Electronic – Time response of LM317 used as current regulator

currentvoltage-regulator

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?

Best Answer

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.