The p-n Junction — Question 9

Question 9

Band bending and electrostatic potential have opposite signs

At equilibrium an abrupt homojunction has p material on the left and n material on the right. The electrostatic potential rises by $V_{bi}=.700\,\mathrm V$ from p to n, and the electron band edge satisfies $E_c=-e\phi+\text{constant}$. The gap is spatially constant.

1. Find the conduction- and valence-band energy changes from p to n in eV.

2. An electron moves from n to p without collisions. Find its electrostatic potential-energy increase and the minimum initial kinetic energy needed to surmount that barrier.

Question 9 - Solution

Band bending and electrostatic potential have opposite signs

(a) Method and calculation

$$\Delta E_c=E_{c,n}-E_{c,p}=-e(\phi_n-\phi_p)=-eV_{bi}.$$$$\boxed{\Delta E_c=\Delta E_v=-.700\,\mathrm{eV}}.$$

(b) Interpretation and further calculation

$$\Delta U_{n\to p}=-e(\phi_p-\phi_n)=+eV_{bi}.$$$$\boxed{K_{\min}=eV_{bi}=.700\,\mathrm{eV}}\quad\text{in the stated classical barrier picture}.$$

Physical check

The flat equilibrium Fermi level coexists with bent band edges. A positive charge would experience the opposite electrostatic potential-energy change.