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Design >> RF Design >> Consideration of noise voltage and current in cascode LNA
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Message started by VINAY RAO on Apr 21^st, 2014, 6:54am

Title: Consideration of noise voltage and current in cascode LNA
Post by VINAY RAO on Apr 21^st, 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.

Title: Re: Consideration of noise voltage and current in cascode LNA
Post by aaron_do on Apr 21^st, 2014, 6:07pm

Hi,

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

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

Title: Re: Consideration of noise voltage and current in cascode LNA
Post by VINAY RAO on Apr 21^st, 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 R_in>R_BIAS. In this case, R_BIASnoise 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 (R_L) >> source impedance (Rs) then we have to consider noise voltage or else in another case (R_L<<R_S) we have to consider noise current” ?

Regards,
Vinay Rao.

Title: Re: Consideration of noise voltage and current in cascode LNA
Post by aaron_do on Apr 22^nd, 2014, 12:29am

Hi Vinay Rao,

for a back-of-the-envelope calculation, you are perfectly correct. i.e. if R_in >> R_BIAS, 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