Question-1: Calculate the minority electron concentration at the junction
edge (or depletion edge) of the p-type at forward bias volatage of V=0.4,
0.55, and 0.7. Assume that the minority electron concentration at equilibrium
is n_{p0} = 1000 electrons/cm^{3} and use the formula n_{p}
= n_{p0} exp(V/V_{T}) where V_{T} = 0.025 V at
room temperature.

Answer:

V=0.4: n_{p} = n_{p0} exp(V/V_{T}) = 1E3 * exp(0.4/0.025)
= 1E3 * 8.886E6= 8.886E9 electrons/cm^{3}.

V=0.55: n_{p} = n_{p0} exp(V/V_{T}) = 1E3 *
exp(0.55/0.025) = 1E3 * 3.585E9 = 3.585E12 electrons/cm^{3}.

V=0.7: n_{p} = n_{p0} exp(V/V_{T}) = 1E3 * exp(0.7/0.025)
= 1E3 * 1.446E12 = 1.446E15 electrons/cm^{3}.

Note: In an NPN bipolar transistor, the forward-biased emitter-base
NP junction has a typical voltage drop of 0.7 Volts. The injected minority
electron concentration at the junction edge in the Base will be about 1.5E15
electrons/cm^{3} if the base doping is N_{A}=1E17 cm^{-3}.