design a semiconductor resistor with specified resistance to handle a given current denisty a semiconductor at T=300k Nd=5×10^15 cm^-3 acceptors are to be added to form a compensated p type material. the resistor is to have resistance 10kohm and handle a current denisty of 50A/cm^2 when 5v is applied

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

energy can cross the boundaries of a closed system in the form of heat and mass

explanation: