Datasheet for 1N5406 diode states "3.0 ampere operation at Ta = 75°C with no thermal runaway".
What does this mean exactly? Can I place these in parallel to increase current capability?
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\$\begingroup\$ My guess from reading the sentence is that it is saying you can operate it with 3A at \$75^{\circ}\$C without experiencing thermal runaway (i.e. the device won't heat itself up and start carrying even more current, causing it to heat up even more...). \$\endgroup\$– Null ♦Commented Jul 11, 2015 at 17:33
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\$\begingroup\$ If Vf vs Tj is negative, as I am guessing is the case on all diodes, I do not see how it can prevent more current from flowing. Unfortunately, they omitted this information. \$\endgroup\$– mcuCommented Jul 11, 2015 at 17:39
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\$\begingroup\$ Well, if the device can dissipate heat well enough it won't heat up while carrying a current. \$\endgroup\$– Null ♦Commented Jul 11, 2015 at 17:41
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\$\begingroup\$ I guess that is what they mean. But wouldn't a lower Vf decrease power dissipation, so a diode is in effect self-regulating? \$\endgroup\$– mcuCommented Jul 11, 2015 at 17:44
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2 Answers
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You are correct about reduction of \$V_f\$ with increasing junction temperature. But, this isn't about forward bias because lower drop in a single diode is not a problem. The diode doesn't determine the current flow. Instead it's about leakage under reverse bias. The datasheet appears to make the claim that under application of 3A forward current at 75C ambient, no significant hot spots will develop. Hot spots can cause failure under reverse bias due to increased leakage.
With increasing temperature, diode leakage will also increase. A hot spot can create a local region with lower resistivity and leakage, which increases loss contribution from reverse bias. A hot spot will cause current crowding, and the device can fail due to increased current in the region with lower resistivity. This is thermal runaway for a single diode.
This is usually more of a problem with high voltage parts and Schottkys since they have higher leakage to start with.
For more than you might want to know about thermal runaway, you might look at this article.
Finally, no, you would not want to parallel these because they will not share current.
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First, it's poor engineering to place diodes in parallel. Instead increasing current capacbility, you will overload just one of them. The catch is that "in parallel" means same voltage across all diodes, but the I-V characteristic will slightly differ between them, so one diode will always take more current than others.
Second, thermal runaway is when increasing temperature will cause more power dissipation on diode and so more temperature. Well, just be sure you stay well within safe operating range.
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\$\begingroup\$ So, what does Fairchild's claim mean and why do they make it? They must have an application in mind where this would be useful. \$\endgroup\$– mcuCommented Jul 11, 2015 at 17:16
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\$\begingroup\$ It has nothing to do with parallel diodes. What they say is don't worry, even with high temperature the diode is robust over most operating range. \$\endgroup\$– user76844Commented Jul 11, 2015 at 17:20
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\$\begingroup\$ Why should I worry about thermal runaway if it is just a single diode? \$\endgroup\$– mcuCommented Jul 11, 2015 at 17:22
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\$\begingroup\$ Because if that would be a "running away" diode, it would heat up itself and burn. \$\endgroup\$– user76844Commented Jul 11, 2015 at 17:23
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\$\begingroup\$ How is this diode different from any others? Vf drops as Tj increases on all of them. Is this just a gimmick of a claim? \$\endgroup\$– mcuCommented Jul 11, 2015 at 17:25