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My Vin=10VDC and my R1=1.25 ohm.

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If I have a 'load' of 1 ohm, what is going to happen?

Initially, will the current be above 1A and settle down to 1A. Or will this actually limit the current to 1A. What is going to happen to the voltage through the circuit. My 'load' will have 1V drop across it, with 1.25V across R1, and ~3V drop across the LM317. Just based off these simple calculations, does that mean my Vin is really only going to be 5.25V?

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    \$\begingroup\$ What does the title have to do with the question? \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 3 '13 at 13:41
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    \$\begingroup\$ It means your calculations are incorrect. If Vin is 10V, Vin is 10V! It cannot change as a function of the output conditions (unless it's a terrible quality source, but for hypothetical analysis we assume it's an ideal source and will always be at 10V) \$\endgroup\$ – Adam Lawrence Sep 3 '13 at 13:45
  • \$\begingroup\$ If the current ever goes above 1A initially, I would like to know roughly how long this would take to drop back down to 1A. Or if the current will go above 1A at all. \$\endgroup\$ – brett s Sep 3 '13 at 13:50
  • \$\begingroup\$ If Vin will always be 10V, where does the rest of the voltage go on its way to ground? Testing this circuit with a current limiting power supply, the voltage coming from the supply is much less than 10V. \$\endgroup\$ – brett s Sep 3 '13 at 13:52
  • \$\begingroup\$ If your 10V supply is limiting the voltage, it's no longer a 10V supply! All of the excess voltage should appear across the regulator if the source doesn't crap out. \$\endgroup\$ – Adam Lawrence Sep 3 '13 at 13:54
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Simplified answers:

There isn't any internal error amplifier compensation for the LM317 so it should try to control the output extremely quickly; as fast as the internal amplifier and Darlington pair will allow.

With a purely resistive load, I wouldn't expect any current overshoot which also means there shouldn't be any voltage overshoot.

Your 3V drop across the LM317 is incorrect. Vout will be 2.25V (1A through the two resistors) and since Vin is 10V, the voltage across the LM317 will be the difference (7.75V) and at 1A current, you'll have 7.75W of dissipation. Better get a heat sink.

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  • \$\begingroup\$ What sort of behavior should I expect if I use an inductive load such as a brushless fan motor? \$\endgroup\$ – brett s Sep 3 '13 at 14:53
  • \$\begingroup\$ If you have reactive elements there, you obviously won't have an in-phase relationship between current and voltage any longer. The formula you're using assumes current and voltage are in phase (i.e. a resistor), and would need to change to deal with reactive elements. \$\endgroup\$ – Adam Lawrence Sep 3 '13 at 17:00

protected by Kortuk Sep 3 '13 at 15:30

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