My primary question (addressed at the end of this post) is that I have a voltage noise at the input to an ADC that I'd like to convert to a noise power. The remainder of this post details how I arrived at this voltage noise and why I'm having trouble converting it to a power level.
I'm attempting to calculate the thermal noise power of a receiver (at the input to an ADC). This is for a radar application where I'll use this noise power plus a minimum SNR to calculate the minimum detectable signal power, which I can use in turn to calculate the maximum range of the radar using the radar range equation. To calculate the noise power at the antenna, I use \$P=kTB\$. To determine the bandwidth, I take the frequency bin resolution of my downstream FFT, which is \$1.953\,\text{kHz}\$. I've used \$T=300\,\text{K}\$, which gives me a thermal noise (per FFT bin) of \$P=-141\,\text{dBm}\$.
The next 3 components in the receiver are an LNA, an RF amplifier and a mixer, which have reported noise figure (NF) and gains (G) of (all in dB):
LNA: NF=1, G=13
RF amp: NF=5.5, G=12
mixer: NF=14, G=-3
This gives me a cumulative NF of 1.7dB. I believe all of this is correct so far. Here's where it gets a bit trickier though. All inputs and outputs prior to the mixer output were matched to \$50\,\Omega\$. The mixer output has a differential impedance of \$200\,\Omega\$ and the IF amplifier that follows it has a high input impedance, and low output impedance as would be expected. The mixer output is AC-coupled to the IF amplifier input. So, I believe the next step is to take the cumulative noise power so far (at the mixer output), which is \$-139.3\,\text{dBm}\$ (\$1.17\times 10^{-17}\,\text{W}\$) and convert it to a voltage noise. \$V=\sqrt{PR}\$, which gives me a voltage of \$48.5\,\text{nV}\$ (\$R=200\,\Omega\$).
The IF amplifier datasheet contains a section detailing how to estimate the output noise voltage. I've followed these instructions (and checked it against their calculator mentioned in the datasheet) and got a output differential voltage noise density of \$90.8\,\text{nV}/\sqrt{\text{Hz}}\$. I multiply this by the square root of my FFT bin bandwidth (stated earlier) to get about \$4\,\text{$\mu$V}\$ of noise voltage added by the IF amp. Then I apply the gain to the mixer output voltage noise (48.5nV) and add that to the \$4\,\text{$\mu$V}\$ I just found. \$23.5\,\text{dB}\$ as a linear voltage gain is \$15\$, which amplifies the mixer output noise to about \$0.7\,\text{$\mu$V}\$. Add that to the 4 gives me \$5\,\text{$\mu$V}\$ thermal noise at the ADC input. I'm slightly less certain about this 2nd part, but I still feel everything is correct. Now here's where I'm stuck. Normally I'd convert this back to a power level using the ADC input resistance. However, it doesn't appear that the ADC I'm using listed the analog input resistance anywhere. At least I can't find it in the datasheet. Is this typical? How can I convert this back to a power level? Do I assume a differential input impedance? For instance, if I were to use \$1\,\text{M$\Omega$}\$, I would have a noise power of \$-136\,\text{dBm}\$. I guess this seems reasonable, but assuming the ADC input impedance doesn't feel very precise. Conversely, if I assume a larger input impedance at some point the noise falls below the initial value of -141dBm. This, of course, can't be correct.