Reverse recovery time of Schottky diodes

Folks,

Have to design a boost converter that creates around 150V from 20V, north of 100 watts. Since this can cause grief with core losses it has to run continuous mode. So I need a diode with no or hardly any reverse recovery time. Now that we have 200V Schottkies here is the puzzler.

This datasheet states 35nsec reverse recovery:

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This datasheet from the same mfg, just a smaller size of the above diode, states ... none at all:

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What gives?

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Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg
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At 0.5 amps, room temp, that thing drops about 0.65 volts, rotten for a Schottky. There may be a PN guard ring that conducts up there, and will have reverse recovery. I've seen that happen, and it can get ugly.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Yeah, not pretty if that's the case. But what about the others slightly smaller brother? They don't state any reverse recovery. Maybe they just forgot?

I can't have much trr because that makes the main switcher FET cook. But I need a diode that can stomach 200V. I don't care much about Vf.

--
Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

I've been looking at some Sic Schottky at infineon. They claim to have "No reverse recovery time" and "Temperature independent switching behavior" Oh, these rate around 6000 volts and up..

Jamie

If you believe that I guess you can believe anything.

Reply to
Maynard A. Philbrook Jr.

Let be clear that up, that was "No reverse recovery charge" not time.

I put in a request at work to have a few sent to me, I have an application where I need a fast switch with at least 500 volts.

Jamie

Jamie

Reply to
Maynard A. Philbrook Jr.

Just keep in mind that these can have a lot of forward drop.

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Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

Well, you DID say you don't care about Vf.

I was thinking SiC, too. I've never heard of the PN guard ring that John was talking about -- what in heck is that for?

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Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

Within reason. My peak current will be around 10 amps because of the lopsided duty cycle. Datasheets of mid-sized SiC diodes list up to 20A repetitive peak but then fail to go there in the Vf graph. Like this one I am considering:

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I assume it'll be close to 4V so it'll burn off almost 5W instead of 1W. But the FET will drop by more than 8W so it's still a 3W net win.

I've searched but found no explanation why 200V Schottky diodes have a trr and 100V ones don't. I can hang two 100V diodes in series and add bleeders but that could be a white-knuckle ride. Have to think about that.

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Regards, Joerg 

http://www.analogconsultants.com/
Reply to
Joerg

Well my very limited understanding is higher voltage means lower doping, or it avalanches. I don't know exactly how lower doping gives longer recovery... but a wider depletion region... starts more hand waving.

George H.

Reply to
George Herold

Supposedly a Schottky diode is an entirely different kind of critter. Instead of having a junction formed of doped semiconductors, it has a hunk of aluminum that's bonded (welded? sintered???) onto a hunk of silicon. If my 30 year old memory of semiconductor physics class serves me here, there is no depletion region.

I think I'm gonna go type "schottky" into Wikipedia & see what pops up...

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Tim Wescott 
Control system and signal processing consulting 
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Reply to
Tim Wescott

This link, second paragraph in the section. Inneresting:

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Tim Wescott 
Control system and signal processing consulting 
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Reply to
Tim Wescott

Probably a byproduct of how they measure it? Big C and dC/dV looks like recovery, but isn't.

Example: curve seems to show 100-200pF around 0V, which will drop to negligible levels (< 50pF) past 10V, or (more hand waving), maybe 2ns.

It's partly your fault for looking at a cheap Chinese product and complaining that there's nothing there... :o)

For example: besides the sparse data in general, there's no conditions specified (dI/dt?!), let alone a test circuit (how much inductance, dV/dt, etc. is also present?).

This one actually shows the circuit:

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Even better, we can see it's a 50 ohm system, so of course it has a time constant on the order of 10ns. This one is rated at "t_rr" < 25ns.

Obviously, for a power inverter of mere ohms impedance (depending on just what time constant and characteristic impedance you've designed it for), "t_rr" will simply be proportionally shorter.

And there's always junction diodes. Recovery isn't /that/ bad in the 200V range, and the capacitance is much lower. There are cases where they perform better, despite common advice against it.

You could even use a beefy ferrite bead as an old school saturable reactor to give more time for recovery.

Tim

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Seven Transistor Labs 
Electrical Engineering Consultation 
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Reply to
Tim Williams

Not far off.

Technically, any metal-semi junction is technically a Schottky junction, but there are two types: ohmic and rectifying. Obviously, the latter is formed intentionally in these diodes, so we call them as such, and leave off the qualifier.

The usual way rectifying contacts are avoided, is by heavily doping the substrate (notated as n+ or p+) and applying a similar type metal on top. Aluminum provides the convenient combination of being both a P type dopant and a suitable metal.

I forget what the deal is with N, if the Schottky junction formed between Al(metal) and n+ is simply irrelevant, or if there's something else at work. But they can do the same with N type material as well, and hence, the ability to make semiconductors at all.

Aluminum is also easily deposited, by vacuum evaporation I think. Probably sputtering too, but I don't know if they bother.

For intentional rectification, I think the most common metal is Pt or PtSi (an intermetallic compound). Probably deposited directly via sputtering, or formed in a diffusion reaction after depositing Pt metal on bare Si (probably evaporated).

Tim

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Seven Transistor Labs 
Electrical Engineering Consultation 
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Reply to
Tim Williams

It's a piece of cake to put them in series *if* the part is adequately specced AND the manufacturer actually 'adheres' to that spec AND your purchasing/production doesn't mix and match in inventory. Or is this a single source item? ....shudder

Maybe four 50V would be easier to work with. By the time you have all the network to make certain the thing doesn't go off one at a time; your 'effective' reverse recovery could eat you alive. Do they put more than one in a package so you have an array from a single wafer?

We used to stack diodes to get up to 150KV, but that took a lot of network and physical positioning to keep from popping them.

Reply to
RobertMacy

"Tim Wescott"

** Definitely one of the better ones.

.... Phil

Reply to
Phil Allison

It's my theory that you can put schottkies in series without any balancing resistors. They leak more and more at higher voltages, so a series string is self-regulating.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Watch out for reverse leakage on Schottky diodes at high voltages. There are some Schottky diodes where a little footnote mentions that the continuous reverse voltage is much smaller than the peak reverse voltage due having a region of thermal runaway. I tried -35V continuous on a tiny 5A 40V Schottky and it cooked itself after being slightly warmed by a heat gun.

Reply to
Kevin McMurtrie

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

Diodes since the 70s or so I think have been fine with this, because purity is better, avalanche is better controlled, and dies are patterened or guard-ringed for better controlled properties. At least, for line frequencies.

By the way, applying full voltage while on the tail of recovery still accelerates the residual charges... this can make a sort of transient second breakdown. Check out a BJT's RBSOA for example (e.g. MJE18008). Obviously this increases power dissipation. I don't remember if it's also a spacially localized phenomenon, i.e., causing spot heating and rapid failure.

Still, not a good idea, since the first one to recover gets to avalanche while the rest come out, and so on. Or for schottky, substitute 'recovery' with junction charge.

Tim

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Seven Transistor Labs 
Electrical Engineering Consultation 
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Reply to
Tim Williams

Schottky diodes don't avalanche like PNs do. As reverse voltage increases, they just conduct more. I suppose they may break over at some voltage, but the practical limit of reverse voltage is power dissipation.

PN diodes have low and fairly constant reverse current vs voltage, up to breakdown, so behave differently in series strings.

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John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    
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Reply to
John Larkin

Thanks Tim, My limited understanding is that there's still a depletion region, it's just all in the semiconductor.

George H.

Reply to
George Herold

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