Cut-off frequency of a LED

Question

I'm trying to solve this problem:

A GaAs LED is at \$ 300 K \$ when the current density is 1000 \$A/cm^2\$ . The width of the active region is \$ 1 µm \$. Assume that, for that current, we are in the strong injection regime, and that the bimolecular coefficient is \$10^{-10} cm^3 /s \$. Calculate the cut-off frequency for that current density.

I know the answer is 12.5 MHz.

I know that the cut-off frequency relates do the carrier's lifetime.

$$f_c=\frac{1}{2 \pi \tau}$$

Right, so now I need to calculate the carrier's lifetime. I know that since this is strong injection we would have:

$$\tau=\frac{1}{B \Delta n_0}$$

where B is the bimolecular coefficient. So now I need to calculate \$ \Delta n_0 \$.

My question now is how can I use the temperature, the current density and the width of the active region to calculate \$ \Delta n_0 \$? I feel that I might need some parameters of GaAs at 300 K, but I'm having trouble understanding what parameters.

Maybe PSE gives better response rather than ESE. As the question is close to semiconductor physics. — Mitu Raj, Nov 09 '20 at 06:06
I have posted there too, no answers and two people voted to close for it being off topic — Granger Obliviate, Nov 09 '20 at 11:29

score 3 · Accepted Answer · answered Nov 09 '20 at 16:53

After a bit of digging, I found this:

Equation 9.17 in this chapter

You can manipulate that equation as -

$$n=\sqrt{\frac{J}{edB}}$$

Given:

\$J=1000A/cm^2\$

\$B=10^{-10}cm^3/s\$

\$d=10^{-4}cm\$

\$e=1.6\times10^{-19}C\$

Therefore -

$$n=\sqrt{\frac{10^{36}}{1.6}}\approx\mathbf{0.791\times10^{18}/cm^3}$$

Calculate carrier lifetime - $$\tau=\frac{1}{B \Delta n}\approx\mathbf{12.64 \text{ ns}}$$

Calculate cut-off frequency - $$f_c=\frac{1}{2 \pi \tau}\approx\mathbf{12.6\text{ MHz}}$$

Thank you so much! I didn't have that equation. That solves it! — Granger Obliviate, Nov 10 '20 at 01:00

Cut-off frequency of a LED

1 Answers1