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The RF LNA of interest is the PGA-105+ from mini-circuits. From only the gain and noise figure I calculated the voltage noise spectral density that would be present at the amplifier input.

If anyone can check my calculations that would be great. My main concern is my formula for available output noise power spectral density(input referred) and the fact that I don't include a bandwidth in my calculations as I am looking for spectral density.

My circuit model of the amplifier is shown in the top left. The noise voltage sources are from the source impedance and the intrinsic, input referred voltage noise seen at the amp input pin.

enter image description here

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    \$\begingroup\$ This should help researchgate.net/profile/Hyem-Saadi/post/… \$\endgroup\$ Commented Jul 8, 2021 at 23:54
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    \$\begingroup\$ An amplifier with a noise figure F (power ratio, not dB) is equivalent to a noiseless amplifier with a noise power spectral density of (F-1)kT added at the input. So a 1.7dB noise figure corresponds to F=1.48 so the equivalent PSD at the input is 0.48kT, or about -177dBm/Hz \$\endgroup\$
    – Tesla23
    Commented Jul 9, 2021 at 0:13
  • \$\begingroup\$ @TonyStewartEE75 That specific TI app note is for OP amps. This is an RF amplifier, so I can't use those equations for the most part. \$\endgroup\$
    – Max W
    Commented Jul 10, 2021 at 21:05
  • \$\begingroup\$ @Tesla23 How would one convert Power spectral density to noise voltage spectral density? Given a impedance matched 50 ohms? \$\endgroup\$
    – Max W
    Commented Jul 10, 2021 at 21:11

2 Answers 2

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An amplifier with gain \$G\$ and noise figure \$F\$ may be represented by a noiseless amplifier with gain \$G\$ and added input noise power spectral density \$N_a\$

schematic

simulate this circuit – Schematic created using CircuitLab

where \$N_a = (F-1)kT_0\$

and \$F\$ is the noise figure as a power ratio: \$F=10^{F_{dB}/10}\$

by convention, \$T_0 = 290K\$ see HP App note p8

You can convert this to a voltage from a source of impedance \$Z_0\$ by $$\bar{e_n^2} = N_aZ_0$$

For your example, with a 1.7dB noise figure in a 50 ohm system, \$F=1.48\$ , the equivalent added noise PSD is \$N_a \approx -177dBm/Hz\$ which is equivalent to an rms voltage of about \$0.31nV/\sqrt{Hz}\$ from a 50 ohm source.

Just be careful as the noise figure will, in general, vary with source impedance, so although this representation is accurate for a 50 ohm source, if you change the source impedance \$F\$ and \$N_a\$ will likely change.

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  • \$\begingroup\$ Isn't it -supposed to be -177 dBm/ Hz not root-Hz? \$\endgroup\$
    – Max W
    Commented Jul 11, 2021 at 22:22
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    \$\begingroup\$ Fixed, thanks - @MaxW \$\endgroup\$
    – Tesla23
    Commented Jul 11, 2021 at 22:50
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    \$\begingroup\$ If you are saying that the noise volts was low by a factor of 3 (about 10dB), then either you have a device with a noise figure of around 0.6dB (probably unlikely) or there is some issue with your measurement. Happy to engage if you post more details. How exactly are you measuring the noise? \$\endgroup\$
    – Tesla23
    Commented Jul 21, 2021 at 7:00
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    \$\begingroup\$ The noise from a 50 ohm resistor at room temp is around 0.9nV/rtHz, feed that through an amplifier with 16dB of gain gives 5.6nV/rtHz. If you are not measuring at least this then either the amplifier hasn't the gain you think, or there is a problem with how you are measuring the noise. \$\endgroup\$
    – Tesla23
    Commented Jul 21, 2021 at 20:12
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    \$\begingroup\$ Is it oscillating? It has gain up to several GHz, how far up can you check for oscillations? What spectrum analyzer are you using? \$\endgroup\$
    – Tesla23
    Commented Jul 21, 2021 at 21:18
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It's in a matched system so the 50 ohm resistor noise is about 0.45 nV/rt(Hz), x gain = 2.8 nV/rt(Hz).

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