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The title says enough, but let’s give a real life example too. Take one of those LM2596 35V boards. What will happen to it if it gets 50V on the input?

Most likely it will burn, but will this voltage propagate to the load? Will it get damaged too? Is there any kind of protection? Does it make a difference if the over-voltage is a spike or continuous?

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    \$\begingroup\$ Use the HV version- some are rated up to 63V. \$\endgroup\$ – Spehro Pefhany Apr 9 at 20:18
  • \$\begingroup\$ It was rated for 40 and not 2 exceed 45V so plan on obeying the recommend limits or get a repuation for bad designs. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 9 at 20:22
  • \$\begingroup\$ It was more of a theoretical question, in case of a mistake or something. \$\endgroup\$ – php_nub_qq Apr 10 at 5:45
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It's in the datasheet:

Stresses beyond those listed under Absolute Maximum Ratings may cause permanent damage to the device. These are stress ratings only, which do not imply functional operation of the device at these or any other conditions beyond those indicated under Recommended Operating Conditions. Exposure to absolute-maximum-rated conditions for extended periods may affect device reliability.

If you check the functional block diagram, an overvoltage will most likely damage the transistors, failing short, so, shorting the IC's input to the output and present the input voltage across the catch diode, the output capacitor and the load.
The catch diode may get damaged (when its rating is less than 50V), the output capacitor will likely blow/explode when its rating is too low.

Typically, the input capacitor is rated 1.5 to 2 times the intended input voltage, so, it most likely will survive.
This input capacitor may also helps shorting the overvoltage when it is a (short) spike. In section 9.2.1.2.5 it is advised to use a low ESR capacitor for this reason.

A low ESR aluminum or tantalum bypass capacitor is required between the input pin and ground pin to prevent large voltage transients from appearing at the input. This capacitor must be placed close to the IC using short leads.

enter image description here

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    \$\begingroup\$ The statement regarding the necessity of the input capacitor to the regulator has much more to do with the switching transients caused by the switching regulator itself than externally applied transients. That is the reason it is specified that this capacitor must be as close to the input pin as possible. There is no doubt that the input capacitor can filter the input supply ripple but in a general case it is usually better to provide separate input transient protection via a clamping type circuit or an L / C type filtering. \$\endgroup\$ – Michael Karas Apr 10 at 9:31
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As power semiconductors often fail short-circuit, there is every danger it can pass the full input voltage to the output.

After an overvoltage spike destroys the power switch, normal input voltage may be destructive to the output.

Sometimes you will see explicit protection, e.g. a crowbar or large zener across the output combined with a fuse on the input. A cheap buck module may omit this...

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Well, the LM2596 will burn out since its internals were not desigend to withstand much more than 35V being applied. Since semiconductors normally fail short, this would either then pass the 50V to the input or short to GND. I would not count on protection since why would you ever apply 50V to a 35V device?

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    \$\begingroup\$ According the LM2596 datasheet operating voltage is up to 40V and abs max voltage is 45V \$\endgroup\$ – Huisman Apr 9 at 20:18

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