Find Voltage Gain: Deriving Vgs=Vds - Small Signal Circuit

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The discussion centers on deriving the equation Vgs = Vds in the context of small signal circuits involving MOSFETs. The key insight is that Vgs, the gate-source voltage, is equal to Vds, the drain-source voltage, when the transistor is properly biased. The relationship is established through the understanding that Vds = VDD - IDRD, where VDD is the supply voltage and IDRD is the voltage drop across the drain resistor. The voltage at the gate is equal to the voltage at the drain due to the direct electrical connection through RG, which has no current flowing through it.

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can someone please tell me how did they derive with the above equation?

my point of view:
Vgs = Vin
Vds = Vout

so how did they even get Vgs = Vds?

below shows the Small signal circuit:
2222-11.png


might be a stupid question to ask. but i am really confuse on when to look at the original circuit, when to look at the AC circuit, when do i remove the load, etc.

Thanks for helping and sorry for asking such a simple question here.
 
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They're using capital letters, so I don't think they're talking about the small signal values, but rather the DC bias values. It's been a while since I've looked at MOSFETS in detail, but assuming that the transistor is biased in such a way that it will conduct, then

VDS = VDD - IDRD

makes perfect sense, because current flows from drain to source across the drain resistor, so the voltage at the drain will be the supply voltage minus the voltage drop across the drain resistor.

The voltage at the gate is equal to the voltage at the drain, because the two are electrically connected by RG. Since no current flows across RG, there is no potential difference across it. So, it's as though the two terminals are connected by a wire.

As far as I can tell, it's as simple as that.
 

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