LNA Noise

What to expect from a terminated LNA

The output of any properly terminated amplifier is simply broadband noise. The power level of that noise equals the input noise power, amplified by the amplifier's gain. The input noise power is in turn the thermal noise generated by the terminating resistor, added to the internal noise of the amplifier (that is, its Noise Figure or Noise Factor). So, let's calculate

Noise contributed by dummy load

As a first order approximation, let's consider the amplifier's noise figure to be near zero. That is, its internally generated noise is negligible -- it is a Low Noise amplifier, after all. So, all we have to do now is calculate the noise generated by your dummy load resistor.

Dummy load as Black Body

Let's consider that load resistor to be a thermal black body. Planck's Law says its noise power (in watts) equals kTB. In this equation, k is Planck's Constant (6.6×(10^(-34))kg (m^2)/sec). T is the physical temperature of the black body, in Kelvins (for lab temperature, you can assume 300K). And B is the bandwidth in which the noise power is being measured.

This is what the Black Body curves look like at various temperatures

<<BlackBodyRadiation` (* Loads the package  *)
BlackBodyProfile[4000 Kelvin,5000 Kelvin,6000 Kelvin,PlotRange->{{0,2.0×(10^(-6))},{0,1.1×(10^14)}}]
MaxPower[300 Kelvin,(Meter/(1420×(10^6)))]

Bandwidth of LNA

Now, since you're measuring noise with an HP 8484 power sensor, whose passband is 10 MHz to 18 GHz, you might assume the bandwidth B to use in this calculation to be about 18 GHz. But, remember that the power sensor is going to be placed at the output of an amplifier with a much more modest bandwidth. Thus, the only part of the load resistor's noise power that counts is that part within the LNA's passband. Let's say your LNA passes frequencies from 1.3 to 1.7 GHz. That's a 400 MHz bandwidth, which is what we'll use for B in Planck's Equation

OK, let's calculate. Noise power Pn (coming out of the load resistor, in the amplifier's passband) is kTb

Clear[k]
k= 6.6×(10^(-34));
kT = 300 (* temp of load in Kelvins *);
bwB = 400 ×(10^6)(* Band width in Hz *);
wattsPn = k×kT×bwB (* watts Noise Power *)
 dBmPn = 10 × Log10[wattsPn]

Converting to logarithmic measure, that's a power level of ≈240 dBm.  This is the noise power that gets amplified by your LNA.

Conclusion

Your LNA has 28 dB of gain, so the -240 dBm of noise, after being amplified, comes out of the amplifier at a level of -212 dBm

gain = 28 ; (* Gain of LNA *)
shouldBe = dBmPn+gain

OK, now the noise you're seeing on the power meter is -36 dBm, which is a whopping 157 dB stronger than the noise you should be seeing. From this, I conclude that, yes, your amplifier must be oscillating.