How to Calculate Johnson Noise Voltage?
Years ago I memorized the standard value of 4.00 nanovoltsRMS per rtHertz for a 1Kohm resistor, at 17 degree (290 degree kelvin). So what?

In 1 Megahertz bandwidth, you'll have 4nV * sqrt(1,000,000) = 4nV * 1,000 = 4uVRMS. For 1K ohm.

In 1GigaHertz bandwidth, you'll have 4nV * sqrt(1,000,000,000) = 4nV * 31,000 = 124uVRMS. For 1K ohm.

For 62 ohm, you'll have 1.00 nanoVoltRMS per rtHertz. In 1 Terahertz bandwidth, your noise voltage will be 1.0nV * sqrt(10^12) = 1.0nV * 1e+6 = 1 milliVoltRMS.

However, can you achieve 1 Terahertz bandwidth?

Consider there are electrostatic fields upon every node in a circuit. These fields store energy, and we call them "Capacitors". You might achieve 1e-15 Farads (inside a silicon layout); the timeconstant of 62 ohms and 1e-15F is 62 fF. Invert that for 15 Tera Radians per second; divide that by 2*pi, to get 2.5 Terahertz.

If you use poli-silicon over thick oxide, you may be able to achieve 2.5TeraHz bandwidth.

Another useful idea, given the electric fields on every node, is the formula often used in switched-capacitor filters, but also very useful in thinking about total integrated noise voltage.

V_total_noise = sqrt( K * T / C)

At C = 10 pF, the total integrated noise is 20.00 microVolts RMS, independent of the bandwidth. We don't know what the Rnoise is; but in a simple R*C node, we know exactly the total integrated noise, so we can easily compute the total jitter if we know the signal Slewrate.

Thus in a delta-signal 24-bit ADC with internal noise of 1uV RMS, we know the total oversampled capacitance (which must be charged up from either the signal source or thru an input buffer with its own noise and power demands) must be at least!!

 C = 10pF * (20 uV / 1uV)^2 = 4,000 picoFarads


System design...............another form of fun.

Consider the challenges of a 1uA voltage regulator, with 5 volts input and 3.3 volts output. That 1uA quiescent current gets allocated to many internal circuits, including a feedback voltage divider; if you allocate 0.1uA to the divider, the resistor value must be 33 Million Ohms. To implement and compare to a 1.25 (or 1.22) volt bandgap in a differential

 Rnoise = R * product/sum = 11Meg * 2/3 = 7MegOhms.


Suppose the capacitance (very distributed, I know) is 1pF. We know the total integrated noise (into the regulator amplifier of our voltage regulator) will be

 Vnoise = 10uVRMS * sqrt(10pF / 1pF) = 31.6 uVRMS


Hence you cannot purchase a thermally-noise-quiet voltage-regulator, unless you provide external capacitors for random-noise-control.

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Here is a 3-stage-amplifier with total gain of 1,000x; we use 3 stages to achieve very high gain precision and stability. The input signal is 1 millivolt; the next-to-last stage is a 10 Hertz discrete passive RC (1uF, 16Kohm); the final stage is a 22-bit ADC. The tool predicts 90+ dB of SNR; I concurrently included Electric Field interference from the EFI Gargoyle (top right controls). Notice on lower right the Thermal Noise is about the same as the Electric Field interference (20uV) and both are huge compared to the ADC's quantization (the ADC has 5v input range; we only use 1/5 of that).

Secret? I edited the default Rnoise of the first opamp down from 1Kohm default to a low-noise value of 62 ohms. And I edited the two gain-set resistor values down to 11 ohms and 100 ohms (gain = Sum/11 = 111/11 = 10X)