blown FET - Page 1
Forum section: electronics / beginners

author: electrolust
author_rank: Unknown
timestamp: February 11, 2017, 01:23:29 am
content: Expermenting with a pair of 2N7002's I released the magic smoke. The packaging looks intact so there's no visual evidence.This thing is supposed to handle Vgs 20V and Vds 60V.I applied Vgs=4.0V and Vds=4.0V without any current limiting and it was fine. Vds was pulling 300mA, whereas the datasheet shows maybe 350-400mA.With Vgs=5.0V and Vds=4.0V, if I have a current limit on the PS of 500mA or thereabouts, it goes into limit mode immediately and Vds slowly rises -- seems reverse exponential, ie the rate of increase gets slower and slower over time like I would imagine how a cap charges. We're talking a second per tenth volt by the time I stopped watching. It never reaches 4.0V in fact it never reaches much above 3.0V. If I crank the current limit to 3A, the FET releases the smoke. I tried this on 2 parts and both behaved the same way. I don't know what the actual current draw is, ie if it's actually pulling 3A or what.The Vgs PS and the Vds PS are isolated supplies in a single unit, running in isolated mode. I have the 2 grounds tied together.Why does the FET fail like this? Shouldn't the FET limit/control the Id on its own, to the amount in the data sheet? In this case, Vgs=5.0V Vds=4.0V the data sheet shows Id=800mA. Datasheet for the exact part I'm using here:https://www.fairchildsemi.com/datasheets/2N/2N7000.pdf

author: electrolust
author_rank: Unknown
timestamp: February 11, 2017, 01:40:50 am
content: I supposed there's a clue in the Absolute Maximum Id spec of 115mA continuous, or 800mA pulsed. Pulsed = 2% duty cycle, 300uS pulse width. I'm not sure how that is consistent with the chart in Fig 1. though. Wouldn't that mean everything above the 800mA line is simply invalid?

author: Audioguru
author_rank: Super Contributor
timestamp: February 11, 2017, 04:16:07 am
content: The chart (graphs) in a datasheet are only for a "typical" device that you cannot buy. Some are more sensitive and others are less sensitive.The text in the datasheet lists minimum and maximum on-resistance that you can use to see that with only 5V for the Vgs then a low sensitivity Mosfet is not fully turned on and gets worse as it heats up so the poor Fet was much hotter that it is allowed to be.

author: David Hess
author_rank: Super Contributor
timestamp: February 11, 2017, 04:29:22 am
content: Besides exceeding the continuous drain current, you also exceeded the 200mW power limit by almost 10 times at 500mA*3V=1500mW. The symptoms you describe are consistent with overheating.

