I'm exploring the AD9625 for an RF tuner application using the 2.5GSPS rate and an IF signal bandwidth of around 1GHz. In order to calculate the tuner's RF chain performance, I need to plug in numbers for NF and IIP3 into the RF cascade equations.

So far, my best estimate is NF=26 and IIP3=35.5, but I would really appreciate some expert guidance on this. I would like to see what ADI recommends and how they were calculated.

One of the tricky parts that I'm never really sure about with these ADC's and the standard NF and IIP3 formulas is the best way to calculate the absolute input level in dBm (which exists in the 50ohm single-ended world) relative to the ADC's 0dBFS. Since this ADC has a 100ohm differential input, then I'm not sure about the correct way to translate to dBm (for use in the standard RF and tuner formulas)?

The method that makes the most sense to me is to calculate ADC input dBm (in the 50ohm world) assuming the use of a lossless balun with a 50ohm single-ended input and a 100ohm differential output. Also, for the purposes of calculating ADC NF and IIP3, it seems I would need to assume that there are no terminating or damping resistors (like the ones used in Figure 33 of the datasheet). Is this a correct approach? I assume that the balun and ADC input resistor dB losses need to be accounted for separately as loss blocks that are placed in front of the ADC in my RF cascade. Am I off base with any of this?

I'm also assuming that once the absolute input level in dBm is calculated (at 0dBFS), then the Noise Figure equation uses -1dBFS (translated to absolute dBm) as the input value?

And, I'm assuming that the IIP3 calculation uses -7dBFS (translated to absolute dBm) as the input value for each of the two tone inputs?

(Note: As a side question, it seems to me that R1 and R2 in Figure 33 of the datasheet are actually double terminating the 100ohm ADC input, because the secondary of an input balun is also acting as a 100ohm termination to the ADC. Am I missing something there?).

Thanks for your help!

After looking through some of my research on this topic some more, I recall another method of calculating the ADC dBm input level at 0dBFS.

The formula is generically "Full Scale Signal Power Level (in Watts) = (Vrms^2)/Rin". Where Vrms is equal 0.3889V (which is the full scale 1.1Vp-p divided by 2 and then multiplied by 0.707). And, Rin is equal to the stated datasheet input resistance value of 100ohm?

If this is correct, then I guess this works for ADC's that have Rin values in the 50 to 100ohm range. I'm not sure what you'd do for ADC's that have more complex or time-varying input impedances (due to the switched capacitor input variations during track and hold)?

So then, in order to calculate this power level in dBm, the formula appears to be "Full Scale Signal Power Level (in dBm) = 10Log [(((0.3889Vrms)^2)/100ohms)x(1000mW/Watt)]", which comes out to be +1.8dBm. Does this reasoning look correct, and does it apply correctly to the AD9625? If so, I'm not sure how this works for ADC's that have more complex input resistance values.

So, I thought I should throw this out as further food for thought on the topic.