How to express Figure of Merit, when Noise Figure less than one

[hari_preetha]

As we know Figure of Merit (FOM) can be calculated using following expression

FOM= Gain(dB) × Band Width (in GHz)/ [NF(dB) - 1] × P_dc(in mW)

From the above expression, we can easily calculate FOM, when NF greater than one and express it.

But my question is, when Noise Figure (NF) is less than 1, denominator term becomes negative. Suppose NF= 0.5 dB, in the denominator [NF(dB) -1]= -0.5, then we have to divide the positive numerator with negative denominator. Finally result will become negative (-) figure of merit.

Can we express Figure of Merit (FOM) in terms of Negative Values, even though the Noise figure is less than 1.

Can you clarify my doubt.

 

[csolis_GT]

It seems that you should use linear values for every specification. The reason 1 is subtracted from NF is because it is always greater than one in linear scale. Therefore, that performs is just how much better than 1 your design is. The same goes for the gain, if you double the gain while maintaining everything else equal, you should have twice the FOM, but if you use dB you will have a much smaller increase in FOM.

 

[hari_preetha]

It seems that you should use linear values for every specification. The reason 1 is subtracted from NF is because it is always greater than one in linear scale.but if you use dB you will have a much smaller increase in FOM.

I am using dB for measuring Noise Figure(NF). For my case, NF(dB) is obtaining as less than 1, then how to express the FOM, when NF is less than 1. So that the denominator term becomes negative, when NF is less than 1. Obviously, FOM becomes negative. How to overcome this difficulty, when NF (dB) is less than 1. How to measure the FOM.

 

[csolis_GT]

Although I have never seen it, if you want to use dB in the NF, you should take the -1 part out of the equation because in dB it can be as low as 0 dB (Divide by NF instead of NF-1)

Can you refer where you got the formula from and that the values you should plug in should be in dBs?

 

[volker@muehlhaus]

As we know Figure of Merit (FOM) can be calculated using following expression

There is no single, unique, fixed definition for figure of merit. Everyone can define a different equation, based on his requirements. Obviously, your equation is made for larger noise figures, and doesn't make sense for very low noise devices.

 

[hari_preetha]

There is no single, unique, fixed definition for figure of merit. Everyone can define a different equation, based on his requirements. Obviously, your equation is made for larger noise figures, and doesn't make sense for very low noise devices.

Yes, as you said it is correct. I have shown equation for the FOM in posting #1, they have used for larger noise figures means that NF is greater than one.
For my case, I am getting the NF less than 0.65 dB. Which equation I have to use for calculating FOM, can you suggest me.

 

[csolis_GT]

You can check examples for LNAs designs.

Pay attention to the units used.

 

[hari_preetha]

You can check examples for LNAs designs.

Pay attention to the units used.

I checked in those LNA design papers also. They have given the same equation as I have posted #1 expression.

 

[csolis_GT]

I checked in those LNA design papers also. They have given the same equation as I have posted #1 expression.

Yes, but the units are [abs] instead of [dB]. In that case you don't have the sign problem as it was intended. You should convert your NF to linear using NF_LIN = 10^(NF_dB/10).

 

[hari_preetha]

You should convert your NF to linear using NF_LIN = 10^(NF_dB/10).

One quire, I have. Where ever [abs] is there that I should convert it linear values =10^{NFdB or Gain(dB)}/10. Can you clarify it.

 

[csolis_GT]

For the NF you have to use \[{10}^{NFdB/10}\] , because is related to a power quantity.

For the gain you have to use \[{10}^{GAINdB/20}\], because it is related to a voltage quantity.