author: radiogeek381
author_rank: Regular Contributor
timestamp: February 11, 2017, 02:13:42 pm
content: Quote from: electrolust on February 11, 2017, 01:23:29 amThe do-it-yourself curve tracer is a great idea. It illustrates lots ofthings, and uses pyrotechnics to get the message across.All the comments so far are on target.SHORT ANSWER:The transistor was operating way way outside its safe operating area.The current started to drop off because it got really hot.LONG ANSWER:Take a look at figure 14. Though you were operating with Vgs = 5V,and the graph assumes Vgs = 10V, the difference doesn' t matter much,given that Vds < Vgs.Now put your operating point at Id = 0.5 A. (And for the moment, imaginethat the solid horizontal lines extended to the left of the dotted Rds(on) limit.)Where is the safe operating area? It pretty much says that a SINGLE pulselonger than 10mS could cause unsafe things to happen for a typical device.What unsafe things?1) At very high currents, carriers arrive at the drain and find a high concentrationof fellow carriers. Some of them find the field created by Vgs provides a betterpath than the path to the drain. A very few will tunnel toward the gate terminal.("But it is insulated!" the crowd shouts. Well, there is some amount of gate-bodyleakage even in normal operation, but in this case that insulator can only do somuch.) Some of these carriers don't make it to the gate and get "stranded" inthe gate insulation, forever to act as a counter-influence on Vgs. (They'll raisethe threshold voltage, or reduce the Idss.)2) But more importantly, high currents bring thermal stresses to the contactsthat connect the device leads into the diffusion areas for drain and source.These heat up, as they are not perfect conductors, and they are surroundedby material that doesn't conduct heat so well. (Chip designers call this "jouleheating.") This "one time" event can damage the contacts permanently.In some cases, it can cause the contact to fail open.3) Finally, you're cooking a chunk of silicon that got to be the way it isbecause somebody else cooked the same chunk of silicon in a very wellcontrolled environment with very very well controlled temperature. Yourenvironment and temperature aren't so well controlled.So, getting back to Figure 14. Your one-time pulse was actually a DCpulse. Elvis left your safe operating area way back at 40 mA for Vds = 5V.--------------------Now, why did you see the voltage drop slowly rise? (the exponentialobservations was good...)Take a look at Figure 3 -- the device is heating up. On resistancerises with temperature. Temperature rises with power dissipation.Power dissipation rises with on resistance.... rinse, lather, reheat.But that doesn't account for all of it. The resistance of the contacts(these are called by some "ohmic contacts" because their effect isnot governed by the carrier concentration in the channel, but by thebulk resistance of the contact material) is going to rise with temperature.Thus Vds rises as the device heats up.There may be many other effects.-----------Now for the final smoke emitting FET event:A FET's max drain current is determined, largely, by the width of thechannel. From the top, it might look like this:SSSSGGGDDDDSSSSGGGDDDDSSSSGGGDDDDSSSSGGGDDDDIn this case, we'd say that the channel (the region under the gate)is 3 units long and 4 units wide. Idsmax is influenced largely bythe WIDTH of the device. To get high Idsmax, we make a widedevice.But designers don't want to make the gate terminal too wide.If they do, the turn-on time can get longer (the gate is driven atits ends, typically, and it has significant capacitance -- for reallywide gates, this can begin to cause the turn on pulse to takemeasurable time to propagate along the gate), and there iseven a pathology caused by the ends of the gate cutting offthe channel before the middle of the gate does.It is unlikely that this device is actually constructed as a single source/gate/drainblob. More likely, it is multiple smaller devices all connected in parallel.Each has its own drain/source metal contacts that lead to the device lead frame.When you exceed the Idss(smoke) limit, one of those contacts will blowbecause it was just too hot. Now the remaining current has to flow throughthe rest of the parallel devices and contacts. Now one ofthemwill blow.and so on and so on, like strands in a cable stressed beyond the breakingpoint.Eventually the drain-source path opens up, and sends its last signal:a smoke signal.-------------------------------I have not mentioned so far, electromigration failure. This is a big problemin high speed digital design, but is a wearout phenomenon. It is not likelytriggered by a one-time event, but overstressing (or even moderately-stressing)a device can lead to early failure. The MTTF by electromigration goesexponentially with temperature. I'm not sure it even happens in discreetdevices like the 2n7xxx, as it also depends on wires being narrower thana critical size.-----------------------------Yet another long post. But this was a great question.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2017, 08:26:18 pm
content: That time constant is your warning that something ain't right...Riddle 1: what happens to a MOSFET when it heats up?Riddle 2: how hot does the 2N7002 get when operated at 200mA and 4V? What about 800mA?Tim

author: Zero999
author_rank: Super Contributor
timestamp: February 11, 2017, 08:59:49 pm
content: Quote from: David Hess on February 11, 2017, 04:29:22 amYes, it's certainly overheating which killed it.As the temperature increases, the on resistance will also increase, resulting in a positive feedback loop, known as thermal runaway.

