[SOLVED] How to estimate TSS of a mixer+LNA given noise figure and voltage conversion gain?

https://www.infineon.com/dgdl/Infin...N.pdf?fileId=5546d4625696ed7601569d2ae3a9158a

I'm evaluating BGT24LTR11 RF transceiver and I want to do theoretical calculation of how well it behaves as a detector and I need to find it's TSS for that. All what's specified in the datasheet are SSNF of 10 dB and voltage conversion gain of +20 dB. I don't have the parameters such as frontend RF bandwidth so I'm not sure if I need to contact Infineon for that.

Any app notes of calculation are appreciated. I've looked at a few but they make use of more parameter, not specified here.

From given parameters (NF, voltage gain), can I somehow gauge how well will the device should work as a receiver other than TSS?

The receive side of this device is an I-Q mixer, from the SSB noise figure if you could cancel the unwnated sideband would be -174dBm (kTB in Hz) +10dB NF + 20dB gain. The I-Q bandwidth is not specified, in its intended application would only need to be very high, about 10kHz.
Given the quadrature and amplitude accuracy specification inwanted sideband cancellation is likely to be as good as non-exitant so the noise figure would increase by 3dB.
Without taking the AM and phase noise of the oscillator into account the noise power in a 10kHz bandwith at the IQ output will be about -100dBm, TSS wwould be about -92dBm.
The AM noise will have an effect even if is likely to be around 30dB higher at 10kHz than the 100kHz offset quoted, this will depend on how good the ocillator is it could be 10dB better, but then it could be 10 worse. This noise power per Hz will keep increasing as the offset decreases, but the bandwidth is also decreasing; you will have to integrate the noise power over the band you want to use. I am guessing you would lose at least another 3dB of sensitivty from local oscillator AM noise.
The above are 'back of envelope' calculations. A fuller ananlysis would take more time than I have to lok at it, but hopefully gives ou some pointers
Peter

G4BCH said:

The receive side of this device is an I-Q mixer, from the SSB noise figure if you could cancel the unwnated sideband would be -174dBm (kTB in Hz) +10dB NF + 20dB gain. The I-Q bandwidth is not specified, in its intended application would only need to be very high, about 10kHz.
Given the quadrature and amplitude accuracy specification inwanted sideband cancellation is likely to be as good as non-exitant so the noise figure would increase by 3dB.
Without taking the AM and phase noise of the oscillator into account the noise power in a 10kHz bandwith at the IQ output will be about -100dBm, TSS wwould be about -92dBm.
The AM noise will have an effect even if is likely to be around 30dB higher at 10kHz than the 100kHz offset quoted, this will depend on how good the ocillator is it could be 10dB better, but then it could be 10 worse. This noise power per Hz will keep increasing as the offset decreases, but the bandwidth is also decreasing; you will have to integrate the noise power over the band you want to use. I am guessing you would lose at least another 3dB of sensitivty from local oscillator AM noise.
The above are 'back of envelope' calculations. A fuller ananlysis would take more time than I have to lok at it, but hopefully gives ou some pointers
Peter

Click to expand...

Thanks for answering. I'm looking at TSS figure you've calculated and can't make sense of it. A really good Schottky diode will have TSS of -60 dBm and this implies this device is 30 dB more sensitive. I'm not sure this makes sense to me.

From the description in section 1 of the data sheet
"The receiver section uses a low noise amplifier (LNA) in front of a quadrature homodyne down conversion mixer
in order to provide excellent receiver sensitivity. "
First the LNA will increase the sensitivity by its gain less the added noise from teh noiswe figure, second this is not a schottky diode detector, it is an IQ detector which gives a linear frequnecy translation to DC of the RF on either side if the local oscillator. As such the sensitivity calculation is just a matter of adding the contributions from various gain and noise sources.
Input noise floor = ktB, -174dBm/Hz
Device gain = 20dB
Device noise figure = 10dB assuming perfect image cancellation
Output noise power/Hz = ktB + gain + noise figure = -174 +20 +10 = -144dBm/Hz
scale for bandwidth your system choice assume 10kHz 10*log(10000) = 40dB
Output noise power per 10kHz = 144 + 40 = -104dBm
As I said before, the actual noise power will be high because the sideband cancellation will not be perfect, if it is anywhere near the maximum youcan assume non existant so add 3dB for the oppsite sideband noise. From a system point of view there will aslo be noise fom the antenna, this will be looking at ground noise. Simplistically you can assume that the ground noise tempereature will be 290K, so the additional noise from the antenna will add another 3dB giving a sensitivity of -98dBm.
The above has neglected the contribution from the loacal oscillator noise and the noise figure of any IQ processing. this is just the noise power output from the each of the I & Q outputs, and is about what I would expect from a device such as this.
Don't forget that your schotky diode detector has no RF amplifier, add the 20dB for this and you are at -80dBm, add another 8dB for TSS and you are at -88dBm. Account for the video bandwidth of the detector, unknown and you will probab;y be not too far different.

Peter