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Design >> RF Design >> Noise contribution in Common source amplifier
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Message started by abhishek dwivedi on Mar 5^th, 2014, 8:46pm

Title: Noise contribution in Common source amplifier
Post by abhishek dwivedi on Mar 5^th, 2014, 8:46pm

Hi,
I got confused about one noise measurement in Commom source topology. It is a common source and load is a resistance Rd. Output noise is proportional to root(Rd) , it means it is proportional is some way. Therefore, if Rd increases, so as noise power. But if we consider inut referred noise , it is actually inversely proportional to Rd.
Can someone explain this ?

Best Regards
Abhishek

Title: Re: Noise contribution in Common source amplifier
Post by carlgrace on Mar 7^th, 2014, 8:48pm

Because the gain is also proportional to Rd. When you refer the voltage noise to the input you divide by the gain squared. Therefore when you double Rd you reduce the input referred noise by root(2). So both the output voltage noise and the gain increase, by the gain wins!

Another way to look at is in the current domain. Larger Rd means lower current noise (by Thevenin). You divide that by gm^2 to get the input referred voltage noise.

Larger R, smaller input referred noise.

Title: Re: Noise contribution in Common source amplifier
Post by abhishek dwivedi on Mar 7^th, 2014, 11:18pm

I understand the math part. I want the intuition behind it. Output noise power increases as Rd increases. But input referred noise power decreases. Why this paradox? I mean if we are to reduce the noise say thermal, then we should reduce the value of Rd. I am getting confused with this input referred noise. I understand physically there is no such thing, it is just a tool of understanding circuit noise, isn't it? But the real thing is output noise power. Please forgive my ignorance, I am learning.

Title: Re: Noise contribution in Common source amplifier
Post by aaron_do on Mar 9^th, 2014, 6:12pm

Hi,

treat "input-referred noise" as a way to consider both the noise and the desired signal. When you increase Rd, you increase the output noise, but because the gain increases, the desired output signal strength increases even more. So your output SNR is higher.

regards,
Aaron