Understanding the formula of the saturation current in a PN-junction diode

Question

According to Wikipedia, the saturation current in a diode is given by the following formula:

$$I_S = qAn_i^2\left(\frac{1}{N_D}\sqrt{\frac{D_p}{\tau_p}} + \frac{1}{N_A}\sqrt{\frac{D_n}{\tau_n}}\right)$$ If my understanding is correct, saturation current is caused by minority carriers reaching the depletion region, getting accelerated by the electric field inside this region, and crossing over to the side where they are majority carriers.

My question is then: how do the doping level, the diffusion coefficient, and the carrier lifetime influence the saturation current?

I would expect that a large doping level will increase the probability that minority carriers recombine before reaching the depletion region, so I understand the inverse proportionality. A large diffusion coefficient means that the minority carriers can diffuse fast and might have a large chance of reaching the depletion region before recombination occurs. However, I specifically don't understand why a short carrier lifetime implies a larger saturation current.

Answer 1

Smaller recombination time means shorter diffusion length: solving the diffusion equation one can easily see that \$L_{diffusion} = \sqrt{(D·τ)}\$. Smaller recombination time effectively makes the carrier distribution more compact, \$Δn_p = Δn_p(0)·\exp(x/L_n); \, Δp_n = Δp_n(0)·\exp(-x/L_n)\$ (in this layout the acceptor-doped region \$p\$ is on the left, the donor-doped region \$n\$ is on the right w.r.t. the pn junction, so both functions are decreasing). From the continuity equation we have \$j_{total}|_{x=0}=qD_p({{dp_n}/dx}|_{x=0}) + (-q)D_n({{dn_p}/dx}|_{x=0})\$, and the exponent derivatives provide you the factors \$1/L_p = 1/\sqrt{D_p·τ_p}\$ and \$1/L_n = 1/\sqrt{D_n·τ_n}\$ which you see in your formula.

Summing up, more compact carrier distribution results in a greater derivative over \$x\$, when we calculate the currents, and consequently wins over a would-be recomb probability increase because of shorter carrier lifetime.