Answer:
I = 0.5mA
L = 0.05cm
A = 0.00001cm²
K = 0.5/(ohm cm)
Na = 1.25*10^16 cm^-3
and the conductivity K of the semiconductor is 0.492
Explanation:
From ohm's law V=IR, when 5V is applied to 10kohm, current I = V/R
I = 5/(10*10^3) = 0.0005A = 0.5mA
Then, the cross sectional area A can found by using A = I/J where J is the current density limited to 50A/cm^2.
A = 0.0005/50 = 0.00001cm²
Considering that the electric field E is limited to 100V/cm, the length of resistor L can determine by using L = V/E
L = 5/100 = 0.05cm
Now, the conductivity K of the semiconductor is K = L/RA
K = 0.05/(10*10^3 * 0.00001) = 0.5/(ohm cm)
The conductivity K of a compensated p-type semiconductor is given by
K = e. up. P = e. up. (Na - Nd)
Then, the mobility is the function of total ionized impurity concentration Na + Nd
If Na = 1.25*10^16 cm^-3 and Nd = 5×10^15 cm^-3,
Na + Nd = 1.25*10^16 + 5×10^15 = 1.75E16 cm^-3
Therefore, the hole mobility from figure mobility and impurity concentration is
(up) = 410 cm²/Vs
Then, the conductivity K is then
K = e. up. P = e. up. (Na - Nd)
K = 1.6*10^-19 *410 * (1.25*10^16 - 5×10^15) = 0.492
This is very close to the value needed to design the semiconductor. From the calculation above, it is found that;
L = 0.05cm
A = 0.00001cm²
Na = 1.25*10^16 cm^-3
