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https://www.tme.eu/Document/e6d16d51cc343d73832bf32cc9d69d1f/p2500a.pdf

How does that "RthA < 4 K/W 1)" work?

"1) Valid, if leads are kept at ambient temperature at a distance of 10 mm from case"

They only spec Ir @ 25 deg C. But there might be a hint to how Ir might increase with Tj:

"Reverse recovery time

IF = 0.5 A through
IR = 1 A to IR = 0.25 A

typ. 1500 ns"

Should it be understood that Ir can reach 1 A?

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  • \$\begingroup\$ What is Ir <-- more detail? \$\endgroup\$
    – Andy aka
    Commented Sep 21, 2022 at 17:23
  • \$\begingroup\$ @Andyaka it is the Leakage current \$\endgroup\$
    – kellogs
    Commented Sep 21, 2022 at 17:32

2 Answers 2

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The datasheet is not very good, so I can see why you're struggling here.

The thermal resistance from the junction to air, \$R_{thA}\$, is less than 4 K/W if you provide sufficient thermal dissipation that the leads 10mm away from the diode case remain at the ambient temperature. The thermal resistance from the junction to the leads is less than 2 K/W. This is a bit of a weird way of specifying it, and in the absence of any clarification from the manufacturer I would probably try to design under the assumption that \$R_{thA}\$ might be more like 8 K/W, i.e. twice the rating they specified.

There is a derating curve for the ambient temperature. You can operate at 100% of the diode's rated current if the ambient temperature remains under 50°C. No information is provided about derating reverse recovery current, but I would tend to assume that the specifications remain generally unchanged until the ambient temperature reaches 50°C.

The reverse recovery specification says that for an \$I_F\$ (current flowing through the diode in the forward direction before it turns off) of 0.5A, the \$I_R\$ (reverse conduction current) may be anywhere between 0.25A and 1A, for a \$t_{rr}\$ (reverse recovery time) of typically 1500 nanoseconds. They do not provide any further information about other forward currents or any temperature derating for this parameter, so the best you can do here is guess.

If the specifics of these parameters are important in your design, I would suggest picking a different diode whose datasheet is clearer. You could also email the manufacturer and ask for more detailed specifications.

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  • 1
    \$\begingroup\$ A similarly encased diode (a bit larger) does mention RthA = 10 K/W without any weird adage, so 8 K/W proposed seems fine here. Still sounds unreal to me, such low RthA... Would you consider this diode a bad choice for reverse fast negative automotive transients? I have emailed the manufacturer, any answer received I shall post it here (i doubt there will be any though) \$\endgroup\$
    – kellogs
    Commented Sep 22, 2022 at 9:11
  • \$\begingroup\$ I would consider it a bad choice just because it is so poorly specified. I can't comment on the actual performance as I'm not familiar enough with automotive transients. \$\endgroup\$
    – Polynomial
    Commented Sep 22, 2022 at 10:59
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Valid, if leads are kept at ambient temperature at a distance of 10 mm from case

The wording is extremely poor indeed!

What this really means is that the "ambient" temperature is defined as the temperature of the leads 10mm away from the case - whatever that temperature may be when the diode is in thermal equilibrium, i.e. everything has warmed up.

For example, if the leads are soldered into the PCB at a distance exactly 10mm from the case, and there are large copper areas to dissipate the heat, then there won't be a big thermal gradient next to the leads, and the temperature of the copper area immediately adjacent to the lead will be what you need to put into the thermal model as the "ambient" temperature.

It doesn't mean that you'll be expected to somehow heatsink the lead to ambient air 10mm away from the case. It'd be entirely impractical, and it would be easier to get a diode in a thermally superior package and make it dump the heat into the PCB better, or use a diode in a package that's mounted to a heatsink.

Should it be understood that Ir can reach 1 A?

\$I_r\$ is controlled by you (or, rather, the circuit you've designed). It is the reverse current that charges the diode capacitance so that the diode turns off. It is not a leakage current.

If the circuit you've designed cannot provide about 1A of reverse current to turn the diode off, then the diode will stay turned on for much longer than 1.5µs.

The diode is a typical slow rectifier diode. It's not even a good diode for transformer rectifier circuits: it will have lots of cross-conduction as the AC voltage changes polarity. This particular diode is best used for circuits where the current flows in one direction for a long time, e.g. for polarity protection or parallel supply isolation.

Real diodes do not turn off immediately. They always need some charge driven as a reverse current to turn off. This particular diode needs about 1.5µC of reverse charge to turn off from full conduction. \$Q=I\cdot t\$, thus 1A*1.5µs=1.5µC. Or, alternatively, it's equal to charging a 1.5µF capacitor from 0 to 1V.

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  • \$\begingroup\$ They do spell 'Ir' the same for both leakage current and reverse turn-off-current in the datasheet, not nice... Ok, I guess i could ensure that current with a properly sized cap just after the diode, but I am not sure how much protection this diode will offer for reverse fast negative automotive transients. Comments ? \$\endgroup\$
    – kellogs
    Commented Sep 22, 2022 at 9:19
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    \$\begingroup\$ @kellogs The diode will stay on for several microseconds in practice. Stay on means it will act like a piece of wire. Less than a microsecond if you can feed it enough reverse current to get it to turn off quickly. Those first few microseconds (or fractions of one) when the diode is transparent must be handled by something else. \$\endgroup\$ Commented Sep 22, 2022 at 17:09
  • \$\begingroup\$ Say, for the ISO 7637 test pulse 1 "Us −75 V to −150 V ::::: Ri 10 Ω :::::: td 2 ms :::::::: tr (1 −0,5 0) μs, would a TVS diode (anode to GND) provide most of this IR current into the rectifier diode, or will the rest of the circuit (which I am trying to protect: linear voltage regulator + MCU) be the faster one to inject that current, which is more likely? \$\endgroup\$
    – kellogs
    Commented Oct 5, 2022 at 19:31

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