For instance, comparing the MIC920 and the BGA2818, it's expected that since the latter is intended for application as an RF LNA and the former is general-purpose, the BGA2818 would out-perform the MIC920 in RF applications. But how could this be proven?

The MIC920 specifies only an input noise voltage; let's assume they intended to write nV instead of V:

noise voltage spec

And the BGA2818 specifies only an input noise figure:

noise figure spec

The "naive" comparison would be (following the Maxim App Note 2895):

BW for my application = 5 kHz

Gain of MIC920 ~ 85dB

Output power noise density: $$V_{IN} = \left( 11 \frac {\text {nV}} {\sqrt{\text {Hz}}} \right) \sqrt{5 \text{kHz}} = 778 \text {nV} $$ $$ P_{IN} = 10 \log \left( \frac {778 \text {nV}} {50 \Omega \cdot 10^{-3}} \right) = -48.1 \text{dBm} $$

$$P_{OUT} = -48.1 \text{dBm} + 85 \text{dB} = 36.9\text{dBm}$$

$$P_{NOUTD} = \frac {36.9\text{dBm}} {5\text{kHz}} = 7.38 \times 10^{-3} \frac {\text{dBm}} {\text{Hz}}$$

$$NF = 7.38 \times 10^{-3} \frac {\text{dBm}} {\text{Hz}} + 174 \frac {\text{dBm}} {\text{Hz}} - 85 \text{dB} = 89 \text{dB}$$

Is this calculation valid?

  • 1
    \$\begingroup\$ Misprint on that data sheet. Should be 11 nV/rt(Hz) noise voltage. And 0.7 pA/rt(Hz) noise current. \$\endgroup\$
    – glen_geek
    Jun 24, 2018 at 3:57
  • \$\begingroup\$ No its wrong - You have put 5kHz into PNF, but not into the 174dBm/Hz. The 5kHz cancels out, so you don't need to consider it. \$\endgroup\$
    – Henry Crun
    Jun 24, 2018 at 5:02

2 Answers 2


At 50 Ω, noise is \$\frac{E^2}{50} + I^2 \cdot 50 = -146\$ dBm/Hz; \$174-146= 27\$ dB noise figure at 50 Ω.

You don't need to consider the 5kHz, it cancels out. At low impedances, you can ignore In, at high impedances you can ignore \$V_n\$. (relative to equivalent noise resistance)

For an amplifier with \$V_n\$ and \$I_n\$ given, you can consider the equivalent noise resistance \$R_n = \frac{V_n}{I_n}\$.

This will be the impedance where noise figure will be lowest. In this case 15kohm.

As you see, it is very noisy at 50 Ω.

Noise power is \$V_nI_n=-171\$ dBm/Hz . Thermal noise is -174 dBm/Hz, so the noise figure would be 3dB at 15 kΩ.

The numbers are only valid at low frequencies, at high frequencies they are different, so you can't really compare.

  • \$\begingroup\$ does that mean that noise figure is in units of dBm/Hz? Is that how you're able to subtract thermal noise density and noise power density? Your -171dBm/Hz is not power, it's power density. \$\endgroup\$
    – Reinderien
    Jun 24, 2018 at 18:28
  • \$\begingroup\$ Noise figure is the ratio of two noise powers. Subtracting dB (subtracting logarithms) is division. The units are milliWatts/Hz. When you divide (mW/Hz) / (mW/Hz) the units are lotally cancelled. No mW, no Hz. Just a pure numeric ratio - dB \$\endgroup\$
    – Henry Crun
    Jun 24, 2018 at 21:55

Simple answer: noise figure requires you define an impedance for interfaces in/out of the circuit.

When designing for predictable noise floors on silicon, must a 50_ohm reference resistance be used? No. Why waste signal power into unneeded "Load resistors".

At times the Loads (or Source-located series-terminations Rs) are needed for stability, but that is a different issue.

At high frequencies in tiny systems (onchip) you often can approach the noise analysis as if you are designing high-speed opamps of near-zero size. In this case, you need not have well-defined interface impedances, unless as stated above there are peaking/oscillation/stability issues.


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