What physical factor does limit the maximum rated current through a semiconducto - Page 1
Forum section: electronics / beginners

author: hkBattousai
author_rank: Regular Contributor
timestamp: May 26, 2014, 08:00:55 am
content: I read diode and transistor datasheets. For example, for a device they say the maximum allowed current is 20A. But they also note that, for a very small amount of time (e.g.; 100us), a 200A pulse can go through without harming the device. This specification confuses me.If 200A can go through even if it is for a very small amount of time, then that means the device is physically capable of conducting that current. I understand that one big limiting factor is HEAT. I would like to know if there are any other factors which limit this rated current.And also I would like to hear answer of the following question.If a device is rated 20A. Can I apply 40A pulse to it with 50% duty cycle whose frequency is high enough? If the only limiting factor is heat, this pulse should yield the same amount of heat and accordingly temperature rise. And if the answer is yes, the following mandatory question arises:Can I apply 200A with D=10%, and 2000A with D=1%, and even 20kA with D=0.1%?

author: tautech
author_rank: Super Contributor
timestamp: May 26, 2014, 08:24:24 am
content: Current is usually specified as DC and that of course is a continuous rating providing it is within thermal specification.Yes you can extend max I(A) with decreased duty cycle and your assumption is correct to the point where the magic smoke escapes.I once heard a story where a 1N4148 in a switching circuit was subjected to 20A and survived.Datasheets often have a SOA (safe operating area) graph and that should be strictly adhered to. Just let those graphs guide you and you will be OK.

author: AndyC_772
author_rank: Super Contributor
timestamp: May 26, 2014, 08:41:36 am
content: Remember that, for a resistive element such as the channel of a switched-on MOSFET, heating increases with the square of the current. So 200A through the channel will generate 100x the heat that a current of 20A would, not just 10x.For a p-n junction, the forward voltage is also a function of current. It's taught as "0.6V" or thereabouts in school, but that's a pretty gross oversimplification especially where significant power is involved.Forward voltage is actually much more like an exponential function of current, which happens to be around 0.6V for currents within the normal operating range of a given device. But, if you push that current way beyond the normal range, even instantaneously, then the forward voltage can increase dramatically. If, say, a diode has a forward voltage of 1V at 10A (ie. 10W dissipation), then it could easily be 2 or 3V (or more) at 100A. So you might get 30x the dissipation for just 10x the current.Other factors may come into play too. I wouldn't want to speculate on how long a bond wire subjected to repeated, massive overcurrent events would last, for example. Electromigration springs to mind as a likely failure mechanism even if thermal shock doesn't break them first.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 26, 2014, 08:55:10 am
content: You can stay within thermal limits by increasing current and decreasing pulse width. Higher currents drop more I^2*R losses of course, so it's a diminishing returns sort of thing. But is there an absolute current limit?Yes. There is some phenomenon (possibly electromigration, but maybe others) which causes breakdown and failure, even without conventional (thermal or avalanche) damage.Example: I once designed a MOSFET inverter board which had rather large voltage/current spikes, to the tune of 80% voltage overshoot (unclamped) out of 700V supply, or current spikes (clamped) ca. 200A out of 70A rated output). To deal with this during prototyping (before refining the layout and ordering new boards), I added clamp diodes. The duty cycle was extremely low, so I first tried 8A diodes. Then 12A diodes. Both failed shorted. 30A diodes survived, however. (The overshoot was addressed in the subsequent rev, which got reasonably clean switching waveforms and overshoot in the 0-20% range, worst only when lightly loaded or unloaded.)Probably, if a diode is specified for a short-period rating, that rating should be treated as absolute for any shorter duration as well. If a short-period rating is not given, probably better off not picking that part...As for your other questions, no, the give-and-take is not linear. If forward voltage were constant, it would be. But voltage drop rises slightly with current due to the ideal diode response, and significantly more due to internal resistance. At very high currents (near, or beyond, the "spooky unreliable failure" regime), internal resistance dominates and power goes as current squared; so duty cycle would have to decrease as the square of current to keep average power constant. (The same scaling law is true of MOSFETs, which are of course resistive in nature.)So, the overall SOA might look kind of like:Duty - Current1 - 1x1/2 - 1.8x (dropping a little due to exp forward drop)1/10 - 5x (dropping faster due to resistance)1/100 - 20x1/1000 - 30x (limiting some due to short peak rating)impulse - 40x (absolute limit)Or whatever the short-period multiplier is. If a diode is rated for 200A for 1us, 10A average, the multiplier is 20x, so it will trend from 1x at 100% duty to 20x as duty --> 0.Tim

author: Psi
author_rank: Super Contributor
timestamp: May 26, 2014, 09:04:11 am
content: Simply put, what limits it is the fact that you '"just can't get the heat away fast enough to prevent the die from getting to a dangerous temperature"The larger the package the easier it is to sink the heat way.There are however some other more complex factors inside the semiconductor that come into it.So you can't just take a 20A mosfet, cool it with liquid nitrogen and push massive gross over current through it.But yeah, the dominant factor is always getting the heat away fast enough.If you look at mostfet datasheets you will notice that, even with good heatsinking, it's really hard to ever reach their max current without the package overheating.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 26, 2014, 09:21:39 am
content: No, it's not just thermal. In my example, there was simply no way overheating, even transient, could have caused the failure observed (the diodes in practice ran little over lukewarm).As long as bonding is sufficient, MOSFETs may very well be capable of more current flow -- assuming enough gate voltage as well. The reason I would guess this is, MOSFETs have a fundamentally different mechanism than diodes: majority rather than minority carriers. Diodes must supply their conduction from thermally generated charges (which is also in part why they turn on and off relatively slowly), while MOSFETs (and schottky diodes, which are also maj. type) are essentially instantaneous (subject to RLC limitations only). That said, the only way I know of to draw more current is by raising the gate voltage; and even then, you get severely diminishing returns past, oh, 20 or 30V (let alone beyond rated Vgs(max)). I don't know offhand what the ultimate physical limitation is, if it's doping concentration or what. Current density of MOSFETs is lower than BJTs and IGBTs though, in any case.In practice, a MOSFET's specified peak current is generally in, or near, the constant current region, where more current simply cannot get through the thing, and the voltage drop is large. It's never useful, because to reach that current, some sizable fraction of rated voltage must be applied at the same time. But your supply voltage is already a sizable fraction, so the available voltage change isn't much (the efficiency sucks). If nothing else, your peak power will be so high that it won't survive more than a few microseconds.Tim

author: tszaboo
author_rank: Super Contributor
timestamp: May 26, 2014, 06:01:51 pm
content: I've read an IRF application note where they stated that with most modern high current transistors the almost the only limititation is the bond wire. If the wire reaches a certain temerature, it dies, otherwise you can send multiples of the rated current through it.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 26, 2014, 11:27:36 pm
content: If that's the one I'm thinking of, that was in regards to the devices that are so freaking massive that there is literally no way to use them to peak ratings. You know, those "300A" transistors ...in a TO-220 package. The package lead will fuse at 100A, let alone the bondwires!The failure of diodes is a different phenomenon; they also fail shorted, whereas one would expect a fused bondwire to fail open.Tim

author: retrolefty
author_rank: Super Contributor
timestamp: May 26, 2014, 11:42:29 pm
content: Well I worked on a large industrial variable speed drive (rectifier module, DC link, output invertor) that used 144 large SCRs driving a 5Mwatt motor. The SCR were in stacks of six in series using hokey puck packaging and the system utilized deionized water cooling. Anyway the current rating were 600A continous. So with the right packaging and proper cooling methods semiconductors can reach increadable continous currents.Looked like this:http://www.powerlabs.org/images/scr.jpgThe package design is such that you can't even test for continuity out of circuit unless the package is squeezed to a certain foot pounds of pressure as they are in service, otherwise they are open circuit and can't conduct.

author: IanB
author_rank: Super Contributor
timestamp: May 27, 2014, 12:21:51 am
content: Quote from: T3sl4co1l on May 26, 2014, 09:21:39 amI don't think you have enough evidence to make this claim. The heat could be localized deep inside the device at a hot spot and be undetectable from the outside.For an analogy, consider welding. The metal can be white hot and molten where the joint is being made, but just an inch away not even red hot. When you have a lot of heat concentrated in a small space for a short time then thermal resistance can act like a very solid temperature barrier.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 27, 2014, 04:01:25 am
content: Quote from: IanB on May 27, 2014, 12:21:51 amIt would be [conveniently enough] impossible to disprove without some sort of microscopic, volumetric, instantaneous thermometer.I have good reason to believe so, just based on design data. The thermal impedance profile of said device tapers off at much longer time periods, ca. 5 microseconds. This is the time constant of the silicon itself. We're talking <100ns pulses, so the amount of delta T, even at extremely high currents, even in an extremely small volume (if it were spot-heating), couldn't possibly be enough to matter.I also have no reason to believe spot heating would occur; these currents are thoroughly into the resistive region, and also into the positive tempco region. There will be localized heating due to spreading (the lateral resistance being comparable to the equivalent junction resistance), but not to a microscopic degree.At ridiculously high currents, I suppose one could actually exhaust the supply of charge carriers and cause an arc discharge (dielectric breakdown), which would be able to melt and destroy silicon within nanoseconds. But this is certainly also not the mechanism, as the diodes survived many cycles (guessing, hundreds to millions) before failure: it was a progressive phenomenon, not a single shot. Which also suggests something gradual like electromigration.Tim

author: IanB
author_rank: Super Contributor
timestamp: May 27, 2014, 04:25:53 am
content: Well it's hard to say what really happened from a distance, but let's just for example deal with hypotheticals. I'm not trying to prove one way or another, just bringing up some theoretical considerations.You mentioned a current spike of 200 A through a diode rated for 8 or 12 A. That leads to an internal power dissipation in the diode about 400x greater than the rating. That seems somewhat considerable.With high powers and short time durations thermal resistance becomes less important than thermal capacitance. In fact at short enough time scales all thermal resistances become effectively infinite because there isn't enough time to conduct the heat away.The thermal capacitance heat equation isMcdT/dt=QwhereMis the local mass being heated,cis the heat capacity andQis the local heat dissipation. WithMcsmall enough andQlarge enough, it is possible to obtain a temperature rise larger than any limit, even in a nanosecond.Now I'm not saying this is what happened in your case, but it is something to be considered. When the total energy is delivered in a short enough pulse no amount of heat sinking can save you.

author: qno
author_rank: Frequent Contributor
timestamp: May 27, 2014, 07:34:07 am
content: It is usually the bondwires that work as a fuse.