High voltage but extreme current limitation - will transistor be damaged? - Page 1
Forum section: electronics / beginners

author: 9voltbrain
author_rank: Contributor
timestamp: February 08, 2024, 09:14:07 pm
content: Something I've been thinking about... Let's say you have a circuit like in the picture below, where the transistors only job is to just to block any current from passing. Since there is no current flowing, the voltage at the transistor should be 500v, far exceeding its rating. But will this really damage it, seeing as the maximum current that can flow through it is 0.01mA?

author: twospoons
author_rank: Frequent Contributor
timestamp: February 08, 2024, 09:30:20 pm
content: May not do any damage (depending on the physical size of the die) , but then its not going to be useful either as you cant actually turn it off with 10x rated Vceo on the collector

author: Manul
author_rank: Super Contributor
timestamp: February 08, 2024, 09:45:36 pm
content: Transistor will start to conduct (breakdown) at some point above it's rating, so voltage is unlikely to reach anywhere near 500V. It might be 75V or 100V, no one really knows.Is it dangerous for a transistor? It depends. For an avalanche rated mosfet for sure it will be no problem. For some others maybe also ok. But some transistors might get damaged. It is important to note that short, momentary current might exceed that 10uA by many orders of magnitude. When breakdown occurs, it might be soft or avalanche. Avalanche meaning that it will continue to conduct while voltage is dropping. Remember, transistors have parasitic capacitance. So that capacitance gets charged. And when avalanche breakdown happens, that capacitance is discharging through transistor junction.Thats what I know, but I'm by no means an expert in unorthodox circuits.

author: JustMeHere
author_rank: Frequent Contributor
timestamp: February 08, 2024, 10:35:13 pm
content: The transistor when fully off is just a high value resistor. So it will divide the voltage with the other resistors.

author: jwet
author_rank: Frequent Contributor
timestamp: February 08, 2024, 10:51:48 pm
content: If you're studying electronics, you might have heard of Thevenin and Norton equivalent sources- this is another way to look at this. A power source can be modeled as a perfect voltage source with an output R or a perfect current source with a parallel R. The Thevenin Source would be what you have- 500V with 50 Megs. The Norton equivalent is a 10 uA current source with 50 megs in parallel. If those two circuits were in a sealed box, you couldn't tell the difference from just the output terminals- if you load them and measure current or measure them open, you'll get same values.The transistor would likely survive a bit of this but it won't be healthy. The leakage of the transistor almost saves you but as others have said, if something gets charged up slowly and then that capacitor discharged into the junction, it will be over. Small changes in temperature would also work against you. Diode type leakage doubles every eight degrees so a 25c rise cold give about 8x leakage, 25 more and you've got 64x.Good question though- might make a good question for a test or extra credit homework question. Have fun.

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2024, 11:06:14 pm
content: Why not draw a single 50M resistor, rather than 5, 10M? It would have made the drawing much easier to read.I say not, it's unlikely most transistors will be damaged by 10µA. Certainly not general purpose ones such as the BC548. It they were that sensitive, more would be killed by ESD, which bipolar transistors are fairly resistant too.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2024, 12:19:17 am
content: Avalanche occurs.Exact breakdown voltage depends on Ic, Ib, and dynamics (a little capacitance, at the right bias condition, leads to aggressive avalanche discharge, which can cause it to discharge very quickly to ~0V), and will be at least somewhat above the rating, but can be quite a lot higher (I've seen older transistors -- probably with looser controlled doping -- break down at maybe 3x the rating). A 50V part will almost certainly break down well below 500V, so will never really "turn off" here.Tim

author: David Hess
author_rank: Super Contributor
timestamp: February 09, 2024, 03:36:52 am
content: That type of operation occurs when you measure Vcbo on a curve tracer and I have never seen it damage the transistor unless the current was too high.

