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Buck regulator with LM2596-adj got damaged.

polashd

Member
I build a circuit with LM2596HV-adj (smd IC). mounted it on a small heat sink. Input 37v dc input cap 100uf, output variable from 1.25v to 15v dc, output cap 220uf +330uf, inductor 100uh (toroidal). I used 2 flyback diodes (fast diode), one across output one between output pin and input pin of ic. Connected a dc motor of 12v 3A. when it runs output dopes to 10v current draw 1.5A max (transformer is not that powerful I guess ). I ran it for 5-10min. IC and transformer gets hot (to touch, not much). Next morning I ran it for 2 min only- the IC got damaged the output became 32+v without load (10v with motor). this is the second time same happened.

I asked chatgpt- got some suggestions – like Overheating, Inductor Selection (recommended 22μH to 33μH), Inrush Current and Load Spikes, Insufficient Input Filtering, Over-voltage Protection, Load Characteristics (Motor), Output Capacitor Insufficiency.
Based on the suggestions I made some changes 1) reduced inductor value around 40uH, 2) added one 100uf capacitor parallel to the existing input capacitor (100uf).

Unfortunately the same thing happened again after 5-6min of running (IC got damaged for the third time).

What is the problem, what to do? Pls help

Nb.: my inductor core is Iron powder T-106-26.
 

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  • LM2596-adj Buck converter diagram.jpg
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Layout is important. Can you please share a picture of your circuit?

You need a really good and short (low inductance) connection between the negative of the input capacitors and the anode of the freewheel diode D5. That jumper JP1 shouldn't be there as the slightest bad connection an will cause all sorts of problems. It should be a solid connection.

During the time when the switch in the LM2596HV is on, current will build up, flowing from the input capacitors, through the inductor and load and back though the jumper into the capacitors. When the switch turns off, the inductor current will flow through the load and D5, and the change in current path is really fast.

Any resistance in JP1 will cause voltage offsets when the switch is on. The current taken will be the load current which could be large when the motor is starting.

Any inductance in JP1 will cause voltage spike when the switch in the LM2596HV turns on or off and the current in JP1 starts or stops.

TI, the maker of the LM2596 has this in the data sheet.

1726048906644-png.147077

 

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Let's see your pcb layout. Layout is the under-appreciated "secret sauce" of power conversion design. Too much stray inductance in series with the schottky diode could cause problems. Note that your schematic shows D1 being connected directly to the IC output pin. I would highly recommend using SMT. Also your input capacitor needs to be up close and tight to the IC input which is shown similarly. And use the ground plane...
 

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