OK, I know how to do this now.

There are 3 main source of noise that need to be calculated:

- Thermal noise of the resistors themselves
- Voltage noise of the op-amp itself
- Current noise of the op-amp, which interacts with the resistors to produce a voltage noise

So first, you want to find the equivalent resistance seen from the inputs of the op-amp looking outward into the circuit, with voltage sources converted to short-circuits (to ground).  For this circuit:
$$
R_\mathrm{eq}=(R_\mathrm{m}+R_\mathrm{s}+R_\mathrm{p})\|(R_\mathrm{f}+R_\mathrm{g})
$$
So for example, if Rs = 100 Ω, Rm = Rp = 1 kΩ, and Rf = Rg = 100 kΩ, then Req = 2.1 kΩ.

To find the thermal noise of this equivalent resistance, use the [Johnson–Nyquist formula](https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise):
$$
v_\mathrm{n}={\sqrt {4k_{\text{B}}TR\Delta f}}
$$
There are online calculators to do this for you:

- [sengpielaudio.com](http://www.sengpielaudio.com/calculator-noise.htm)
- [Daycounter, Inc.](http://www.daycounter.com/Calculators/Thermal-Noise-Calculator.phtml)

For example, with Req = 2.1 kΩ, at 27 °C, with an audio bandwidth of 22 kHz, the resistors would contribute 0.87 μV<sub>RMS</sub> = −121 dBV input noise.

Then find the voltage and current noise of the op-amp in the datasheet.  Typically:

- If \$R_\mathrm{eq}\$ is small, you want a BJT-input op-amp, which has lower voltage noise (0.7-5 nV/√Hz), but higher current noise (500-4000 fA/√Hz).
- If \$R_\mathrm{eq}\$ is large, you want an FET-input op-amp, which has lower current noise (1-10 fA/√Hz), but higher voltage noise (3-15 nV/√Hz).

To convert the spectral density \$\tilde v\$ (in nV/√Hz) to a voltage (in V<sub>RMS</sub>), you need to multiply it by the square root of the bandwidth:
$$
v_\mathrm{RMS}=\tilde v \cdot \sqrt{\Delta f}
$$
So for example, if the op-amp is a TLC071, with equivalent input noise voltage density of 7 nV/√Hz, the voltage noise of the op-amp contributes 7 nV/√Hz ⋅ √(22 kHz) = [1.04 μV<sub>RMS</sub>](http://www.wolframalpha.com/input/?i=7nV/sqrt(Hz)*sqrt(22+kHz)+to+microvolt) = −120 dBV.

The resistor noise and op-amp noise are similar levels, which means they'll combine to about 3 dB higher, or −117 dBV.  To calculate their combination exactly, since they're uncorrelated, you need to use root sum squared:
$$
v_\mathrm{total}=\sqrt{{v_\mathrm{R}}^2+{v_\mathrm{OP}}^2}
$$
So √(0.87<sup>2</sup>+1.04<sup>2</sup>) = 1.36 μV<sub>RMS</sub> = −117 dBV, as estimated.

The current noise is probably irrelevant for an FET-input op-amp, so we can skip to calculating the output noise:  Just multiply the input noise by the gain of the amplifier.  However, you need to multiply by the "*noise* gain", not the signal gain.  [To find the noise gain of the amp](http://electronics.stackexchange.com/a/282646/142), convert your existing sources into short circuits and put a test voltage source right in series with the non-inverting input of the amp:

[![Differential amplifier with noise source in series with non-inverting input][1]][1]

So the op-amp will do whatever it takes for the inverting input to equal the non-inverting input.  There will be one current path:
$$
I=\frac{V_\mathrm{out}}{R_\mathrm{f}+R_\mathrm{m}+R_\mathrm{s}+R_\mathrm{p}+R_\mathrm{g}}
$$
and this is related to \$V_\mathrm{t}\$ by:
$$
V_\mathrm{t}=I(R_\mathrm{m}+R_\mathrm{s}+R_\mathrm{p})
$$
combining and solving:
$$
\frac {V_\mathrm{out}}{V_\mathrm{t}} = \frac {R_\mathrm{f}+R_\mathrm{m}+R_\mathrm{s}+R_\mathrm{p}+R_\mathrm{g}}{R_\mathrm{m}+R_\mathrm{s}+R_\mathrm{p}}
$$
So in our case, this is a noise gain of 96.2× = +39.7 dB, and our input noise of −117 dBV becomes −77 dBV at the output.  (A TINA simulation gives 137.5 μV<sub>RMS</sub> = −77 dBV, for comparison.)

## More detailed steps

There are several extra steps you can do to make your calculation more accurate:

To calculate the effect of the op-amp's current noise, take the current noise and multiply it by the equivalent resistance calculated earlier.  For the TLC071, this is 0.6 fA/√Hz.  So, combined with \$R_\mathrm{eq}\$ of 2.1 kΩ, we get 0.00126 nV/√Hz.  Obviously this is much smaller than the op-amp's voltage noise, so it will have no effect on the result in this example.  In cases with large \$R_\mathrm{eq}\$, it will have an effect. You can calculate it this way and combine it with the other sources as shown above:
$$
v_\mathrm{total}=\sqrt{{v_\mathrm{R}}^2+{v_\mathrm{V}}^2+{v_\mathrm{I}}^2}
$$
Also likely to have an effect is the bandwidth of your measurement equipment.  The previous measurements assume a brickwall filter at 22 kHz, but brickwall filters can't exist in reality.  You can correct for the fall-off of a real-life filter by calculating the equivalent noise bandwidth (ENBW).  Here's a table of [ENBW Filter correction factors vs order](https://gist.github.com/endolith/7d2b2b08466976a95732).  See also [Why are there two sets of ENBW correction factors?](http://electronics.stackexchange.com/questions/281155/why-are-there-two-sets-of-enbw-correction-factors)

In fact, voltage noise of the op-amp is not actually a constant.  It varies with frequency, so is better written as \$\tilde v(f)\$.  You can calculate it more accurately with numerical integration.  See [Noise and what does V/√Hz actually mean?](http://electronics.stackexchange.com/a/280943/142)


  [1]: https://i.sstatic.net/YCkVR.png