# Maths of carrier densities in p-n junction w/ bias.

by Silversonic
Tags: bias, carrier, densities, junction, maths, w or
 P: 130 Hi, I'm trying to recreate some formulae of my professor's but I'm hitting a dead end. Starting with the following equation for the electron carrier density in the conduction band of a semiconductor: $n = N_c exp(\frac{-(E_c - E_f)}{kt})$ $N_c$ is just a constant, $E_f$ is the electron fermi energy and $E_c$ is the energy of the bottom of the conduction band. Considering a p-n junction with no bias, then the whole system is in thermal equilibrium and the bottom of the conduction bands are separated by the contact potential $eV_0$. The fermi energies of the p and n type semiconductor line up (become equal) and so we get $E_{fn} = E_{fp}$. We can use the above formula for the carrier density $n$ to show $n_p = n_n exp(\frac{-eV_0}{kT})$ My problem is now when we consider a bias. The whole system is taken out of thermal equilibrium and the fermi energies of the p and n junctions separate by an amount $eV$, but I guess we consider separately the p and n junctions to be in thermal equilibrium with themselves. So here we have $E_{fn} = E_{fp} + eV$. Also the bottom of the conduction bands are separated by an amount $e(V_0 - V)$. Hence we have $n_p = N_c exp(\frac{-(E_{cp} - E_{fp})}{kt})$ $n_n = N_c exp(\frac{-(E_{cn} - E_{fn})}{kt})$ Where the $n,p$ subscripts indicate whether we're talking about the p or n part. But, using $E_{cp} - E_{cn} = e(V_0 - V)$ and $E_{fn} = E_{fp} + eV$ I again just end up with $n_p = n_n exp(\frac{-eV_0}{kT})$ But I'm told it's meant to be; $n_p = n_n exp(\frac{-e(V_0 - V)}{kT})$ Can anyone see the problem? Here's an image to indicate the situation: http://imageshack.com/a/img191/8208/cvmy.png Edit: Hmm, I think I might understand this, but I'm not sure. In a completely separate situation, when considering an intrinsic semiconductor under a pump light source the semiconductor itself was outside of thermal equilibrium but the electrons and holes were in thermal equilibrium with themselves, so the conduction band electrons could be described by Fermi-Dirac statistics. The same might be happening here. Electrons are being injected in to the conduction band of the p-type semiconductor from the n-type and it (the p-type conduction band electrons) eventually reach a thermal equilibrium that can be described by Dirac statistics. So the p-type conduction band electrons have a Fermi Energy (completely different from $E_{fp}$) that is equal to $E_{fn}$. Using this then gives me the desired answer.