kT/C: the noise you cannot design away

Noise spread over frequency

Thermal noise is the random motion of charge in every resistor. Its power spectral density is the noise power in each 1 Hz slice of frequency: 4kTR in V²/Hz, where k is Boltzmann's constant and T the temperature in kelvin. It is white, the same at every frequency, and its square root in V/√Hz is the density that SPICE prints.

A capacitor across the resistor makes a low-pass filter with its corner at f_-3dB = 1/(2πRC). Noise below the corner reaches the capacitor, and noise far above it is shorted away by the capacitor. The rms (root-mean-square) noise on the capacitor is the square root of the density squared, added up over every frequency. Noise sources are random, so their powers add and their voltages do not.

Read the two curves together. The larger resistor makes more noise per hertz, but its corner sits ten times lower, so it passes ten times fewer hertz. On the lower panel both running totals rise to the same 64.37 µV.

Why the resistor drops out

The step that removes R is the noise bandwidth: the width of an ideal brick-wall filter that passes the same noise power as the real one. For a single pole it is π/2 times the −3 dB frequency.

1. Density

   The resistor gives 4kTR V²/Hz. For 10 kΩ at 27 °C that is (12.87 nV)² per hertz.

2. Noise bandwidth

   One pole at f_-3dB = 1/(2πRC) passes noise like a brick wall of width (π/2)·f_-3dB = 1/(4RC). For 10 kΩ and 1 pF: 25 MHz.

3. Multiply

   v_n² = 4kTR · 1/(4RC) = kT/C. R cancels. For 1 pF at 300.15 K: (64.37 µV)².

4. Check on the simulator

   The lab prints 64.37 µV for 10 kΩ, and the hook's run gives 64.37 µV for 1 kΩ and for 100 kΩ.

The table below changes one thing at a time on the simulator. Resistance does nothing, while capacitance and temperature act exactly as √(kT/C) says they should.

Bits cost capacitance

Every switched-capacitor circuit and every sample-and-hold freezes this noise on its capacitor at the instant the switch opens. The frozen value is random from sample to sample, with the rms value √(kT/C). An analog-to-digital converter (ADC) with N bits over a range V_FS has steps of V_FS/2^N, one LSB (least significant bit). A common budget keeps the sampled noise under half an LSB.

Solve √(kT/C) = LSB/2 for C to get C = kT/(LSB/2)². Each extra bit halves the LSB, so C grows four times, and the circuit that charges it in the same time needs about four times the current.

So a lower switch resistance buys speed, which matters for settling, but it never buys lower sampled noise. Only the capacitor sets that, as the two benches below show.

R1 in out 1k
C1 out 0 1p
noise v(out) V1 dec 100 1 1T

R1 in out 10k
C1 out 0 4p
noise v(out) V1 dec 100 1 1T

Try it

Run it: vn is the simulated total and ktc is √(kT/C) from the same run. They agree to 0.002 %; the rest is the integration over 100 points per decade. Open Plots to see the density that vn integrates.

Now change R1 to 1k, then 100k, and predict vn before each run. Then change C1 to 4p (and 1p in the ktc line), and add .temp 125. Only the last two move the answer.

Check yourself

* the noise on 4 pF behind 50 k, rounded to whole microvolts
V1 in 0 DC 0 AC 1
R1 in out 50k
C1 out 0 4p
.control
noise v(out) V1 dec 100 1 1T
dc V1 0 0.1 0.1
let one = 1+0*v(in)
let uv = floor(1e6*noise2.onoise_total+0.5)*one
meas dc vn_uv FIND uv AT=0
.endc
.end