Thermal noise voltage measured by an impedance matched ADC

Question

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?

Answer 1

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 ⟨V2ADC⟩=2kTRdf. Is this a correct conclusion?

Yes, the most important thing to know when finding noise in analog systems is finding all of the noise sources, then adding the noise by the sum of the squares.

In the real world, however, a simple model like this will not suffice for ADC's. Because most ADC's now use switched capacitors, it complicates the noise figures. One way of calculating that I've used is to find the amount of noisy bits at the input of the ADC, and then compare that with INL and DNL. If you must find what the SNR is then look at ADC noise frequency graphs, and use that to find noise figures.

In a real world analog system the gains will make a difference, so you'll need to keep track of which noise sources are gained up and which are not. (and there will be gains, because how many of us have used only an ADC without a preamp/amp and filters?)

Impedance matching has nothing to do with thermal noise in the real world. Thermal noise is dependent on temperature, and the noise temperature is dependent on other much larger drivers of temperature like:

• How much power is dissipated in the part, the power dissipated in the ADC by the incoming signal will be insignificant when compared to the power dissipated in the ADC by the input signal. The input impedance for most ADC's is high, beyond 100k, so the currents at the inputs are low.

• Other thermal factors like ambient temperature, PCB temperature, and thermal junction resistance will determine the junction noise temperature of the PCB.

• The ADC input is not best modeled as a simple resistor, if it is, the noise temperature would be determined by it's surroundings. I would use 25C or 30C for the temperature.

There are two good resources:

Noise Reduction Techniques in Electronic Systems by Henry Ott

Noise: The Three Categories -- Device, Conducted,and Emitted. by Bonnie Baker

Answer 2

Is this a correct conclusion?

No, for two opposing reasons.

In theory, an amplifier can have a noise figure of less than 3dB, because it is not in thermal equilibrium, and the relationship between input impedance and noise only holds for a system in thermal equilibrium. So if you could build an ADC that acted like an amplifier, you could get a very low noise figure.

But, life is not so kind. In practice, ADCs have lots of excess noise. I'm not enough of an expert to tell you why -- but it is, it has been, and it probably will be, even when we're fending off Romulans and Klingons.

I'm sure that part of it is because the comparator needs to be very wide band compared to the sampling rate, and because sampling aliases the noise into baseband. I'm sure that another part of it is because of necessity ADCs are mixed-signal devices, so a monolithic ADC needs to use a process that is a compromise between good digital performance and good analog performance. Yet another part of it is because digital circuits are noisy, and an ADC puts a sensitive analog circuit right next to a digital one.