Recently, I’m studying some RF basics. When comes to the noise performance of a wireless system, noise figure (NF) is frequently used as a characterization term. It is defined as

$NF = \frac{SNR_{in}}{SNR_{out}}$

Noise figure is a measure of how much the signal-to-noise ratio (SNR) degrades as the signal passes through a system. The NF of a noiseless system is equal to 1. In reality, the finite noise of a system degrades the SNR, yielding NF > 1. Compared to the SNR , which I use it all the time to characterize the ADC performance, the characterization of NF applied to a ADC seems rather complicated to me. Hence, a note about the NF of ADC is written here based on my study. I found two reference are quite helpful, which are from Maxim and EETimes.

Q1: How to determine the NF of ADC?

Steps to determine NF of ADC (Maxim)

Just as the above figure shows, the steps to determine the NF of ADC are as follows:

1. Calculate the full-scale signal power in dBm (dB refer to mW) knowing the peak-to-peak voltage Vp-p and input terminating resistance, Rin.                     $P_{sig} = 10 log({\frac{V_{rms}^2}{R_{in}}*1000}) = 10 log{\frac{V_{rms}^2}{R_{in}}} + 30 dB$  ( Note that $V_{rms} = \frac{V_{p-p}}{2\sqrt{2}}$)
2. Measure the SNR of the ADC in Nyquist bandwidth (usually 0.5dB or 1dB below fullscale)
3. Normalize the converter’s noise power to a 1Hz bandwidth by simply subtracting 10log(fs/2) from the SNR value
4. Calculate the input noise of the converter, which is the theoretical thermal noise floor limit, kTB (-174dBm at room temperature)*.
5. Subtract kTB from the normalized Nyquist band noise power

Q2: How comes kTB?

The noise voltage across the matched input terminal is $V_n^2 = 4kTRB$. The power available from the source is $P_a = \frac{V_n^2}{4R}$. Hence, the available noise power from the source is $P_a = kTB$, which is independent of impedance for non-zero and finite resistance values.

An example of calculation:

Given an ADC with a fullscale voltage of 2Vp-p, an input terminating resistance of 200Ohm, a sample rate of 65MS/s, and an SNR of 69dB (for a -1dBFS input level), the noise spectral density of the ADC is

$NSD_{ADC} = +4dBm - 1dB - 69dB - 10log(65e6/2) = -141.1dBm/Hz.$

The resulting NF is

$NF = NSD_{ADC} - kTB = -141.1dBm - (-174dBm) = 32.9dB.$

Since an ADC does not provide gain (just numerical quantization), the output signal is the same as the input signal plus the quantization noise.  With this in mind, the NF of ADC becomes

$NF = Noise_{out} - Noise_{in}$.

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