In searching the internet for thermal noise on an inductor and I'm not finding much on the basics for a real lossy inductor. I'm interested in thermal noise for a fixed inductor in a surface mount package. Here is a diagram TDK provides for their surface mount inductors that shows a loss model for the inductor..

TDK inductor model

At point 2 in the above diagram, is the Johnson-Nyquist noise the following?

\$\overline{v^2_n} = 4k_BTR1 + 4k_BTR2\$

I'm working on building my first LNA for a hobby project and wanting a basic understanding of noise added by inductors. I know there are other sources of noise for a LNA. This question is about the noise contribution by the inductors in particular.

  • 1
    \$\begingroup\$ Where's the bandwidth factor? \$\endgroup\$
    – jonk
    Commented May 19, 2019 at 17:43
  • 1
    \$\begingroup\$ Is this inductor near some sort of shield? that will cause losses. Do you need to include shield resistance (and skin depth) in your design? \$\endgroup\$ Commented May 19, 2019 at 21:40

3 Answers 3


I want to edit @Bimplerekkie's answer, but cannot.

From a noise perspective, the coil will act like the equivalent circuit schematic. This means that at very low frequencies R1 will be shorted out by the equivalent coil -- I would say that at very high frequencies it would be shorted out by the equivalent cap, but I suspect the model breaks down above resonance.

At any given frequency, it's safe to calculate the effective resistance of the circuit -- i.e., reduce it to an inductor (or cap, above resonance) either in parallel or series with an appropriate resistance. The noise characteristics of that resistor at that frequency and around it will be valid.

This sort of thing will work for just about any passive circuit that's all at a constant temperature -- if you burrow deeply enough into it you'll find out that it's a consequence of the laws of thermodynamics.

And note:

All of this equivalent resistance circuit stuff doesn't necessarily apply to active circuits. The "equivalent resistance is equivalent noise resistance" rule works for a circuit whose elements are in thermodynamic equilibrium. It's a stretch, but by definition an amplifying element that has current flowing through it is not at thermodynamic equilibrium, because energy is flowing from the voltage source through the amplifying element; this is why you can build LNAs that have noises temperatures below ambient, and even ones (through clever use of transformer feedback) that have controllable input or output impedances.


You're correct in thinking that the noise can be evaluated like that. The ideal L1 and C1 in the model are noiseless, all the noise originates from R1 and R2. For an inductor these resistors will have quite a low value so their noise contribution is small compared to other elements in a typical LNA circuit.

I am quite sure your formula

\$\overline{v^2_n} = 4k_BTR1 + 4k_BTR2\$

isn's correct as it doesn't show the frequency dependency caused by L1 and C1 and also noise cannot be added like that as the sources are uncorrelated.

In general experienced LNA designers do not worry about the noise of the inductors as they already know it can be neglected. That is not to say it is not a good exercise to do once.

Usually in a properly designed LNA the active element (NPN, NMOS etc.) will be the most dominant noise contributor.

  • \$\begingroup\$ When you say there's a frequency dependency, this is because R1 and R2 themselves change as the frequency changes? Or is it because of a bandwidth factor that needs to be considered, as @jonk suggests? \$\endgroup\$
    – acker9
    Commented May 19, 2019 at 18:18
  • \$\begingroup\$ It is both! To properly specify noise, I either use spot noise, that's the noise (voltage, current or power) at a certain frequency, for example at 1 kHz. Or integrated noise and that's the noise (again voltage, current or power) in a frequency band, for example 1 kHz to 1 MHz. \$\endgroup\$ Commented May 19, 2019 at 19:03
  • \$\begingroup\$ The parallel resistor would be high in value relative to the reactance of the inductor, not low. It would act as a current noise source in parallel with the inductor. \$\endgroup\$ Commented May 19, 2019 at 19:07
  • \$\begingroup\$ @KevinWhite That was what I thought too. If higher inductor values have a larger Rp, it seems like the noise of inductors goes up as the inductance goes up for a given frequency, correct? Which, it seems, the lower the frequency for a NMOS-based LNA, the greater the inductance needed, and the greater the noise from said inductors. \$\endgroup\$
    – acker9
    Commented May 19, 2019 at 20:49
  • \$\begingroup\$ True, although I've never encountered a situation where the noise from the inductor was significant compared to other components in a circuit. \$\endgroup\$ Commented May 19, 2019 at 21:21

Late answer, but I just want to note that the thermal noise from any passive circuit component is really easy to calculate if you know its complex impedance function \$Z(f)\$. You should probably already know it since you have already understood how that linear component is affecting your circuit. In fact the original Nyquist paper already worked this out, and the resistor is just a special case.

The voltage noise (appearing in series) is frequency-dependent, with the following spectral density:

$$ v_n^2(f) \ \mathrm df = 4 k T \operatorname{Re}(Z(f)) \ \mathrm df $$

Here Re() means taking the real part. (spectral density meaning: if you want to know the total variance of voltage noise over a frequency band \$f_1\$ to \$f_2\$, then just integrate the spectral density over that frequency range)

Alternatively you can represent it as a parallel current noise:

$$ i_n^2(f) \ \mathrm df = 4 k T \operatorname{Re}(1/Z(f)) \ \mathrm df $$

Those formulas are much easier to use than trying to separately consider each resistor's independent noise contribution and then trying to figure out how the noise propagates out into the external circuit. You would end up just finding the above result. In fact the above formulas are even correct when you don't know any equivalent circuit, for example you might have a ferrite bead with a messy \$Z(f)\$. The underlying physical reason why only the total impedance matters (and not any internal details) is the fluctuation dissipation theorem.

In your case you'd see that the voltage noise at extremely low and extremely high frequencies is only \$4kTR_2\$, since the \$R_1\$ is being shorted out by either the inductor or capacitor. But at intermediate frequencies, the noise will shoot up, and will reach its peak of \$4kT(R_1 + R_2)\$ at the LC resonance frequency. I'll leave it to you to work out the full formula for yourself.

Interestingly, at low frequencies, there is that useful range where the inductive impedance (\$\operatorname{Im}(Z)\$) is rising linearly with frequency, and yet the noise contribution from \$R_1\$ is only rising quadratically with frequency:

$$ v_n^2(f) \approx 4kT \left(\frac{(2\pi f L)^2}{R_1} + R_2 \right) \qquad \text{(low freq, $2\pi f \ll L^{-1}R_1, \sqrt{LC}^{-1}$)} $$

So, perhaps counterintuitively, a higher \$R_1\$ actually produces less voltage noise at low frequencies, all other things being equal. So when you see on those datasheets that \$R_1\$ is very large, like several kΩ, that actually is a good thing for the most part. :-)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.