Why is the diffusion length defined for minority charge carriers only?

Question

As far as I could infer from the solar cell literature, when talk is about the "diffusion length", only the minority carriers are concerned.

1. Is there a diffusion length defined for majority carriers?
2. In what way does the diffusion length of the majority carriers differ from the minority carrier diffusion length?
3. If the answer to question 1. is No: Why isn't it deemed relevant enough to have a definition for it?

Answer 1

tl;dr: The diffusion length of majority carriers is on the order of 1000 km so it can be ignored as a useful device parameter.

We can show this by looking at the rate question for bimolecular recombination. At constant temperature the law of mass action says that product of electron and hole concentration of any doped or undoped semiconductor is a constant,

$$ n_0 p_0 = n_i^2 $$

Carrier concentrations can be perturbed by injection or by absorption of light, $n = n_0 + \Delta n$ and $p = p_0 + \Delta p$, for electron and hole concentration, respectively.

The recombination rate is then,

$$ R = -\frac{dn}{dt} = -\frac{dp}{dt} = B n p $$

We want to solve this equation for the lifetime of each carrier.

Let's make some assumptions clear so far:

• All recombination is bimolecular. Moreover, for a carrier to recombine it must find a carrier of the opposite charge, so there are no monomolecular or higher-order recombination processes.

Substituting in,

$$ R = \left( n_0 + \Delta n \right) \left( p_0 + \Delta p \right) $$

$$ R = B \left( n_i^2 + n_0\Delta p + p_0 \Delta n + \Delta n\Delta p \right) $$

More assumptions:

• Perturbation is very low $\Delta n \ll (n_0 + p_0)$ compared to the majority carrier concentration.

• Because the perturbation is low the excess carrier concentrations are equal $\Delta n(t) = \Delta p(t)$.

• We can ignore the last term in the brackets $\Delta n \Delta p \approx 0$

So we can now write the recombination rate as a constant term plus a perturbation term which is time-dependent,

$$ R = B n_i^2 + B \left( n_0 + p_0 \right)\Delta n(t) = R_0 + R_{pert}(t) $$

Solving for excess carrier concentration caused by the perturbation,

$$ \Delta n(t) = \Delta n_0 \exp \left(-B(n_0+p_0)t \right) = \Delta n_0 \exp \left(-t/\tau \right) $$

Where $\Delta n_0 = \Delta n(t=0)$ is the excess carrier concentration at $t=0$ when the applied perturbation is turned off.

Finally, the lifetime is,

$$ \tau = \frac{1}{B\left( n_0 + p_0 \right)} $$

Let's assume n-type semiconductor with doping concentration $N_D$,

$$ \tau_p = \frac{1}{B\left( N_D + p_0 \right)} \approx \frac{1}{B N_D} $$

The lifetime of the holes (minority carriers) is only dependent on the availability of free electrons from donor atoms.

Let's assume "GaAs-like" parameters, $B=5\times10^{-10}$cm$^3$/s and $N_D=1\times10^{-16}\text{cm}^3$, the minority carrier lifetime is then,

$$ \tau_p = 20~\text{ns} $$

The intrinsic carrier concentration of GaAs is, $n_i=2\times 10^{6}$cm$^{-3}$, therefore the hole concentration at this doping level is $p_0 = n_i^2 / N_D = 4\times 10^{-5}\text{cm}^{-3}$.

If we associate the "majority carrier lifetime" with this $p_0$ value,

$$ \tau_{\text{maj}} = \frac{1}{B n_i^2 / N_D} = 1.6~\text{million years} $$

In terms of diffusion length, using diffusivity $D=200\text{cm}^2/s$,

$$ L = \sqrt{D\tau_\text{maj}} = 1000~\text{km} $$

This is way larger than the dimensions of any semiconductor device. So for all intents and purposes, the diffusion length is infinite.