Finding the conduction current density

[PhysicsTest]

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?

 

[Steve4Physics]

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

 

[PhysicsTest]

I am not sure if the question in the link is correct, because if you see

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.

 

[Steve4Physics]

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”