Inductive Degenration

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Hurt_it_Circuit

Joined Oct 2, 2012
53
I am trying to understand how Inductive degeneration helps with noise reduction in a LNA design. I have attached a picture of the schematic I am looking at. My confusion comes from the location where Vout is taken. Since inductors have a zero voltage drop across them why are taking Vout at this location isn't this simply the rail voltage? Also the reason why inductive degeneration is still not clear to me. All I have read about inductive degeneration is that it lowers noise figure but it is not clear to me why this is the case?
 

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AnalogKid

Joined Aug 1, 2013
11,055
In theory, inductors have no resistance and so have no voltage drop - *at DC*. An inductor's impedance is directly proportional to frequency. At 1 GHz, a 1 uH inductor has an impedance of 6.28 Kohm. In your schematic each inductor is doing double duty, passing the static current of the amplifier with allmost no power loss and acting as the load impedance of the collector.

ak
 
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t_n_k

Joined Mar 6, 2009
5,455
With respect to noise figure:
The understanding of this issue isn't trivial. Although, an "obvious" advantage of using a predominantly inductive (reactive) degenerative component in the CMOS source path is the reduction in resistance generated noise that would arise from a predominantly resistive degenerative element of equivalent "impedance" at the operating frequency placed in the source path.

There are other matters at play which lead to complexity in the design process - such as the need for good input matching conditions & stability at the operating frequency.

I would refer you to a very useful (2005) Berkley lecture summary by Prof. Ali M. Niknejad produced as a pdf. You can obtain the document here:

http://rfic.eecs.berkeley.edu/~niknejad/ee142_fa05lects/pdf/lect14.pdf
 
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