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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.

There was this vague idea that struck about using a zener diode. I am not sure how I can implement this and is this actually practically viable? Circuit

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closed as unclear what you're asking by The Photon, uint128_t, Andy aka, dim, placeholder Jun 23 '16 at 19:13

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ 1. What is the relationship of the schematic to the circuit you are designing (the schematic looks like a standard buck converter with constant-voltage output, but you ask about a constant current source)? 2. What is the schematic of the actual circuit you designed? 3. What inductor part number did you use? 4. What was your idea about using a zener? \$\endgroup\$ – The Photon Jun 23 '16 at 17:07
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    \$\begingroup\$ @VoltAmpereWatt your understanding about Zener diodes is completely wrong. Or at least your description is. I afraid your background is insufficient to design power supplies. \$\endgroup\$ – Eugene Sh. Jun 23 '16 at 17:18
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    \$\begingroup\$ Also, maybe you're not understanding this: D1 in your schematic is a schottky diode, not a zener. \$\endgroup\$ – The Photon Jun 23 '16 at 17:27
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    \$\begingroup\$ If you want to know how a zener works, do some research (say, on this site or Wikipedia). If you still have specific questions, come back and ask a new question. But this is not a chat site, we can't just answer broad questions like "how does a zener work" in comments. \$\endgroup\$ – The Photon Jun 23 '16 at 17:29
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    \$\begingroup\$ Oh, with regular inductor you mean a SMALL SIGNAL inductor, designed for filters at several mA? Can you already guess what went wrong, maybe? You're trying to push a proverbial elephant through a garden hose. Components have specifications, which are there for a reason. \$\endgroup\$ – Asmyldof Jun 23 '16 at 17:34
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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: -

enter image description here

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!

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