Why does the thermal noise of this large resistor not contribute to the input noise for this JFET-input inverting amplifier?

Question

I'm interested in building a low-frequency, low-noise, AC-coupled preamplifier (mostly for measuring the noise of very quiet linear regulators). Several existing designs that I've seen use a matched JFET pair with blocking cap and large resistor to ground at the input. I'll use this paper as an example. I've copied the full schematic below.

The authors state that the thermal noise contributed by RA1 can effectively be ignored above the low-frequency cutoff of \$100\,\text{mHz}\$, assuming a sufficiently low-impedance source. Specifically they say:

The values of the AC coupling network components (CA1, RA1) are such that the pass band extends down to a few mHz and that the contribution of the thermal noise of RA1, assuming much smaller source impedances, is negligible with respect to SVEQ for frequencies larger than \$100\,\text{mHz}\$.

SVEQ is the power spectral density of the equivalent input voltage noise accounting for JFET noise, noise from the load resistances (\$R_D\$) and the equivalent input voltage noise of IA1.

I don't understand this. Why does the source impedance of the noise source matter? Does this mean that if the input were left open, RA1 would contribute significantly to the equivalent input noise?

This amplifier is constructed as an inverting amplifier, with gain \$-R_2/R_1\$. Why does the resistor thermal noise voltage (equal to about \$7.4\,\mu V/\sqrt{Hz}\$ at \$300\,\text{K}\$) not get amplified according to \$-R_2/R_1\$?

Answer 1

Keep in mind the application: you're evaluating voltage regulators. They have low source impedance Rs. The input filter, filled in with the source that was missing in the schematic in the question, looks as follows:

^{simulate this circuit – Schematic created using CircuitLab}

At frequencies of interest, Ca1 is a short, and Rs||Ra1 ≈ Rs. The downstream circuitry "sees" the source impedance Rs, while Ra1 is made irrelevant.

Answer 2

The Johnson (thermal) noise of a resistor is $$ E_t = \sqrt{4kTR \Delta f} $$ Where:
k = Boltzmann' constant = 1.38e-23
T = temperature in °K
R = resistance
\$ \Delta f \$ = bandwidth in Hz

At room temperature (33°C), \$ E_t \approx 0.13 \sqrt{R} \quad \text{in } \; nV/\sqrt{Hz} \$

The real resistance at the input will be the parallel combination of R_A1 and the impedance of the driving source. If the source impedance is low, then the total resistance at the input will be low, thus, the thermal noise will be low.

If the source is a capacitor, such as a piezoelectric hydrophone, best noise performance is if R_A1 is a high resistance, often above 1MΩ as seen in the simulation below. This is why high impedance preamps are desirable for piezoelectric sensors.

Answer 3

I just want to add that there is a very easy way you can prove to yourself, formally, that the noise of that resistor isn't important.

We can do this by the use of the Blakesley source shift theorem (nothing new, it's >100 years old).

The theorem can be easily understood with 2 pictures:

Pictures taken from Structured Electronic Design by CJM Verhoeven et al. (2004)

For your situation, it's easier if we model the noise of the resistor to ground with a current source in parallel. Let's just call it In for now.

In the following schematics I'm already listing all the steps to derive the input noise with respect to Ra1.

^{simulate this circuit – Schematic created using CircuitLab}

For the last step, there's an extra noise current source in parallel with the voltage source, so it drops out.

To find the input-referred noise due to this resistor, assuming its current noise is \$ I_n = \frac{4kT}{R_{A1}}\$, then the input referred expression is simply:

$$ V_{n,R_{A1}}^2 = \frac{4kT}{R_{A1}}R_s^2 + \frac{4kT}{R_{A1}}\left(\frac{1}{\omega C_{A1}}\right)^2 $$

From this expression, we can easily see that the bigger \$R_{A1}\$, the more we can disregard its noise. Therefore, it's not that it doesn't matter because it belongs to the biasing or high-pass filtering; it doesn't matter because we make it not matter by making it big.

Also, this explains the importance of your source impedance being low compared to RA1, otherwise your input referred noise also increases.