I need to estimate the temperature at which p-n junction made of silicon lose it rectifying characteristics. (\$N_A=N_D=10^{15}\,cm^{-3}\$)
\$E_G\$ is independent of the temperature and are 1.12 eV for Si. Intrinsic carrier concentration at room temperature (T=300 K) is \$n_i^{Si}=10^{10}\,cm^{-3}\$.
I try to solve this problem with this argument: p-n junction stops working when concentrations of electrons and holes equalize.
It happens when \$N_D(N_A)\approx n_i=\sqrt{N_c N_v}exp(-E_g/2KT)\approx T^{3/2}exp(-E_g/2KT)\$. The maximum temperature is \$T_{Si}\approx 650 K\$ (result).
I tried to replace the values in this previous formula to see if I get the Nd value but I did not get this value, or anything of the same order of magnitude:
\$ T^{3/2}exp(-E_g/2KT)=650^{3/2}exp(-1.1/(2\times 8,617\times 10^{-5}\times 650))\approx 0,901 \neq N_D\$
What is escaping me in reasoning?
Best Answer
Some terms are missing in \$n_i\$... $$ n_i = \sqrt{N_cN_v}e^{-E_g/2kT} = 2\left (\frac{2\pi k}{h^2}\right )^{3/2}(m_e^*m_h^*)^{3/4}T^{3/2}e^{-E_g/2kT} $$
Note that, in non-degenerate semiconductors, effective density of states in conduction band \$N_c\$ and that in valence band \$N_v\$ are expressed as (see Principle of Semiconductor Devices) $$ Nc = 2\left(\frac{2\pi m_e^* kT}{h^2}\right)^{\frac{3}{2}} $$ $$ Nv = 2\left(\frac{2\pi m_h^* kT}{h^2}\right)^{\frac{3}{2}} $$
Now, using effective mass \$m_e=1.08m_0\$ and \$m_h=0.81m_0\$ given in the reference, and at 650 K, it will yield, $$ n_i=3.29\times 10^{15}\ (cm^{-3}) $$ which is about \$10^{15}\ (cm^{-3})\$.