author: radiogeek381
author_rank: Regular Contributor
timestamp: February 14, 2017, 02:35:53 am
content: Quote from: T3sl4co1l on February 11, 2017, 08:26:18 pmThis is another neat one. For years I've carried around a constant from"Reference Data for Radio Engineers" (a friend once called it "Reverence Datafor Radio Engineers") This was "55 degrees-C Watt per square inch"Never mind the goofiness of the mixed systems of units. This was a "rule"to estimate the temperature rise over ambient for a widget dissipating powerin "free air". It is, of course, baloney, as the Stefan-Boltzman law would suggest.But RD4RE is a really really good source. Is the rule "reasonably close?"By that estimate, our 0.8W widget connected to a 2.5x2.5cm heat sinkshould rise about 44C above ambient. (The purely metric versionof the constant is 0.035 C-W/meter ) At an ambient temperature of 23we end up with a device temperature at its surface of 67, call it 70C.Let's try the Stefan Boltzman model.s = 5.6e-8 * W / (m2K4)Let's assume perfect emissivity:A = 0.025 * 0.025 = 0.6e-3 m2Ambient temperature (Tc) is 296K.T4= P / (As) + Tc4When I do this in python:>>> P = 0.8>>> Tc = 23 + 273.15>>> A = 0.025 * 0.025>>> s = 5.6e-8>>> T = (P / (A*s) - Tc**4)**0.25>>> T - 273.1577.77204661393222T = 78CHOLY COW!!!! So my stupid little linearized model gets withinshouting distance of the Steffan-Boltzman 4-th power model.Now what happens at 1.5WStefan-Boltzman model: 160CRDRE model: 108CSo, there you are, when the going gets tough, the linearized modelisn't as useful as we'd hope. Still, I'll keep it around when I can't dothe fourth root of a bazillion with the tools at hand.this is fun.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 14, 2017, 08:32:45 am
content: Hmm... radiation is the wrong model to use here: convection dominates pretty hard.Effectively, the conductivity of space goes as the 4th power of whatever's hottest. Or if the temperature difference is small, it goes as the 3rd power of the difference. Around room temperature, we should expect radiation to be pretty crappy!Convection is somewhere between proportional (heat diffusing through solid material or stagnant fluid) and square-law. It depends heavily on geometry, obviously, so, refer to datasheets where you can (i.e. when choosing heatsinks), or else use wave-of-the-hands estimates, then measure it in practice.So as for the numbers!Radiation isn't a temp rise, it's the radiative power output at absolute temperature. (Riddle: does the inside of a red-hot sphere still glow red, even when it's sealed closed so that light cannot escape it? YES! It definitely always glows!) Heat loss is the difference in radiation, so if you want to know temp rise, you need to difference the powers, not the temperatures.That looks like...P_diss = P_radiated - P_absorbedP_radiated = power given off by hotter object, P_absorbed = power taken in from the surroundings. Assuming infinite surroundings here (which is reasonable for an object sitting in a room). Note that the area is that of the smaller object, too.P_radiated = (T_hot)^4*A*sP_absorbed = (T_cool)^4*A*sP_diss = (T_hot)^4*A*s - (T_cool)^4*A*s= [(T_hot)^4 - (t_cool)^4]*A*sP_diss / (A*s) = (T_hot)^4 - (t_cool)^4(T_hot)^4 = P_diss / (A*s) + (t_cool)^4T_hot = [P_diss / (A*s) + (t_cool)^4]^(0.25)Which is what you have there (except for the minus in the later step, a typo?), so your math checks out.If anyone wanted some illumination (pun intended?) where this came from, here it is.As for convection, the figure of 55 C/W in^2 is ballpark of course, also, hrm, those units don't smell right. It's per area? So you get more C's with more area? No, that's gotta be area on top... 55 C in^2 W^-1. That makes sense!That's a lot smaller than the figure I remember. There may be two reasons for this...1. Radio Engineers may be concerned about somewhat higher temperatures than the engineers of today. A ceramic resistor or glass vacuum tube at 200C gets great thermal efficacy!2. If that's counting both sides of a surface (such as for sizing heatsink plates), the figure is halved automatically. (That doesn't help any for figuring out envelope temperatures, though!)The number I heard is 150 C in^2 W^-1.I've measured dissipation pretty well in line with this figure -- for PCBs, and shiny metal enclosures in mixed conditions (namely: bottom face stagnant, sides open to convection, top face exposed but somewhat stagnant). Obviously, the combination of poor emissivity (minding that radiation is a lot weaker than convection is, around room temperature, but it's still a helpful ~10% or so) and partially obscured surfaces doesn't help, but it's also not unrepresentative of real conditions, either.As for area...How much area on the PCB is actually dissipating power, anyway? If you make the power planes the size of the board, is it the whole board? No, that doesn't make sense, what if it's the size of the universe?(Another fun math riddle:Suppose you have a sheet of metal, which has a simple linear heat dissipation rule: the heat lost from a point on the surface, is proportional to the temperature at that point.The material has some bulk thermal conductivity, as well.A heat source, of given radius and temperature, lies in the center of the sheet.What is the temperature on all points of the sheeta. when the sheet is finite (sheet radius > source radius)b. infinite (sheet radius --> infinity)?Note: you can't actually solve this, analytically speaking -- but you can identify a "special function" which does. And, yes, such a simple problem is already one of those kinds of "unsolvable" problems! Cool, huh?)Tim