📘 Analog Circuit Design · Ch. 5 Noise

RC Noise Bandwidth: where 4kTR becomes kT/C

Why the 4, the R and the π all vanish, and why a single pole's noise bandwidth is π/2 times its 3 dB corner, shown four ways until it looks obvious.

📑 Index
Hand-written notebook page 26: 4kTR is a PSD, TIA SNR versus R, RC low-pass rms noise versus R, and adding a load resistor at Vout.
Figure 0 · Notebook p.26, the three noise questions this page grew out of.
#📍 Question✔️💡 Note
#1📐 4kTR is V²/Hz, not Vn✅PSD, not rms
#2📈 TIA: R↑ raises SNR✅Pay R in noise, earn R² in signal: power SNR ∝ R
#3🔁 RC LPF: Vrms if R↓✅Unchanged, area is kT/C
#4🔌 Add Rload at Vout⚠️Reduces spot noise only; see §2
#5🧮 "uncorrelated noise cancels at √N"❌Never cancels; adds in power. Node sees 4kT(R₁∥R₂), the Thevenin R
V̅o² / Δf = 4kT [ R₁·(R₂/(R₁+R₂))² + R₂·(R₁/(R₁+R₂))² ] = 4kT·(R₁∥R₂)

Each resistor's noise reaches Vo through the other's divider, and the two add in power. Nothing cancels; the divider just attenuates faster than the noise accumulates. The √N rule is only the special case of N equal resistors in parallel.

Every real node has capacitance, so "does adding Rload reduce the noise?" has two honest answers. Name which one you mean.

#📏 Quantity🧾 Units🔌 Add Rload🔑 Set by
#1Spot noise, below the RC cornerV/√Hz↓ drops√(4kT·R∥)
#2Noise over a fixed band below fcV rms↓ drops√(4kT·R∥·B)
#3Total integrated noise, 0 to ∞V rms→ unchanged√(kT/C)
Below the corner the node is resistor-limited (flat 4kTR). Integrated through the corner it is capacitor-limited (kT/C, R drops out). Lower R lowers the density but widens the band by the same factor.

No approximation is thrown away. Every constant cancels exactly.

#🪜 Step📐 Expression✂️ What cancels
#1Resistor PSD4kTR  [V²/Hz]
#23 dB cornerfc = 1/(2πRC)
#3Noise bandwidth = ∫|H|² dffc·∫0∞ dx/(1+x²) = fc·π/2arctan(∞) = π/2
#4Plug in fc(π/2)·1/(2πRC) = 1/(4RC)π meets π
#5Total noise4kTR · 1/(4RC) = kT/C4 meets 4, R meets R
Two bandwidths, don't mix them: the 3 dB corner 1/(2πRC) is the real, gentle roll-off. The noise bandwidth 1/(4RC) is a fictitious brick wall that passes the same noise power. It is π/2 ≈ 1.57× wider than the corner, because the 1/f² tail keeps leaking noise past fc.

Figure 1 · Linear f: slide the brick wall until the areas balance

Figure 1. Coral: area the wall adds, w − arctan(w). Teal: tail the wall drops, π/2 − arctan(w). The arctan terms cancel, so they match only at w = π/2. The catch: the teal tail runs to infinity off the right edge, so the eye is not convinced. That is why Figures 2 to 4 exist.

Figure 2 · θ = arctan(f/fc): the RC curve goes flat

x = tan θ  ⇒  dx/(1+x²) = cos²θ · sec²θ dθ = dθ
Figure 2. 0 to ∞ in frequency squeezes into 0 to 90°, and fc lands at 45°. The RC is now a flat bar of length π/2. The brick wall becomes a sec²θ horn. Both coral and teal are finite and fully on screen.

Figure 3 · Quarter circle vs triangle

Figure 3. Frequency lives on the vertical tangent line x = 1; each f is seen from O at θ = arctan(f/fc). A thin slice df makes a skinny triangle of area ½·df. Pulled back inside the unit circle it shrinks by the distance squared, 1 + (f/fc)², which is exactly the RC's |H|². So the brick wall counts the whole triangle (½·w) and the RC counts the quarter disk (π/4). Equal area: ½·w = π/4, so w = π/2. Every area here is half the noise-bandwidth units.

Figure 4 · Rolling the arc: arc length = wall height

Figure 4. Sector area = ½·radius·arc and triangle area = ½·base·height, both with a 1 in front, so equal area means height = arc length. Roll the circle up the wall and the 90° end touches down at exactly π/2·fc. Two different maps live on this chart: rolling (arc length → height, 90° → π/2) proves the equality; projection from O (90° → ∞) is where the teal tail slice from 57.5° to 90° comes from.
50% 50%
Figure 5 · Single-pole noise split. 0 to fc: arctan(1) = π/4. fc to ∞: π/2 − π/4 = π/4. Exactly half. In Figure 3 that is the 45° line splitting the quarter disk.

Steeper filters leak less past their corner, so the brick-wall equivalent hugs f3dB. Ranked by excess noise bandwidth (ENBW/f3dB − 1):

#🎛 Filter📏 ENBW/f3dB📊 Excess

Butterworth n: ENBW/f3dB = (π/2n)/sin(π/2n). Two identical real poles: ENBW = 1/(8τ), f3dB = √(√2−1)/(2πτ).

#❓ Prompt🗣 Say this
#1Why kT/C?4kTR times the noise bandwidth 1/(4RC): the 4 and the R cancel exactly. It is energy over capacitance, no approximation.
#2Why π/2?The RC tail past the corner is worth 57% more width: ∫dx/(1+x²) = arctan → π/2. Geometrically, a right triangle on a unit base must be π/2 tall to match a quarter circle.
#3Add Rload: more or less noise?Spot noise drops as 4kT·R∥. Total integrated noise on a capacitive node stays kT/C. Name the measurement bandwidth.
#4Uncorrelated sources?They add in quadrature, growing as √N. They never cancel; correlated signals grow as N, which is why averaging wins.
#5TIA: bigger R always better?Power SNR ∝ R while resistor noise dominates. But R also lowers bandwidth, and at high f the op-amp en·Cin noise takes over.