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**Selected Problems Tutorial # 3**

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1. The concentration of conduction electrons in pure Si at 300K is 1016 m-3. Assume that by doping the lattice with phosphorus, you want to increase this number by a factor of a million (106). What fraction of the silicon atoms must you replace with phosphorus atoms? (At room temperature, the thermal agitation is effective enough so that essentially every phosphorus atom donates its “extra” electron to the conduction band).

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Answer: concentration of phosphorus atoms (1016 m-3) x (106) = 1022 m-3 = np concentration of silicon atoms in a pure silicon lattice NA = Avogado’s number 6.02 x 1023 mol-1 A = molar mass of silicon 28.1g/mol = density of silicon = 2330 kg/m3

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= 5 x 1028 m-3 The ratio of the concentration of Si atoms and the concentration of P atoms is the fraction of Si atoms that need to be replaced with P atoms This says that if only one silicon atom in five million is replaced by a phosphorus atom, the number of electrons in the conduction band will be increased by a factor of 106.

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**2. Solution A Si sample contains 1016 cm-3 In acceptor atoms.**

The In acceptor level is 0.16 eV above Ev, and EF is 0.26 eV above Ev at 300 K. How many (cm-3) In atoms are unionized (i.e. neutral)? Solution Na = 1016 cm-3 Not all of them are ionized. The concentration of ionized atoms is given by Naf(Ea), where f(Ea) is the probability of electrons being found in energy state Ea.

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**The concentration of unionized atoms is**

1016 – x 1016 = 2.1 x 1014 cm-3

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Lecture #3 OUTLINE Band gap energy Density of states Doping Read: Chapter 2 (Section 2.3)

Lecture #3 OUTLINE Band gap energy Density of states Doping Read: Chapter 2 (Section 2.3)

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