A pn-junction is formed in silicon by doping one side with acceptor atoms at con

Question

A pn-junction is formed in silicon by doping one side with acceptor atoms at concentration Na = 10^19 cm^(-3) and another side with donor atoms at Nd = 10^18 cm^(-3). For the n-type side of the pn-junction at room temperature: 1. What is the hole and electron concentration on that side? 2. What is the resistivity of a silicon on that side (you can use approximate values for mobility from the graph)? 3. Assuming hypothetically that there is an electric field 100 V/cm, what would be the drift current density?

Explanation / Answer

1. e= nd= 10^18 cm^(-3) p= ni^2/e = (*1.5*10^10)^2/10^18 = 225 2. conductivity= nqu = 10^18*1.6*10^-19*800= 128 S/cm resistivity = 1/conductivity = 7.81*10^-3 ohm.cm (am taking mobility=800, if u have other values, u can keep dat) 3. J= conductivity*E = 128*100 = 12800 A/cm^2

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