by temperature, as higher temperatures can increase the movement of charge particles.

Why is conduction current density important?

Conduction current density is important because it helps us understand the flow of electricity through different materials. It is also a crucial component in the study of electromagnetism and plays a role in various technologies, such as power generation and electronic devices.

How is conduction current density related to Ohm's Law?

Conduction current density is directly proportional to the applied electric field and inversely proportional to the material's resistance, as stated in Ohm's Law. This means that as the electric field increases, the current density also increases, while an increase in resistance leads to a decrease in current density.

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