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Interesting: Will SAW Help Improving the Chain NF?

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Ruritania

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Hi, all,

Let's consider a 50 ohm to 200 ohm impedance transformation SAW filter, and it has the insertion loss of 1dB. If it is inserted between a 50 ohm antenna and a 200 ohm LNA, will it help with the chain noise figure, or the noise figure of SAW+LNA will be better than LNA alone (refer to 50 ohm)?

Please let me know your calculation, if you're interested.

Thanks,
Ruri
 

kspalla

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the noise figure of SAW+LNA will be better than LNA alone (refer to 50 ohm)?

No

Added after 2 minutes:

LNA NF opt impedance normally of the same order but we will have matching circuit with little compromise on NF.
I suggest LNA +SAW model unless the LNA is subjected to high power input.
 

darcyrandall2004

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Interesting: Will SAW Help Improving the Chain

I would have thought that the Saw filter will actually worsen your NF by 1db corresponding to the insertion loss. However the Saw filter provides other benefits such as minimising the bandwidth thereby improving the mds, reducing the chances of spurs etc. The gain of the LNA will actually reduce the NF contribution of other following cascaded stages. Therefore it is best that the LNA be placed at the start of the cascade just after the antennae.
 

Ruritania

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kspalla said:
the noise figure of SAW+LNA will be better than LNA alone (refer to 50 ohm)?

No

Added after 2 minutes:

LNA NF opt impedance normally of the same order but we will have matching circuit with little compromise on NF.
I suggest LNA +SAW model unless the LNA is subjected to high power input.

Hi, thanks for your reply.

During the impedance transformation, the SAW filter actually has voltage gain, although it introduces 1dB power loss. The voltage gain can be calculated as:

Vout^2/200 = Vin^2/50*L => VG=1.7825 ~ 5dB

Obviously this gain will help with the noise figure. Agree?

Thanks,
Ruri
 

biff44

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I think I see what you are saying. If you have an unmatched system (50 ohm input sources, and 200 ohm input impedance amplifier that has a 1 dB NF in a 200 ohm system) then you are throwing input power away with a big reflection loss.

|ρ|=(200-50)/(200+50) = 0.6

So transmission "thru" this mismatch is |t|=[1-(0.6^2)]^0.5 =0.8

20 LOG 0.8 =-1.9 dB

So with the mismatch, your effective NF should be 1+1.9=2.9 db

With a 1 dB loss saw that corrects the mismatch exactly, you should have a NF=1+1=2 dB

Looks to me like the SAW wins.
 

    Ruritania

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kspalla

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I agre with biff44.
However
Simple matching circuit from 200ohm to 50 ohm may take insertion loss not more than 0.2 to 0.3dB than 1.9dB as mentioned above calculations.

That why I mentioned "we will have matching circuit with little compromise on NF"
 

Ruritania

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biff44 said:
I think I see what you are saying. If you have an unmatched system (50 ohm input sources, and 200 ohm input impedance amplifier that has a 1 dB NF in a 200 ohm system) then you are throwing input power away with a big reflection loss.

|ρ|=(200-50)/(200+50) = 0.6

So transmission "thru" this mismatch is |t|=[1-(0.6^2)]^0.5 =0.8

20 LOG 0.8 =-1.9 dB

So with the mismatch, your effective NF should be 1+1.9=2.9 db

With a 1 dB loss saw that corrects the mismatch exactly, you should have a NF=1+1=2 dB

Looks to me like the SAW wins.

Thanks,
Ruri
 

biff44

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Absolutely. The best NF would be a low loss matching network from 50 to 200 ohms. (In reality, the optimum would be a matching network for the best "noise match" impedance, whatever that is )
 

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