Thanks Dave, that was a really nice video on opamp input noise!

But, from the POV of a physicist speaking, I have to mention that the explanation of the nV/sqrt(Hz) unit was a bit dodgy. Especially the "this is not divided by..." is mathematically not correct, since it is exactly that ("divided by" equals "per"). This voltage over root-Hz does indeed mean that the noise voltage is determined as a typical density function over a range of sqrt(Hz) units. It is also not a nV reference to 1 Hz, but it gives the noise voltage caused by operating the opamp over a bandwidth of 1 sqrt(hz), which incidentally translates to operating it over a 1 Hz frequency range. So in practice it boils down to what you said, but seen for units of this sort in general, it should be mentioned. Therefore you calculations should actually read "noise voltage = 10 nV/sqrt(Hz) * sqrt(f-max - f_min)" to be mathematically sound.

More generally, this e_n is actually a function of frequency, which in case of most data sheets is just given by a few values at key frequencies. So for this sample data sheet:

e_n(10 Hz) = 10.3 nV/sqrt(Hz)

e_n(100 Hz) = 10.0 nV/sqrt(Hz)

e_n(1kHz) = 9.6 nV/sqrt(Hz)

So if you were to operate the opamp in a typical audio range of e.g. 20 - 20 000 Hz, we would have to calculate the total expected noise level E_n by integrating this curve. Since this density function is given per root-hz, we have to integrate the square of the curve, i.e. integrate e_n(f)^2 df, in the required frequency range, and then take the root of the result.

E_n = sqrt( Int_a^b { e_n(f)^2 df } ),

where Int_a^b denoted the integration from frequency a (=20 Hz) to b (= 20 kHz).

Lacking the full noise density function, we can now either do the integration on a forward step function (1), or more realistically interpolate the function linearily in the V^2/Hz space (2).

(1) E_n^2 = Int_a^b{ e_n(f)^2 df } = Int_20Hz^100Hz{ e_n(10 Hz)^2 df } + Int_100Hz^1kHz{ e_n(100 Hz)^2 df } + Int_1kHz^20kHz{ e_n(1kHz)^2 df } =

= e_n(10 Hz)^2 * (100-20)Hz + e_n(100 Hz)^2 * (1k-100)Hz + e_n(1kHz)^2 * (20k-1k)Hz = 1.85 muV^2

E_n = sqrt(1.85 muV^2) = 1.36 muV

(2) Now we interpolate e_n(f) linearly between the key frequencies from the data sheet:

E_n^2 = Int_a^b{ e_n(f)^2 df } = Int_20Hz^100Hz{ [e_n(10 Hz)^2 + (e_n(100 Hz)^2-e_n(10 Hz)^2)*(f-10Hz)/(100Hz-10Hz)] df } +

Int_100Hz^1kHz{ [e_n(100 Hz)^2 + (e_n(1kHz)^2-e_n(100 Hz)^2)*(f-100Hz)/(1kHz-100Hz)] df } + Int_1kHz^20kHz{ e_n(1kHz)^2 df } =

= e_n(10 Hz)^2 * (100-20)Hz + (e_n(100 Hz)^2-e_n(10 Hz)^2)/(100Hz-10Hz) * 4000

+ e_n(100 Hz)^2 * (1k-100)Hz + (e_n(1kHz)^2-e_n(100 Hz)^2)/(1kHz-100Hz) * 405 000 Hz^2

+ e_n(1kHz)^2 * (20k-1k)Hz

= (8216.53 + 86472 + 1.75104e6) nV^2 = 1.85 muV^2

E_n = sqrt(1.85 muV^2) = 1.36 muV

As we can see for this case the two methods yield basically the same result, since the third frequency range dominates the terms.

I just wanted to add this, in case some of your viewers get confused when they encounter similar units.