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I was reading about the self-biased common-source stage and I understand that it has an inherent 'feedback' mechanism against variations in threshold voltage. For example, if Vth decreases, current would like to increase, but the voltage at the drain and hence the gate will also decrease, so current falls back down, correcting itself.

enter image description here

My question is, what's the point of the feedback Rf resistor? Current to the gate will always be zero, why do we need a Rf resistor?

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  • \$\begingroup\$ Presumably you are applying an input signal (AC coupled) to the gate. Rf (combined with stage gain and the Miller effect) form the input impedance. \$\endgroup\$
    – user16324
    Jul 12, 2020 at 13:08
  • \$\begingroup\$ If implemented in BIPOLAR, the R_base_collector has a special purpose; a rather stable current is drawn. \$\endgroup\$ Jul 12, 2020 at 13:20
  • \$\begingroup\$ @BrianDrummond I see. So it's just to modify the input impedance? \$\endgroup\$ Jul 12, 2020 at 13:42
  • \$\begingroup\$ No, it is to provide the input impedance. Consider : the higher the input impedance, the better. But if it was infinite you would have no bias. \$\endgroup\$
    – user16324
    Jul 12, 2020 at 13:44

2 Answers 2

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enter image description here

If you didn't have \$R_F\$ (i.e. it were shorted out), the input impedance to the gate would be \$R_D\$ and, the gain from gate to drain would be unity. Not much of a circuit without \$R_F\$ really.

In the circuit above I've added \$R_{IN}\$ and the gain will be approximately: -

$$\dfrac{V_D}{V_{IN}} = -\dfrac{R_F}{R_{IN}}$$

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  • \$\begingroup\$ Ah I see. So if we left out Rf and replaced it with a short, the MOSFET would still be in saturation, however our input and output nodes would now be shorted, so pointless as an amplifier. By including Rf, we still allow for the MOSFET to be in saturation yet we also allow the output node to move now. \$\endgroup\$ Jul 12, 2020 at 15:24
  • \$\begingroup\$ Hmm. I'm still confused actually. If we supply the input with an ideal votlage source (infinite impedance) and the gate has high impedance too, then no current will ever flow through Rf, so doesn't that make it redudant and the same as short for both AC and DC? \$\endgroup\$ Jul 12, 2020 at 15:26
  • \$\begingroup\$ @AlfroJang80 An ideal voltage source has zero impedance. \$\endgroup\$
    – Andy aka
    Jul 12, 2020 at 15:45
  • \$\begingroup\$ Ah. That would make more sense. Apologies, I messed that up. \$\endgroup\$ Jul 12, 2020 at 16:57
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The resistor RF sets the DC operation point voltage at the gate to the same value as the voltage on the drain. Hence the name of the structure: "self-biased". If its value is much higher than RD, then you can even neglect it for AC analysis.

OFF, but might be interesting: A similar structure is often used for broadband common source LNA stages, where the transconductance of the transistor provides the broadband 50Ohm impedance through the feedback. In such cases resistor RF also determines the gain.

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