See this circuit:

enter image description here

Applying KVL around the gate-source-GND-gate loop, I get Id = (Vg - Vgs)/Rs. I know the typical argument begins with, if Id increases then Vs increases, but I am not sure of how it continues.

There is something else I don't understand. The reasoning is based on the previous KVL equation, but it can be further simplified down to Id = Vs/Rs since Vgs = Vg - Vs. Well, if we examine Id = Vs/Rs, an increase in Id corresponds to an increase in Vs, and that's all there is to it. No negative feedback. Usually the argument is based on the non-simplified equation Id = (Vg - Vgs)/Rs, I am not sure why since in reality Vg cancels out.

How does Rs stabilize the amplifier?


3 Answers 3


The gate to source voltage, \$V_{GS}\$, controls the transistor. As this voltage increases, the channel becomes more conductive, and more current flows through the drain and source.

The voltage across a resistor is given by Ohm's law:

$$ v = I R $$

So, as more current flows the drain and the source, more current flows through \$R_S\$, and, by Ohm's law, the voltage across \$R_S\$ increases. Since the gate voltage isn't changing, the gate-source voltage decreases, since the source is now closer to the gate.

There will be one point at which this is stable. If the transistor is too much off, there won't be enough current in \$R_S\$, and \$V_{GS}\$ will be high enough to turn the transistor on more, increasing the current in \$R_S\$.

If the transistor is too much on, there will be too much current in \$R_S\$, and \$V_{GS}\$ will be low enough to turn the transistor off more, decreasing the current in \$R_S\$.

If the gain of the transistor is infinite, then as long as the input is not pushing the output into the supply rails, then the gate-source voltage will be constant, right at the transistor's threshold voltage: \$V_{GS} = V_{th}\$. Since the transistor's gain is infinite, it has unlimited ability to correct any deviation from this through the feedback mechanism mentioned.

The gain of the circuit approaches \$R_D/R_S\$ as the transitor's gain increases. Or put another way, as \$R_D/R_S\$ decreases, the transistor's gain becomes less relevant to the gain of the whole circuit. This is how the simplifications you mention are made. That is, almost all of the change in gate voltage appears as a change in \$V_{R_S}\$, and only a negligible amount as a change in \$V_{GS}\$.

  • \$\begingroup\$ To be clear, the "gain" of the transistor is its transconductance gm. This quantity can NEVER be infinite. And it is wrong to say that "the ideal gain would be RD/RS". This sentence is misleading. In contrast, this is a simplified expression which does not apply for low RS values. \$\endgroup\$
    – LvW
    Oct 17, 2018 at 9:05
  • \$\begingroup\$ @LvW True, but isn't any idealized model of a component a simplification with certain assumptions that have no practical effect on the circuit under many but not all applications? In a circuit such as this, Rs exists precisely to render deviations from the simplified model negligible: the statement is not misleading but rather embodies explicit intent of the circuit. \$\endgroup\$
    – Phil Frost
    Oct 17, 2018 at 13:08
  • \$\begingroup\$ Yes - full agreement to your first sentence. But I never would use the term "ideal gain" for the ratio RD/RS. It is, rather, a simplified expression that holds for relatively large RS values only (large if compared with 1/gm). To me, the term "ideal" sounds a bit misleading, that`s all. \$\endgroup\$
    – LvW
    Oct 17, 2018 at 14:50

How does Rs stabilize the amplifier?

\$V_{GS} = V_G - V_S \$

\$V_S = I_D R_S \$

\$\Rightarrow V_{GS} = V_G - I_D R_S \$

Now, \$V_G\$ is constant so, if \$I_D\$ increases, \$V_{GS}\$ decreases.

But, if \$V_{GS}\$ decreases, \$I_D\$ decreases.\$^1\$

Thus, an increase in \$I_D\$ acts to decrease \$I_D\$. This is the hallmark of negative feedback.

\$^1 I_D \approx K(V_{GS}-V_{TH})^2 \$

  • \$\begingroup\$ But since Vg is constant, if (Vgs = Vg - Vs) decreases then Vs increase, which means that Id increases... \$\endgroup\$ Aug 4, 2013 at 23:12
  • 1
    \$\begingroup\$ @DouglasEdward, if V_GS decreases, I_D decreases. Agree? Thus, if I_D increases, V_GS is reduced, thus reducing I_D. Negative feedback is self-correcting in this way. \$\endgroup\$ Aug 5, 2013 at 2:16

Id increases => Vs increases => Vgs decreases => Id decreases

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    \$\begingroup\$ Welcome to EE.SE. While all contributions are welcome, they are expected to be of high quality and substantial content. Unfortunately your answer doesn't add anything not already mentioned in the existing answers from 5 years ago. \$\endgroup\$ Oct 17, 2018 at 7:19

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