I am using LM2576T as a regulator in my design. I made a circuit by the datasheet, but I haven't connected the diode (I haven't got any at the moment). Could it be a reason for switching regulator to overheat? I've tried it with low load currents (100-200 mA) with heatsink installed and the regulator is still overheating.

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Datasheet : Switching regulator datasheet

  • \$\begingroup\$ Is it even putting out 5V? \$\endgroup\$
    – JRE
    Jul 17, 2020 at 8:14
  • \$\begingroup\$ Without that diode, the Vout pin may go -ve several hundred volts when th eswitch turns off, which won't do the regulator any good. Replace it when you add the diode. \$\endgroup\$
    – user16324
    Jul 17, 2020 at 10:12
  • \$\begingroup\$ The diode is a required part of a switching regulator. It is not optional. It's not for ESD protection or something like that. The regulator doesn't work without it. \$\endgroup\$
    – user253751
    Jul 17, 2020 at 10:32

2 Answers 2


From the datasheet:

8.1.5 Catch Diode

Buck regulators require a diode to provide a return path for the inductor current when the switch is off. This diode must be placed close to the LM2576 using short leads and short printed-circuit traces.

Because of their fast switching speed and low forward voltage drop, Schottky diodes provide the best efficiency, especially in low output voltage switching regulators (less than 5 V). Fast-recovery, high-efficiency, or ultra-fast recovery diodes are also suitable, but some types with an abrupt turnoff characteristic can cause instability and EMI problems. A fast-recovery diode with soft recovery characteristics is a better choice. Standard 60-Hz diodes (for example, 1N4001 or 1N5400, and so forth) are also not suitable. See Table 3 for a Schottky and soft fastrecovery diode selection guide.

It would surprise me if your "regulator" is even regulating without the catch diode.

Whatever your circuit is doing, it isn't operating as a proper buck converter. You should stop messing with it. Install a proper diode before you use it again, or else you may need to replace the chip as well as install a diode.


The diode is fundamental to the device operation. Without it the circuit stores energy in the inductor and then, instead of it being directed to the load with the help of the diode, it splashes that energy all over the output transistor and probably damages it as well as making it get hot: -

enter image description here

If you have an example circuit, then don't test it it until it's fully built (unless you understand or have the knowledge that you can).


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