Diode VR != VRM

[ACharnley]

Hi,

I purchased some diodes (datasheet attached). On the website description it stated a VRM of 40V and I went off that, and then upon arrival they have a VR of just 20V. Usually I find they are the same especially with popular diodes like the 1N5819.

The datasheet states 40V with < 0.5V duty. Am I right in saying then if 4 were used in a bridge rectifier with a resistive load then it could sustain 40V, whereas if it were charging a capacitor the duty would be greater once the power is turned off (and until the capacitor discharges)?

Thanks!

Andrew

 

[ACharnley]

As an example here it is listed

https://uk.rs-online.com/web/p/rectifier-schottky-diodes/1714691/

 

[ronsimpson]

This is a Schottky diode which is not like a "normal silicon" diode.
**broken link removed**
Data sheet page 2 top right: Reverse current verses reverse voltage:
Schottky diodes leak current much more than non Schottky. The reverse current is very related to voltage and temperature.
If you could hold the diode at 20C and 20V you might get 90uA of current. and 20C and 40V = 300uA
Schottky diodes heat up by two modes:
1) forward voltage X forward current = watts.
2) reverse voltage X reverse current = watts.
Look more at reverse power:
If the inside if the diode was 75C: 20V = 2mA and 40V = 5mA = 200mW This 0.2 watts heats up the diode causing even more leakage and more power.
At 125C the diodes is very leaky.
I think the data sheet is saying that at 40V the reverse power alone (with out forward losses) is too much for the diode to survive. ("0.5 duty cycle")

Schottky diodes are complicated. You might be OK using the parts at 40V for a while. This is not something I would do in production. They will not all survive. On a hot day they will get too hot. I would not use the part above 20 volts.

If you are pushing lots of forward current the diodes will run hotter and thus the leakage will be worse.

 

[ACharnley]

Thanks Ron, so the gist of it is these are really 20V diodes being mis-sold as 40V?

 

[dknguyen]

Thanks Ron, so the gist of it is these are really 20V diodes being mis-sold as 40V?

I normally see max DC blocking V and the max blocking Vrms in the datasheet like https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwi71PGDrYPdAhUxMn0KHTDXBpAQFjAAegQIChAC&url=https://www.diodes.com/assets/Datasheets/Ds30538.pdf&usg=AOvVaw0LkpSNDSme84GkDEtX2U76

yours is really weird

 

[ACharnley]

I thought I was wising up to marketing ploys but it continues to get me.

I'm actually dealing with 30V so maybe they'd survive. I'll compare the reverse power as you say Ron.

dkngyen, that's what I'm used to.

Edit: and here's the real 40V version.

**broken link removed**

 

[ronsimpson]

Another way of saying it:
>normal 40V diode will hold back about (40 plus a little) volts with out a problem. At some point maybe 43 volts (peak) it explodes.
>A Schottky diode does not have a "explode" point. It just leaks more and gets hotter and hotter until it explodes. Probably your diodes will hold back 50V for 1% duty cycle. There is not a clear don't cross this line. The line is very vague. The total heat and temperature is much of the kill function.

If the forward current is very low you can push the voltage.

 

[tomizett]

That's interesting Ron - not something I'd ever had to consider.
Thanks for the information.

 

[Pommie]

I've learned something new too.

Mike.

 

[throbscottle]

Is this why ultrafast rectifiers exist?

 

[Nigel Goodwin]

Is this why ultrafast rectifiers exist?

No, it's because fast rectifiers are required for switch-mode use, conventional rectifiers rapidly overheat and go S/C if you try to use them to replace fast rectifiers.

 

[throbscottle]

I was thinking in terms of the same problem, but with schottky rectifiers as being discussed, not conventional ones.

 

[MrAl]

Hello,

For Schottky diodes you could use the 10+10 rule which is based on percent of voltage rating 5 to 100 percent and temperature steps of 50 C. This rule can be stated as follows.

If you have a certain leakage current at 5 percent of voltage rating then at 100 percent voltage rating you have 10 times that leakage current.
If you have a certain leakage current at 25 degrees C then at 50 degrees C higher (75 degrees C) you have 10 times that leakage.

So when we go from 5 percent of voltage rating to 100 percent the leakage goes up by 10 times, and when we jump in temperature by 50 degrees C the leakage jumps up by 10 times (which means at 125 deg C we get 100 times more leakage current then at 25 C).

It's not perfectly linear like that but it's close enough to get a good idea what happens.