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I want to start building a tube headphone amp from the details on DIY Audio Projects, but I'd like to understand what I'm doing there too.

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Source: DIY audio projects.

While I get most of it, I can't understand why the LM317 is placed after the output MOSFET and not before it. Can anyone shed some light on this choice?

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  • \$\begingroup\$ What do you mean "before"? \$\endgroup\$ – diverger Nov 7 '14 at 10:49
  • \$\begingroup\$ I would've thought the LM317's output should be connected to the MOSFET drain (before it, that is). In the provided schematic, the regulator has it's input connected to the MOSFET source. I'm missing the logic of this. \$\endgroup\$ – CatalinM Nov 7 '14 at 10:54
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The LM317 in this circuit is not being used as a voltage regulator. It's being used as a constant-current load. It's set up with a resistor in such a way as to always draw a constant amount of current from the MOSFET. In that configuration, the current flows through the LM317 In-to-OUT-to-ground.

But if the LM317 was set up in its normal way as a voltage regulator, then you're right, the supply would go to IN and the MOSFET would be connected to OUT.

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  • \$\begingroup\$ Thanks! Small side question though: why does the MOSFET need a constant-current load? \$\endgroup\$ – CatalinM Nov 7 '14 at 11:06
  • \$\begingroup\$ @CatalinM The LM317 takes the place of an ordinary resistor as a load for the MOSFET. Using a constant-current load is one way to make an emitter-follower stage more linear (this circuit is the MOSFET equivalent of a bipoolar emitter-follower). Consider what happens if you replace the LM317 with a resistor and then the MOSFET's conductance varies. The constant-current version will allow a greater signal swing before clipping. \$\endgroup\$ – John Honniball Nov 7 '14 at 11:10
  • \$\begingroup\$ @John Honniball: If put LM317 at the drain of MOSFET, then the source will be connected to C3 than to ground? Where the bias current go? \$\endgroup\$ – diverger Nov 7 '14 at 11:11
  • \$\begingroup\$ @diverger If the LM317 is in the drain side of the MOSFET, then it's no longer an emitter-follower. It won't work in that configuration. C3 and source will be grounded; no output. \$\endgroup\$ – John Honniball Nov 7 '14 at 11:14

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