author: Jwillis
author_rank: Super Contributor
timestamp: February 09, 2024, 06:11:31 am
content: Quote from: David Hess on February 09, 2024, 03:36:52 amI agree. For something to do and personal knowledge I ran some tests on the a few SS9013 because of the low VCEO (20V) and ran them up to 60V. The Vce broke down at a bit above 20V and was fairly consistent between several samples. With a 1M collector resistor they didn't do much. No measurable change during breakdown. With 200 ohm down to 100 ohm they got fairly warm but not damaged. Below 50 ohm they promptly cooked and died. As long as the current didn't exceed the the maximum of 500mA nothing happened to the transistors over repeated cycling. They just acted like 20V blocking diodes until the breakdown occurred.Interesting experiment a bit of smoke. Fun to blow things up once in a while.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2024, 07:45:44 am
content: Larger transistors (Ic >= 1A?) typically fail from a single avalanche discharge event, even with no external impedance connected, more or less because their entire junction capacitance gets discharged through a tiny (100s nm?) channel; but again, this only happens when all conditions are present to trigger the effect. Other than that, for most if not all transistors, the current in question here should be too low for that breakdown mode.Most things can handle at least a little avalanche, before either breaking down catastrophically at higher currents (but without overheating) (or being biased into such a condition as above), but I haven't seen any GaN parts yet that tolerate any avalanche at all; blue/white LEDs for example typically fail suddenly at 20-30V, no matter how slowly you approach it.IGBTs are another example; they're rarely if ever rated for any avalanche energy at all (and with good reason, being four-layer devices..!), but are regularly rated in terms of breakdown -- Vces (C-E sustain voltage), meaning, biased under, say, a fraction of a mA. They're likely to latch up or fail outright if driven harder, but for small currents, yeah, you absolutely can measure it.BJTs are mostly the same way, as Vcbo transitions to Vceo depending on base current or wiring, and Ic. With base open, high avalanche currents are not a good idea, but with it grounded or reverse-biased, there can be some energy handling. (If nothing else, a switching type BJT must handle dynamic recovery, i.e. the C-B junction turning off, which acts like a rising avalanche voltage during the turn-off slope. RBSOA plots are often given for these types, which show what V and I are safe for what time scale.)BJTs have been used as zeners or TVS from time to time; a decidedly off-label use, but maybe not uncommon in the 70s before TVS proper came along. Not sure if there were specific types for this or it was more of a shoot-from-the-hip idea? Although the only example that comes to mind is for input protection / clamping in several Heathkit oscilloscopes.The main downside of BJT breakdown is the noise, either from poorly controlled breakdown compared to a proper regulator zener, or the jaggy avalanche I keep mentioning.Note that single events aren't repetitive events: even if something like a BJT or rectifier happens to survive a single avalanche event of nearly any current, it's no guarantee it always will. This is the main concern, at least as a design matter, distinguishing TVS from zeners: zeners are simply never rated for surge, or not the same way TVSs are anyway, while TVS obviously are. They might be identical dies, but without the datasheet saying so, you never really know for sure.Repetitive avalanche is a known hazard for many types; modern MOSFETs rarely if ever handle it, more or less a consequence of the gate oxide being too close to the avalanche site (along with other implementation details). Over time, charges get trapped in the gate oxide, until gate oxide failure occurs. Older types had repetitive as well as single-pulse avalanche ratings (e.g. most or all? HEXFETs); modern (trench and SJ) types usually have single-pulse only. Older generation also used comparable I_AS to I_D(max), whereas modern types it's usually a fraction thereof. (But to be fair, you rarely if ever use modern MOSFETs anywhere close to full ratings, so as to keep conduction losses down; which means, even if you do make use of avalanche (and, you probably shouldn't, but for those rare times when you do?), it's probably at comparable to the load current, so, this is fine. There are definitely some parts that use a suspiciously low I_AS (lower than the load current you would use) though.Speaking of old types: possibly, modern "linear" type MOSFETs are just the old planar design, rebranded. Though there's really just IXYS/Littelfuse making them, I think, and they only give single-pulse in their datasheets.I also suspect TI's NEXFETs might handle repetitive, thanks to a unique wrap-around design that puts more distance between the channel and avalanche site; though it might just be that the pitch is fine enough that it's affected anyway. In any case, they aren't testing repetitive avalanche, so the safe assumption is they do not, in fact, handle it.Tim

author: magic
author_rank: Super Contributor
timestamp: February 09, 2024, 08:08:37 am
content: Quote from: JustMeHere on February 08, 2024, 10:35:13 pmIt's not a resistors and will not behave as such.More like a diode, with some annoying extras.