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I'm a little stuck on a subtlety of RF/microwave design that I find a bit baffling. It has to do with the convention of using of noiseless resistors as loads in circuit noise analysis. I have not yet found a satisfying answer for this issue. I also realize this question is heavy on the electrical engineering side, but I am hoping physicists will have a better understanding of the core processes at work here.

It is a well-known phenomenon that thermal agitation of charge carriers in an impedance $Z$ in electrical equilibrium leads to a voltage spectral density ($V^2/Hz$) of the form:

$\sigma^2_v(f)=4k_BTR$ (assuming $hf<<k_BT$)

where $k_B$ is Boltzmann's constant, $T$ is the temperature in Kelvin, $f$ is frequency in $Hz$, $R=Re\{Z\}$ , and $h$ is Planck's constant. Then if we have a circuit bandwidth $B$, the mean-square voltage measured ideally (that is, with an infinite input impedance, noiseless voltmeter) across the terminals of a noisy resistor is just: $4k_BTRB$

Source impedance voltage divider[1]

This thermal voltage source is then put in series with the noisy resistor. Considering a Thévenin equivalent circuit with noiseless load impedance $Z_L$ and noisy impedance $Z_S$, we can use linear circuit theory to solve for the output voltage across $Z_L$ and, in the case where $Z_S=Z_L^*$, the power delivered is the power available from the source, e.g. power transfer is maximized. The value for this quantity (normalized to a one-Hertz bandwidth) is

$P_{avs}=k_BT\approx-174~~dBm/Hz$.

This quantity is used time and again in the design of radio systems and the noise at the output is typically just calculated using cascade analysis of the noise figures and gains/losses of each component. This makes sense to me when ultimately there are amplifiers in a receiver that are going to make the input noise dominate the output noise. However, I am a little confused that no sources that I can find even consider the noise of the load itself.

It is apparent to me that considering the noise of the load resistor is necessary to satisfy the second law of thermodynamics (otherwise we have $Z_S$ deliver net power to $Z_L$). However, this would mean the load resistor receives $k_BTB$ from the noisy source and also "sees" an additional $k_BTB$ from "self-dissipation" of its own voltage thermal voltage. Therefore, in a passive conjugate-matched system, it would appear that the noise power seen across our load would actually be:

$P_n = 2k_BT = -171~~dBm/Hz$.

I cannot find any mention of this in the literature, and to me it implies that, were I to use the $-174~~dBm/Hz$ power spectral density convention that any passive, conjugately-matched system will have a noise figure of $3~dB$. Am I in error, or is this just a case (non-amplified input) that the community does not care about with traditional radio receivers?

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  • $\begingroup$ that "self-dissipation" noise you are referring to is seen (or rather "sloshed" back and forth between the source and load) by the source resistance, your nominally matched but noiseless detector sees noise density only as $k_BT$ $\endgroup$ – hyportnex Oct 15 at 17:32
  • $\begingroup$ and that is because the noise density between matched resistors that are in thermal equilibrium depends only their common temperature $\endgroup$ – hyportnex Oct 15 at 17:37
  • $\begingroup$ I agree that a noiseless detector will see $k_BT$, but considering a non-cryogenic physical detector, it seems slightly odd that there's such a large insistence on $k_BT$ when even in a matched environment we will see more noise than that. The fact that I have never seen any discussion of a noisy load resistor at all is bizarre to me. Two matched resistors $R$ in parallel (one source, one load) can be considered a single resistor $R/2$ connected to an open, so the voltage variance measured by an ideal detector would be $4k_BTBR/2=2k_BTBR$... $\endgroup$ – NubbleWumps Oct 15 at 17:54
  • $\begingroup$ In practice you have a matched low noise/high gain amplifier before the detector, and that is the reason why you do not see the detector's own noise $\endgroup$ – hyportnex Oct 15 at 20:14

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