Receiver sensitivity with temperature - RF Cafe Forums

 Post subject: Receiver sensitivity with temperature

Posted: Fri Nov 16, 2007 6:38 am 

Hi ALL,

We have a debate in my company on how to calculate the receiver sensitivity with temperature.

Sensi = kTB + NF + C/N

C/N is demodulator requirement, and does not vary with temperature for a digital implementation

NF is the noise figure, and is referenced to 290K according IEEE standard (Friis proposed this temperature in 1944). NF varies with temperature of course.

My point is that, when you compute the sensitivity at different temperature, the "T" of kTB should remain equal to 290K, otherwise you account twice for the thermal noise change, once in the NF, and once in kTB.

What is your opinion on that?

Cheers, Patrikc

 Post subject:

Posted: Fri Nov 16, 2007 2:42 pm 

Hi Patrikc,

Here is my take on the issue at hand.

Looking at the equation

Pin_mds(dB)=kT(dBm/Hz)+NF(dB)+B(dBHz)+C/N(dB)

we must first note that the bandwidth and the required C/N at the input of the demodulator usually do not change with temperature. Therefore, the only two terms which are in question are the NF and Pin_mds(or sensitivity). The NF is defined as the SNR(dB) at the input of the system minus the SNR(dB) at the output of the system. The NF may also be derived from a total integrated input referred voltage noise relative to the voltage noise produced by the source resistance (typically 50 Ohm). In the latter case, the integrated input referred noise does change with temperature, but the reference voltage noise is still calculated using the 290K number (~0.895nV/sqrt(Hz) for 50 Ohm @290K). Therefore, the NF does increase with temperature.

The term kT may be assumed only if the input impedance of the system is matched to the source resistance. This term should also be treated as a constant much like the noise of the source resistance. Then, the input referred voltage noise of the system will account for the noise floor variation inherently because of the kT noise associated with it.

In short, you are correct in your statement.

-J