# how to select schottky diode on this device?

this device is buck regulator click here to see the datasheet i need to use this to reduce the voltage from 10V to 5V, and i use the schematics on figure 1.1.2A but the value of the diode is not specified anywhere, how to select the correct value of this schottky diode?

Schottky Diode Selection The external freewheeling diode supplies the current to the inductor when the high side NMOS switch is off. To reduce the losses due to the forward voltage drop and recovery of diode, a Schottky diode is recommended. The maximum reverse voltage rating of the Schottky diode should be greater than the maximum input voltage, and the current rating should be greater than the maximum load current.

OldSteve:

and i don’t understand it,

The maximum reverse voltage rating of the Schottky diode should be greater than the maximum input voltage,

if i want to convert 10V to 5V shuld i place a diode of 10V in the output??? this does not make sense because if i understand the reason this diode is placed there is to limit the voltage from pin FB to stay below 5V, so i shuld choose diode for 5V.

and about the current, since the load isn’t connected in series to the diode, i shuld not consider the load current.

KobiAflalo: i did read this paragraph and i don't understand it,

if i want to convert 10V to 5V shuld i place a diode of 10V in the output??? this does not make sense because if i understand the reason this diode is placed there is to limit the voltage from pin FB to stay below 5V, so i shuld choose diode for 5V.

and about the current, since the load isn't connected in series to the diode, i shuld not consider the load current.

Rather than make up your own rules, since you don't understand the working of the circuit, why not just do what the datasheet recommends?

The diode will carry the load current, so ensure it's current rating is larger than the load current. And for long diode life and proper circuit operation, the reverse-breakdown voltage of the diode should be larger than the input voltage. What's so hard to understand? If you use a diode with a 5V reverse-breakdown voltage, there's a good chance that it will fail prematurely. Just do as it says and you'll have no problems.

I don't see why I should repeat what the datasheet says. It makes it extremely clear. If you had a better understanding of the circuit's operation, you wouldn't be asking these questions.

If someone doesn't like answering questions then he should [u]not[/u]answer questions, simply not trouble himself - he should find something else to do.

[quote author=Runaway Pancake link=msg=2493085 date=1448381839] If someone doesn't like answering questions then he should [u]not[/u]answer questions, simply not trouble himself - he should find something else to do. [/quote]

What he said. ;)

If someone asks for advice, then ignores it,,, why should they have asked in the first place?

I'll endeavour to provide informative reasons for component choice. :

When the unit switches off, the inductance of the load will endeavour to maintain current flow and if there is no freewheel diode the voltage may rise enough to destroy the circuit. Therefore the diode has to be rated to full load current as this is the value that will flow through it at the instant of switch-off.

Reverse breakdown voltage is the voltage rating of the diode when exposed to the maximum circuit voltage. If your input device fails the full supply voltage will be "seen" by the diode so in your case the diode should be rated to have a reverse breakdown in excess of 10 volts.

KobiAflalo: i did read this paragraph and i don't understand it,

if i want to convert 10V to 5V shuld i place a diode of 10V in the output??? this does not make sense because if i understand the reason this diode is placed there is to limit the voltage from pin FB to stay below 5V, so i shuld choose diode for 5V.

and about the current, since the load isn't connected in series to the diode, i shuld not consider the load current.

No, that's not at all what the diode does. That's a schottky diode, not a zener diode.

See the wikipedia article for Buck Converter https://en.wikipedia.org/wiki/Buck_converter

When the switch is turned off, the entire load current flows through that diode. Additionally, the diode must be able to hold off the entire input voltage comfortably, since it's exposed to that when the switch is turned on.

First he didn’t bother to read the datasheet thoroughly before posting, then he ignored it’s information even when it was pointed out to him.
I said nothing wrong, despite what you think.

Perhaps next time, he’ll read the datasheets thoroughly, as he should have done this time.

And you didn’t contribute anything to the thread - only answered to have a go at me. Perhaps you should find something else to do. At least I told him where the information was.

OldSteve: Perhaps you should find something else to do. At least I told him where the information was.

I have plenty else to do. For one thing, I work. At any rate, I won't be [u]anywhere[/u] near 18 posts per day or 1300 posts in 2 months before I figure to do so.

[quote author=Runaway Pancake link=msg=2493836 date=1448410081] I have plenty else to do. For one thing, I work. At any rate, I won't be [u]anywhere[/u] near 18 posts per day or 1300 posts in 2 months before I figure to do so. [/quote] The fact that you work doesn't make you better than me. I'm disabled, and have plenty of time on my hands, so I choose to fill much of that time here. So what? You obviously went researching me just to find another reason to have a go at me. Let it go.....

OldSteve i really appreciate your assistance, i did read this section in the datasheet and i didn't understand it fully because english is not my native language and i just need to make sure because i'm new in this board design thing and afraid to mess up. please understand

KobiAflalo: OldSteve i really appreciate your assistance, i did read this section in the datasheet and i didn't understand it fully because english is not my native language and i just need to make sure because i'm new in this board design thing and afraid to mess up. please understand

No worries. Just rest assured that the datasheet does know what it's talking about and you'll be fine. Maybe I should have been a bit more patient, (others seem to think so). I had several similar episodes in a row, testing my patience a bit.

DrAzzy explained it well. The diode will be exposed to the input voltage when the chip's internal switch is 'on', and will have to pass the full load current when the switch is 'off ', hence the datasheet recommendation.