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Consideration of noise voltage and current in cascode LNA (Read 801 times)
VINAY RAO
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Consideration of noise voltage and current in cascode LNA
Apr 21st, 2014, 6:54am
 
Hello all,

As shown in cascode LNA, RBIAS is usually made large enough so that it adds very less noise current. But if we calculate noise voltage (sqrt 4KTR) then it is proportional to resistance value. Won't this large RBIAS couples large noise voltage to M1?

Why noise current is considered here instead of noise voltage? Along with input signal voltage, doesn't this RBIAS noise voltage also appears in Gm through an amplification factor gm*Qin?

Regards,
Vinay Rao.
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aaron_do
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Re: Consideration of noise voltage and current in cascode LNA
Reply #1 - Apr 21st, 2014, 6:07pm
 
Hi,


when you say it is a noise source of 4kTRBIAS you forgot that its actually a noise source of 4kTRBIAS in series with a resistance RBIAS. So if the gate input impedance is 50 ohm (for example), then there will be a voltage drop at the gate from RBIAS equal to 50/(50+RBIAS).

In the end it doesn't matter whether you consider it a voltage or a current since you will get the same answer.


regards,
Aaron
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VINAY RAO
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Re: Consideration of noise voltage and current in cascode LNA
Reply #2 - Apr 21st, 2014, 11:29pm
 
Thanks a lot Aaron. It makes sense. Still I want to clarify about the following.

1> If suppose we consider the frequency f<<f0, then situation arises as Rin>RBIAS. In this case, RBIAS noise voltage dominates right? So here, is it logical to consider noise voltage rather than noise current?

2>Can we generalize it by saying “Whenever, driving load (RL) >> source impedance (Rs) then we have to consider noise voltage or else in another case (RL<<RS) we have to consider noise current” ?

Regards,
Vinay Rao.
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aaron_do
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Re: Consideration of noise voltage and current in cascode LNA
Reply #3 - Apr 22nd, 2014, 12:29am
 
Hi Vinay Rao,


for a back-of-the-envelope calculation, you are perfectly correct. i.e. if Rin >> RBIAS, then you can ignore the resistor divider and just say that the noise voltage appears directly at the gate. The opposite is true for current. If you really write out the actual equations, then you will get the same answer using voltage and current.


regards,
Aaron
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