Gain spec for an LNA design

S21 > 10dB, find the minimum voltage gain required.

I think that S21 is equal to 2*Vgain

So if that is right:
20log(2*Vgain)=10dB
2*Vgain=10^0.5
Vgain=1.58

That does not seem like much voltage gain. Did I do this right?

[quote="snafflekid"]S21 > 10dB, find the minimum voltage gain required.

I think that S21 is equal to 2*Vgain

S21 means voltage gain, why not S21 equal Vgain ?

S21 is the ratio (output port voltage)/(input port voltage) and I agree this seems like the definition of voltage gain. But the voltage at the input port is not the voltage coming from the source.

There is also the definition |s21|2=(power delivered to load)/(power available from source). From this definition I get S21=2*Vgain. I show this in the attachment

The question came up in this post also https://www.edaboard.com/ftopic60916.html

I am not sure my reasoning is correct. Can someone confirm before I go to an interview and am wrong? :D

No. This is wrong.

S21 = 10db means that you have 10x power gain -> sqrt(10) voltage gain

Hi
If LNA's load impedance is equal to the s-parameters reference impedance i.e. Γ(load)=0 then voltage gain is equal to
Volt gain = S21/(1+S11)
which is reduces to
Volt gain = S21 if LNA is perfectly matched at input.

But in your case minimum voltage gain is obtained when LNA is badly matched (S11 ≈1) and equal to S21 / 2 ≈ 4 dB = 1.58.

Regards, pavel.