I have a question regarding the thermal noise voltage measured by an impedance matched ADC.
In the figure above, I have a simplified circuit with a source impedance (on the left) and an ADC (on the right). I have modeled the ADC as a matched load resistance along with an ideal voltmeter (this is perhaps oversimplified to the point of being inaccurate, but I'm trying to break down the problem to the minimum complexity. Please let me know if there is a better model.).
I fully understand the typical thermal noise treatment of this problem.
In the figure above, the source impedance is replaced by an ideal resistor in series with an ideal thermal noise voltage source, with \$\langle V_{th}^2\rangle=4kTRdf\$. In the typical treatment, the noise from the load is ignored, and the question is, how much noise power is dropped across the load from the source? (The answer of course is \$kTdf\$).
However, this is not the question I am after. My question is, what voltage is measured by the ADC? If only thermal noise from the source is included, then we have \$\langle V_{ADC}^2\rangle=kTRdf\$. But, in reality, we also have thermal noise from the load resistance.
In the figure above, the thermal noise from both source and load are included. In this case, the noise voltages should RSS, so that the voltage measured by the ADC should be \$\langle V_{ADC}^2\rangle=2kTRdf\$.
Is this a correct conclusion? If so, I would interpret this result as the ADC having a noise factor of 2, or noise figure of 3dB, since the measured noise power is twice \$kTdf\$. Does this mean that an impedance matched ADC fundamentally has a minimum noise figure of 3dB?