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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.

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    \$\begingroup\$ What are the names of the variables in the given equation? It seems several of your questions would be resolved by simply providing definitions. \$\endgroup\$ Jun 15 at 16:23
  • \$\begingroup\$ @TimWilliams My apologies. $q$ is the charge of an electron, $A$ is the surface area, $n_i$ is the intrinsic carrier concentration, $N_D$ is the donor doping level, $N_A$ is the acceptor doping level, $\tau$ is the carrier lifetime and $D$ is the diffusion coefficient. \$\endgroup\$
    – Surzilla
    Jun 15 at 20:50
  • \$\begingroup\$ So how do the doping level, diffusion coefficient and carrier lifetime influence \$I_S\$? (Note: you need backslash-dollarsign to do math mode on this stack.) \$\endgroup\$ Jun 16 at 2:24
  • \$\begingroup\$ @TimWilliams Thanks, I was struggling with the math notation in the comments. I know how they influence \$I_S\$ by looking at the formula but I'm trying to develop a physical intuition, specifically for why a large \$\tau_p\$ and \$\tau_n\$ decrease the saturation current. \$\endgroup\$
    – Surzilla
    Jun 16 at 7:23

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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.

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  • \$\begingroup\$ Thank you! That makes sense. \$\endgroup\$
    – Surzilla
    Jun 17 at 13:38

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