Approximations for the intrinsic carrier density of silicon at room temperature?
I'm not actively designing semiconductor devices, so take this with a grain of salt or two.

The value you are citing was or is (depending on who you ask) the accepted value for 300 K (seems to be a usual measure for "room temperature" in semiconductor science).

Note however that you will find documents saying the value is \$1.5\ 10^{10} \frac{e^-}{cm^3}\$ down to \$8.72\ 10^{9} \frac{e^-}{cm^3}\$ or the latest I'm aware of is \$9.65\ 10^{9} \frac{e^-}{cm^3}\$.

I'm wondering a bit what kind of problem you stumbled upon to get an error of 9 trillion percent just by varying \$n_i\$. \$n_i\$ is highly temperature dependent, that is true and should be considered for each problem.

Now the question if engineers use these approximations and if that is a good practice or not is opinion based. Actually the value you quoted is based on a measurement by Sproul and that makes it (if the measurement was done carefully) a better candidate than a number you get from an equation which is based on assumptions or simplifications or more or less an empirical formula.

I'd say we use approximations, but usually we calculate a security margin in our products, and these should cover for the variance of the real world and errors in approximations. If you are designing cutting edge technology with a low security margin you probably find that the models used so far are not exact enough for your needs and then you start developing your own models or make measurements to get to the point you need. It is often an iterative process, make a prototype, see how it performs, find why it doesn't perform as expected, make it better, start again until requirements met.



Problem:



My try at solving the problem (taken from the comments to my first answer):

(a) p-type silicon as \$N^-_a &gt; N^+_d\$

(b) 
\$p_0 \approx N^-_a - N^+_d \approx 5 \times 10^{18}\frac{e^-}{cm^3}\$ 

\$n_0 = \frac{n_i^2}{p_0} = \frac{(10^{10} \frac{e^-}{cm^3})^2}{5\times10^{18}\frac{e^-}{cm^3}} = 20\frac{e^-}{cm^3}\$

And if I use a different \$n_i\$: 

\$n_0 = \frac{n_i^2}{p_0} = \frac{(8.27\times10^{9}\frac{e^-}{cm^3})^2}{5\times10^{18}\frac{e^-}{cm^3}} \approx 13.7\frac{e^-}{cm^3}\$

or

\$n_0 = \frac{n_i^2}{p_0} = \frac{(1.5\times10^{10}\frac{e^-}{cm^3})^2}{5\times10^{18}\frac{e^-}{cm^3}} \approx 45\frac{e^-}{cm^3}\$

So - sure I get some different values - but not by a trillion percent. Don't know how you calculated it.