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19 February, 00:40

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

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  1. 19 February, 04:05
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    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
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