Noise Floor/Device Temperature Problem

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SUMMARY

The discussion centers on calculating the noise temperature of a Low Noise Amplifier (LNA) using the noise power equation Pn = 10*log10(kTB). The user converted -100 dBm to -130 dBW and derived an overall system noise temperature of 362.3 K, leading to an LNA temperature of 262.3 K after subtracting the antenna temperature. The community confirms the method is valid but emphasizes the distinction between physical temperature and antenna noise temperature, suggesting that the latter is the relevant factor in this context.

PREREQUISITES
  • Understanding of noise power equations in telecommunications
  • Familiarity with decibel (dB) and decibel-watt (dBW) conversions
  • Knowledge of Low Noise Amplifier (LNA) characteristics
  • Basic principles of antenna noise temperature
NEXT STEPS
  • Study the derivation and application of the noise power equation in RF systems
  • Learn about the significance of antenna noise temperature in communication systems
  • Explore the impact of environmental factors on LNA performance
  • Investigate methods for measuring and optimizing LNA temperature
USEFUL FOR

Engineers and technicians in telecommunications, RF system designers, and anyone involved in the analysis and optimization of Low Noise Amplifiers and antenna systems.

ashah99
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Homework Statement
Please see an attachment below.
Relevant Equations
Pn= 10*log10(kTB)
dBm = dBW - 30 [dB]
Hi everyone, I would like to get some help with the following problem. I'm not sure if my answer is feasible. Thanks.

1634318611581.png


My attempt at a solution:

I used the noise power equation: Pn= 10*log10(kTB) and converted the -100 dBm to dBW by subtracting 30 dB to get -130 dBW. Then I tried to find the overall system noise temperaure, T, and then I subtract the antenna temperature to find the temp of the LNA.

Pn= 10*log10(kTB) = Pn= 10*log10(k) + Pn= 10*log10(T) + Pn= 10*log10(B)
-130 dBW= 10*log10(1.38e-23 J/K) + 10*log10(T) + Pn= 10*log10(20e6 Hz)
10*log10(T) = -130 dBW - (-228.6 dB/J/K) - 73 dB/Hz = 25.6 dB/K
T = 10^(25.6/10) = 362.3 K
T_LNA = T - T_antenna = 262.3 K

This seems quite high for LNA device temperature. Could someone help to see if my work is right or if I'm totally off-track or misinterpreting the problem?
 
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Why should the temperature of the dish introduce noise? It is a metal sheet with negligible resistance.
 
Actually 100K is very cold so I think they mean Antenna Noise Temperature, not physical temperature. I think your method is OK but have not checked the numbers.
 
tech99 said:
Actually 100K is very cold so I think they mean Antenna Noise Temperature, not physical temperature. I think your method is OK but have not checked the numbers.
My assumption was antenna noise temp as well. Would you kind as to check if I’m on the right path?
 

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