# RE: Sun noise

```A quick comment:

My 85cm offset feed dish with 0.6 dBNF/40dBG preamp directly connected to
the feed produces about 1/2 to 3/4 s-unit rise...about the same rise I see
from looking at the trees to cold-sky!  My s-meter is very approx. 4-5
dB/s-unit.  I think that is close, considering how difficult making s-meter
reading on noise.  In other words enough rise that is definitely there, not
my imagination!

Ed - AL7EB

At 02:15 PM 12/21/2002 -0500, Tom Clark, W3IWI wrote:
>
>
>> How much sun noise should i expect on 2.4 GHz with the
>> following set-up:
>> - 90 cm offset dish
>> - +/- 0,2 dB estimated loss in adapters
>> - 0,7 dB NF preamp
>> Merry Christmas and 73's
>> Eric
>> EA5GIY
>
>Eric -- let's calculate it from first principles:
>
>Your dish is 90 cm and the wavelength is 13 cm. Therefore your dish has a
>beamwidth of a bit more than 13/90 radians = 8.3 degrees. For a circular
>aperture, Airy's criterion would predict the diffraction limited beam to be
>1.22 times this large, or about 10.1 degrees in diameter. To be simple, I'll
>call this 10 degrees.
>
>The radio sun at 13 cm, depending on sunspot count, is about 40,000 Kelvin
>and is about 1 degree in diameter. Therefore the sun fills about (1/10)^2 or
>~1% of the beam. Therefore, the sun should contribute about (1/100)*40000 =
>400 degrees Kelvin if you dish has perfect (100%) aperture efficiency.
>
>Since I know nothing about the quality of the feed and dish, I will guess
>that you actually have about 50% aperture efficiency (meaning that half of
>the power is in the main beam and half is scattered into sidelobes).
>Therefore the sun probably contributes ~400/2 or about 200 Kelvin.
>
>You say the preamp has 0.7 db NF and you have 0.2 dB connector loss. You do
>not say how long the short piece of coax between the feed and the preamp is,
>so I will guess that it adds another 0.1 dB of loss. Therefore your real
>Noise Figure is 0.7+0.2+0.1 = 1 dB.
>
>Many graphical and slide rule devices make the next calculation be easier,
>but we go back to first principles to find that the correspondence between
>Noise Figure and Temperature is
>      T = 290 * [10^(Noise Figure/10)-1] Kelvin
>Which, for NF=1, is 75 Kelvin as an estimate of your system temperature.
>
>So this says that the sun should have S/N = 200/75 = 2.66 times the noise in
>the receiver. Therefore if you have a true power detector on the output of
>the receiver and set it to 1 volt = receiver noise off the sun, it should go
>up from 1 volt to (1+2.66) =3.66 volts when pointing at the sun, or a
>Yfactor of 3.66:1.
>
>You can check this by making the dish look at large trees with many leaves.
>These trees will contribute about 200-250 K of noise, but since they fill
>the entire ~10 degree beam, they will also contribute about the same noise
>as we predict for the sun.
>
>Another idea is to have a 90 cm wide person stand in front of the dish
>completely covering the dish. The person will contribute about 300-320
>Kelvin of noise. If you happen to transmit, this makes the person bring a
>new meaning to the words "dummy load" ;<}
>
>Hope this helped -- seasons greetings de Tom W3IWI
>
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>

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