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I built this buck converter circuit using the LM2596T-5 (5 V OUTPUT, 12 V INPUT) using the exact components mentioned in the datasheet's first page (schematic below), except for the diode.

I used the 1N5822 because the 1N5824 wasn't available at the store: the only difference between those two lies in the maximum current they can handle, which isn't a problem, since I'll only be driving a load that draws 2 A maximum.

The regulator works just fine with loads that draw anything below 1 A. Once we go past that however, the voltage drops significantly (even reaching 1 V if I insist on drawing more current.)

I immediately suspected that the inductor might be the root of my problem here (33 µH inductor, see pic below.)

I tried swapping it with another inductor of a higher value (1 mH inductor) but the problem persisted.

This probably means that the inductor is getting saturated and is no longer storing sufficient energy for proper operation.

I confirmed this when I changed the 1 mH inductor with a 100 µH inductor in a different package, which in turn, raised the bar to 1.2 A before voltage drop.

How do I deal with this exactly?

Unfortunately the inductors that are available in the stores near me do not have any sort of reference numbers or manufacturer names written on them (only inductance value codes) so I cannot see their datasheets, which is quite a limitation.

I've read in a different post that this issue might be caused by the ON/OFF pin when left floating, but I did check if that was the cause and it definitely wasn't.

Would making an air-core inductor be a valid option? As far as I know, those cannot get saturated, since they have no core, and if so, then would regular connection wires (covered with plastic or silicon) do the trick?

Note: The soldering traces might be the issue here (since I am not very good at soldering) but that still doesn't dismiss the fact that the current is dependent on the inductor.

LM2596 data sheet here

enter image description here

33 µH:

enter image description here

100 µH:

enter image description here

1 mH:

enter image description here

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    \$\begingroup\$ Your inductors are not suitable as their current rating is too low. Find an inductor with a current rating of around 3A and 33uH. An air core inductor will bee physically too large. If you can find a suitable ferrite or powered metal toroid core you can wind your own. There are calculators and examples on the web. You might be able to harvest something suitable out of a PC power supply. \$\endgroup\$
    – Kartman
    Sep 21 '21 at 0:09
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    \$\begingroup\$ If you cannot obtain an inductor that suits easily then an air cored wind your own is doable but probably too large long term this webpage provides a formula. A closer spaced scramble wound coil will have usefully more inductance and work as well. Any insulated wire will work. Wire should be thick enough to have under 1 ohm resistance - but the lower the better \$\endgroup\$
    – Russell McMahon
    Sep 21 '21 at 2:31
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    \$\begingroup\$ Saturation current can be calculated of you know the core characteristics BUT the ones you show are "just far too small". The core you provided a limk for is designed for 16 kHz operation and the LM2596 operates at about 150 kHz. The core MAY still be OKish at that frequency. HOWEVER why not wind an air cored coil as per the link I provided. If the resistance is low (litz or other wire) "it will work. || As a starting test - wind a 30 turn coil scranble wound loosely on two of your fingers - loosely enough that you can get it off. It should work. Less turns probably OK. Experiment. \$\endgroup\$
    – Russell McMahon
    Sep 21 '21 at 12:55
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    \$\begingroup\$ Easy way how to find if you have inductor in saturation is to measure/scope a current through it. If you dont have a current probe, just put a small resistor (100 mOhm) in series with inductor and scope the drop. The shape must be almost linear, without roundings. \$\endgroup\$
    – user208862
    Sep 21 '21 at 18:07
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    \$\begingroup\$ @MicroservicesOnDDD I do appreciate the effort you've put into your answer, and I apologize for the misleading part number ( which I will fix right away ), at first I was a bit confused by your answer, but now that you have edited it, it definitely answers my question. \$\endgroup\$
    – A.H.Z
    Sep 22 '21 at 23:26
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You're probably right about the inductor being the problem. According to the datasheet 33uH might not do it -- 22uH or 15uH is called for. Try this inductor, Chilisin Electronics BWVS00808040150M00 (3.6A Isat, 15 µH Shielded Inductor 3.2A 50mOhm SMD) from Digikey for 48 cents.


EDIT:

I apologize -- I read the datasheet wrong because you said that you had a LM2596T-12, but you really have a LM2596T-5 (based on 12V in, 5V out, and your diagram, expressing the same, and having the LM2596T-5.0 part number on it).

Here is the relevant graph:

enter image description here

You'll see that I'm using a margin of an extra half amp of power, and the intersection falls at L40, which is 33uH at 3.5 Amps min continuous power. In that case, you need a beefier inductor -- my choice is Pulse Electronics Part# PA2729.333NL, 32.9 µH Shielded Wirewound Inductor, 6.7A Max continuous, 18.5mOhm Max, having 8.4A Saturation current limit, at $3.31 for one at Digikey.

An unshielded inductor may work, but may cause EMI or other problems.

If it still doesn't work, then you may have a layout problem, and try again from the guidance given in the datasheet, then come ask us if it still doesn't work.

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  • \$\begingroup\$ I have tested an Air core inductor and it works better than the other inductors i've tried so far, ( ignoring the fact that even so i cannot go beyond 1.6A either because of resonant frequency or EMi , i did fix my layout so I can be certain that it isn't a power loss problem) , as for the inductors you have suggested, unfortunately, I have no way to obtain them in my region, but I might consider buying ferrite cores and experiment with them. \$\endgroup\$
    – A.H.Z
    Sep 22 '21 at 23:29
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    \$\begingroup\$ @A.H.Z -- Andy aka gave an answer on how to measure inductance with just a multimeter, if you have a range of known inductance references. How to measure coils without inductance measuring device?. Just know that the inductor you think is 100uH is probably 10uH, as it is marked "100" and 100uH is usually marked "101". \$\endgroup\$ Sep 23 '21 at 14:46

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