Electronic – Op-amp noise figure

The question which noise sources need to be taken into consideration depends on how severe they are. Your question indicates that you are interested in noise generated at the op amp and not noise generated by interference from neighboring circuits (internal/external noise).

In order to make things comparable, all noise is referred to the op amp's input (RTI). In theory, I guess any point in your circuit might work as long as you refer all noise sources to that point, but it is common practice to act as if all noise sources were directly at the input pins. Sources include noise in the resistors, noise generated by current flowing into the op amp's input pins and noise that may be considered as a voltage between the inputs pins.

There is a very good discussion at this Q&A-style source and also in this nice article from 1969 (!), both authored by Analog Devices' staff.

Without re-typing everything in these sources, here are some rules of thumb:

Noise in the resistors becomes bad when the resistor values are high (some 100k or some 1M) and when the circuits are designed for high bandwidth since the noise is proportional to \$ \sqrt{4k \cdot T \cdot B \cdot R}.\$

You can try to minimize R, you can try to limit the bandwidth B if possible, you can put the circuit in liquid nitrogen (low temperature T), but you can't go for a low Boltzmann constant, because Boltzmann is dead (quote stolen at Analog Devices).

Current noise, i.e. noise generated by current flowing into the op amp inputs, will be converted to a noise voltage by the resistors around the input (\$R_f\$, \$R_g\$) and amplified by the circuit's gain. This is one of the reasons why one prefers op amps with very low input currents especially for high-ohmic circuits.

Voltage noise results from a real op amp's inability to completely null the voltage between the input pins.

All noise sources can be combined as the square root of the sum of their squares since they are independent of each other, which will work only if all sources are RTI.

Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.

However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).

But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.

So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.

EDIT: re: updated question.

(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.

(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.

So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.

Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.

These results are the best you can do, there is no way round them.