I was helping a friend to design a constant 3A current source using an LM2576 to drive a load. Problem is that the 100uH inductor keeps burning every time the circuit is switched on. One option is to swap the regular inductor with a coil inductor. However, it is expensive and we cannot afford it yet.
closed as unclear what you're asking by The Photon, uint128_t, Andy aka, dim, placeholder Jun 23 '16 at 19:13
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The chosen regulator (LM2576) is probably operating in continuous conduction mode on normal and full loads so the inductor will be seeing a peak current that might be about 20% higher than full load current. If full load current is 3A (as per the circuit in the question) then the inductor has to be able to handle 3.6 amps without too many losses.
Losses include copper resistive losses and core saturation.
One option is to swap the regular inductor with a coil inductor.
This makes no sense because all inductors are coils. Maybe because it "looks like regular resistor with color bands" you were fooled into thinking it wasn't a "wound/coiled" device. It is.
Those "regular resistor with color bands" are unsuitable for your application. This one: -
Is 150 uH but only rated to currents of 0.15 amps. It has very fine windings under the paint/enamel and has a dc resistance of 5.4 ohms i.e. it will fry!