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delete reference to incorrect circuit
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Dave Tweed
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In the ideal world, it can't — but the voltage can still change.

Your diagram shows a source follower (common-drain). The source voltage will track (with an offset) the gate voltage, even though the channel current is held constant.

In a common-source circuit, the output voltage is a function of the effective resistance of the current source (in parallel with any load resistance). With an ideal source, that resistance is infinite, resulting is just two output states. But in any real circuit, the effective resistance is very high, but finite. This high load resistance creates a high voltage gain for the amplifier stage.

In the ideal world, it can't — but the voltage can still change.

Your diagram shows a source follower (common-drain). The source voltage will track (with an offset) the gate voltage, even though the channel current is held constant.

In a common-source circuit, the output voltage is a function of the effective resistance of the current source (in parallel with any load resistance). With an ideal source, that resistance is infinite, resulting is just two output states. But in any real circuit, the effective resistance is very high, but finite. This high load resistance creates a high voltage gain for the amplifier stage.

In the ideal world, it can't — but the voltage can still change.

In a common-source circuit, the output voltage is a function of the effective resistance of the current source (in parallel with any load resistance). With an ideal source, that resistance is infinite, resulting is just two output states. But in any real circuit, the effective resistance is very high, but finite. This high load resistance creates a high voltage gain for the amplifier stage.

Source Link
Dave Tweed
  • 178.3k
  • 17
  • 242
  • 418

In the ideal world, it can't — but the voltage can still change.

Your diagram shows a source follower (common-drain). The source voltage will track (with an offset) the gate voltage, even though the channel current is held constant.

In a common-source circuit, the output voltage is a function of the effective resistance of the current source (in parallel with any load resistance). With an ideal source, that resistance is infinite, resulting is just two output states. But in any real circuit, the effective resistance is very high, but finite. This high load resistance creates a high voltage gain for the amplifier stage.