# Cut-off frequency of a LED

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. Nov 9, 2020 at 6:06
• I have posted there too, no answers and two people voted to close for it being off topic Nov 9, 2020 at 11:29
• Are you sure that it's 12.5 MHz Nov 9, 2020 at 15:34

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! Nov 10, 2020 at 1:00