A uniformly-doped n-type Si is in equilibrium and room temperature. The probabil

Question

A uniformly-doped n-type Si is in equilibrium and room temperature. The probability in finding a hole at Ev + 20kT is 0.021. It is known that a donor concentration of 10^14=cm^-3 was applied. For this problem,use the following values: EG = 1.12eV , ni = 10^10=cm^-3, kT = 0.026eV . Reminder: Some of the answers in the latter questions might depend in your answer in the previous questions. For this problem, wrong answers due to cascading effect will still be marked as wrong even if the formula used is correct. Double-check your answers.  

(a) What is the energy difference between Ef and Ei? Round off in the nearest 0.001 place.
(b) What is the electron carrier concentration? Express in scientific notation with three significant digits.
(c) What is the hole carrier concentration? Express in scientific notation with three significant digits.
(d) Acceptors are present in the semiconductor. What is its concentration? Express in scientific notation with three significant digits.
(e) If the temperature decreases in such a range that the change in ni is still negligible, will the resistivity of the material increase or decrease or will not change?
(f) If only donors are present to achieve the electron carrier concentration solved in (b), will the conductivity of the material increase or decrease or will not change compared to the previous configuration? Assume that the temperature did not change.  

Explanation / Answer

A. The energy difference between EF and Ei is given by the formula,

EF - Ei = KT * ln ( n0 / ni )

= 0.026 * ln ( 1014 / 1010 )

= 0.026 * ln (104 )

= 0.026 * 9.2103

= 0.2394 ev

B. The expression for electron carrier concentration is given by,

n0 = ni exp( (EF EI) / kT )

n0 = 1010 exp ( 0.2394 / 0.026 )

C. The expression for hole concentration is given by,

ni = n0 exp( (EF EI) / kT )

n0 = 1014 exp ( 0.2394 / 0.026 )

D. Acceptor concentration can be given as,

Na- = Na / ( 1 + 4 * exp (Ea - EF) )

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