# Effect of Rs in a CS amplifier with degeneration

I was studying the topology of CS amplifier with degeneration (a resistor connected from source to ground and I have two question about this topology:

1. It is clear for me that Rs will decrease the gain (cons.) and increase the output resistance (cons.).

I know also that the input capacitance will decrease (pros.). Is there any other motivation behind using this series resistor Rs?

1. They also mention that Rs adds noise, while adding an inductor instead will not add noise and will give the same (pros.) as the resistor. I know that resistors give thermal noise, but why inductors don't? Is it because they don't dissipate real power? And is this the only difference between adding an L or an R? ## 1 Answer

HaneenSu, you forgot to mention the most important effect caused by such a resistor Rs: Negative feedback!

Here are the main effects of this (current controlled voltage) feedback:

• Input resistance increase
• Decrease of gain, but the reduced gain value is less dependent on transistor parameters (primarily determined by RD and RS)
• Bandwidth increase
• Improved linearity (less total harmonic distortion)
• Small influence on output resistance.

Dont forget that the dc drop across RD determines the DC operational point of the amplifier stage. An inductor has a DC resistance that is negligible.

• Also Rs sets the inpit bias point when input signals are 0V referenced i.e. it self-biases the transistor. +1 – Andy aka Nov 1 '15 at 10:24
• @Andy : yes, self-bias is a big pro. And contrary to the question, an inductor won't do that... – Brian Drummond Nov 1 '15 at 11:35
• Isn't the input resistance in both cases infinite? Can you clarify please? Also, can you please explain how we consider it as a feedback element even though the Rs terminals are not tied to input and output nodes (gate and drain respectively) ? – HaneenSu Nov 1 '15 at 19:00
• The input impedance is large but not infinite. However, I agree that for FETs this property of negative feedback is not too important. Rs provides feeback because the output CURRENT (Id=Is) produces a VOLTAGE at one of the terminals of the differential input (source). An output current increase leads to a voltage reduction at the input. – LvW Nov 2 '15 at 7:12