I'm a computer engineering student preparing for Analog Circuits in the spring. My textbook Jaeger's Microelectronic Circuit Design uses an approximation for the intrinsic carrier density for silicon at room temperature of

$$ 10^{10} \frac{e^{-}}{cm^{3}}$$

Now I'm doing fine solving the problems on my own and taking notes except for this approximation. It never defines room temperature but most websites I've found define room temperature as somewhere on the range of 293 Kelvin to 298 Kelvin.


By using their approximation I solved for what temperature they consider room temperature by means of numerical solving in mathematica using the equation above using the numbers they gave for boltzmann constant, material dependency of silicon and the approximation for the intrinsic carrier density.

I obtain 3 solutions, two extraneous and one real:

$$T = 299.707-41.3659j$$ $$T = 299.707+41.3659j$$ $$T = 305.226$$

Now 305 Kelvin does not seem very far from 298 Kelvin but for certain problems the difference can be many order of magnitude off if I don't use their approximation. Is this a bad approximation? The one thing I've gathered from this chapter is that these equations and this process is extremely temperature dependent. For one problem I calculated I had a 9 trillion percent error by not using their approximation. Do many engineers use approximations like this in practice? Is this a good or bad practice to be following. I can't help but think I'm doing something wrong by using an approximation such as this.

  • 1
    \$\begingroup\$ Generally an error in estimating \$n_i\$ doesn't have too much effect on understanding device physics. Do you have any particular situation in mind where it would matter? \$\endgroup\$
    – The Photon
    Commented Dec 20, 2017 at 2:02
  • \$\begingroup\$ You could read 10^10 (an approximation) as somewhere between boundaries 0.95*10^10 and 1.05*10^10 otherwise the approximation would be written 0.9*10^10 (or 9*10^9 or 1.1*etc). For your comfort, you might want to translate both of those boundaries (or at least the lower one) to a temperature. If it's comfortably lower than your definitions of room temperature there's no need to look for further error. \$\endgroup\$
    – user16324
    Commented Dec 20, 2017 at 10:11
  • \$\begingroup\$ 9 trillion percent error for using a slightly different ni value? I have never come across such a case. You probably made some other mistake, or you are working with a very very temperature sensitive device, in which case you shouldn't be using this approximation. Anyway, room temperature was always defined to be 300 K in the classes I took. \$\endgroup\$
    – Matt
    Commented Dec 20, 2017 at 14:26

1 Answer 1


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.


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

(a) p-type silicon as \$N^-_a > 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}\$


\$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.

  • \$\begingroup\$ link Here is the problem. By using 300F and the equations they gave me I calculate n to be 9.74153*10^(-7) e^(-)/cm^3 but they calculate it to be 20 e^(-)/cm^3 which is a 2.05306*10^9 absolute % error. I don't think I made a mistake because these exact equations I use obtain correct results in different problems. The only difference here is their approximation that ni=10^10 \$\endgroup\$ Commented Dec 20, 2017 at 21:53
  • \$\begingroup\$ The p I obtain is exactly the same and I use the same equation and numbers to obtain it. The only difference is their approximation. \$\endgroup\$ Commented Dec 20, 2017 at 21:54
  • \$\begingroup\$ @jakemckenzie I've added a solution how I would have probably solved it. \$\endgroup\$
    – Arsenal
    Commented Dec 20, 2017 at 22:56
  • \$\begingroup\$ I found the source of my error, I was using the material dependent parameter for gallium arsenide not silicone. After recalculating it I get an ni of 6.7354*10^9 with an n0 of 9.0462 which is much more in line with what I should be getting. Thank you for your time. \$\endgroup\$ Commented Dec 21, 2017 at 1:49

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