hi,
There is a high speed version of the 1N400x ie: UF400x.
E

The UF400x series is perhaps a bit marginal for the output diode for 1 A output. They are rated for 1 A but will run moderately hot and I'd be more concerned with adequacy of heat spreading foil on the PCB. There are many other ultrafast diodes available. There are UF versions of the 1N5400 series. I've never looked for 2 or 3 A rated ultrafast in surface mount. I would recommend staying with a Schottky if a suitable type could be found. I'm quite confident they exist, but I can't make a recommendation offhand.

 

Yes, indeed, do not use IN400x series or the 1N4148.
A suitable high speed diode of at least double the output current is needed. And what is your inductor? Just 100uH is not enough to go by. What is it's current and frequency rating?

 

Yes, indeed, do not use IN400x series or the 1N4148.
A suitable high speed diode of at least double the output current is needed. And what is your inductor? Just 100uH is not enough to go by. What is it's current and frequency rating?

current is around 500mA

 

current is around 500mA

There for a start is a bit of a problem. If the inductor is rated at 500mA and you want 500mA to 1Amp as specified earlier, not going to work.

Have you tried using the on line calculator?
https://webench.ti.com/power-designer/switching-regulator
It is worth creating an account to use the calculator.
For instance, a 1Amp 5V supply requires an inductor with 3Amp current .

Here is an example output and is has recommended part numbers.
The full calculator does a lot more. This is just a bit off it.

 

Also consider the layout.

Is it on a PCB? or breadboard?
Have you looked at the wave forms to make sure it's stable and operating correctly?

"leave no stone un-turned"

 

Yes, it is at the input for reverse polarity protection and has nothing to do with the problem.

The first sentence was towards the thread starter.

The second was towards you.

 

There for a start is a bit of a problem. If the inductor is rated at 500mA and you want 500mA to 1Amp as specified earlier, not going to work.

Have you tried using the on line calculator?
https://webench.ti.com/power-designer/switching-regulator
It is worth creating an account to use the calculator.
For instance, a 1Amp 5V supply requires an inductor with 3Amp current .
View attachment 163887
Here is an example output and is has recommended part numbers.
The full calculator does a lot more. This is just a bit off it.

thanks for the link . it is a good referene for now and future also

 

Also consider the layout.

Is it on a PCB? or breadboard?
Have you looked at the wave forms to make sure it's stable and operating correctly?

"leave no stone un-turned"

it is on PCB. yes have checked the waveforms and it is stable.

 

it is on PCB. yes have checked the waveforms and it is stable.

Indeed it should be on a PCB that gives adequate consideration to ground impedances and other potential couplings. So it was a good choice. I am thinking that if the series diode is still getting way to hot that possibly a series resistor plus a pair of capacitors, as a voltage dropping means so that the regulator does not need to handle such a high input voltage. And the 2 caps. say 0.1 and 100, could provide an adequately low source impedance. The alternative may be to try an inductor rated for a higher load current.

 

Unless the inductor is much too small, I doubt if it has anything to do with the failure of the output diode.

If the inductor were to saturate, the current limiting in the switcher IC would "trip." This would result in reduced output voltage. With 48 V in and 5 V out, the ripple current in the inductor will be fairly low (duty cycle will be around 12-13%; at 52 kHz this equates to about 120 V·µs, so about 1.2 A p-p ripple current with a 100 µH inductor. With only 500 mA output current, this puts it into discontinuous mode which of course means the duty cycle will be further reduced. If the inductance were substantially reduced, it would push the converter further into discontinuous mode which would increase the peak to average current ratio in the diode, but I really doubt that would be any sort of issue with a diode rated for 3 A average current.

The schematic and components look OK to me in general terms.

Because of the comparatively low switching frequency, the reverse recovery loss from an ultrafast diode isn't going to be great. A Schottky diode would still have somewhat lower forward voltage, but once you get up over 60 PIV rating, the forward voltage will be higher, so all in all an ultrafast won't be very much higher loss than a Schottky.

I had a quick look on Digi-Key and there are several 2 or 3 A Schottky diodes rated 70 to 100 V available.