Switching by (ab)using a BJT - can it work? - Page 1
Forum section: electronics / beginners

author: acolomitchi
author_rank: Contributor
timestamp: July 24, 2016, 07:52:36 am
content: Question is:how much can one stretch a BJTin pulsedmode before it lets the magic smoke go?I sorta tried to figure out and it seems that it may work beyondthe SOA specified for continuous DC, but... maybe I'm wrong, please do tell me if so.Ok, to make the thing more concrete, let's say a2N5551in a circuit like that:The input is square pulses, with a 50% duty cycle, assume D1 ideal and don't mind if I'm just dumping the capacitor (normally, used as a bootstrap to switch on the upper side NMOS). I'll assume the max frequency being low enough to allow a 99% charge on C1 (period would be 2*R1*C1, letting C1 charge in 5*R1*C1).At input up, assuming saturation, the max current through Q1 will be (ignoring Vce(sat)) 12V/10R=1.2A - twice over the ratedcontinuousmax Iccurrent for 2N5551.However, the max power rating (Total Device Dissipation on page 2) is 625mW. With a Vce(sat)of 0.35V, 1.2A, I'll have a max power per cycle of 0.42W, which is comfortable less than the rated max. Besides, the current goes with Vcc/R*exp(-t/(R1*C1)), which means the instant power to dissipate decays - after log(2)=0.69*R1*C1, the current through Q1 goes below the max rated currents for DC and will stay there for the rest of 9.3*R1*C1 (up to 10*R1*C1). That is,iffQ1 survive repeated surges.True, Q1 should get hotter, with the "Thermal Resistance, Junction to Ambient" (page 2) of 200C/W, 0.42W should go with 84C over the ambient temperature, but the transistor is rated up to 175C so itshouldhandle it OK.(generally, the Vce(sat)goes down with increased temperature, while the hFEgoes up - the latter has even a diagram at page 4. So what worked when cool should work when hot as well).Now, the R1 resistor. The max power through it will be 144/10=14.4W. But... the current is not continuous, goes with an I0*exp(-t/(R1*C1)). Which means the R11 will have to dissipate Vcc2/R1*exp(-2*t/(R1*C1)). Making the average over 5*R1*C1 (the time in which R1 is active):* the total energy dissipated in 5*R1*C1 is Vcc2/R1*(1/2-exp(-10)/2)=(144/10)/2=7.2 J* the average power is 7.2 J/5 = 1.44W - under the 3W rating.(and then other 5*R1*C1 of inactivity, R1 may cool a bit before the next surge).So, I'd expect R1 to get warmish, but not quite hot.Iff the above is correct and neither Q1 or R1 goes vapours, then a switching freq of 1/(10*R1*C1) = 178kHz should be possible.Before I'll get my hands and try it, what do you think, can Q1/R1 withstand the abuse?

author: Dave
author_rank: Super Contributor
timestamp: July 24, 2016, 11:40:16 am
content: Soooo... What exactly is going to discharge the capacitor after it gets charged by the first pulse?

author: acolomitchi
author_rank: Contributor
timestamp: July 24, 2016, 01:04:21 pm
content: Quote from: Dave on July 24, 2016, 11:40:16 amRiiiiight! I simplified the thing a bit too much.Let's try again. This time it's the PNP under abuse, same max current, voltage and thermals vals as the 2n5551.Will it be able to take the abuse? (again, the C1 discharge part doesn't make any sense in real usefulness terms; I'm just interested in Q1)

author: Dave
author_rank: Super Contributor
timestamp: July 24, 2016, 07:25:06 pm
content: Well, now you have a slightly different situation, because the VEC_Q1is now going to be VEB_Q1+ VCEsat_Qlvl1, so roughly 0.9-1.0V.Why don't you just try to shove everything into SPICE? Should give you results good enough for a ballpark estimate.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 24, 2016, 09:26:06 pm
content: SPICE may give you results that are a little too inline with expectations. Namely, the hFE is usually pretty high to begin with, and the hFE may not roll off correctly at high Ic.2N5551 is down to 50 hFE at only 70mA Ic. Roughly speaking, 22mA into the base of Qlvl1 will allow it to draw up to maybe 200mA.Curiously, there's no base current maximum rating, so it's not obvious how well Q1 will survive this. Probably not well.So we put about 200mA into Q1, and, it's not obvious how much additional current it will draw. The 200mA will go B-E into C1 just fine (for the ~1.5us it takes to charge up), and it'll probably draw around 300-400mA additional collector current, for a total emitter current around the maximum 600mA rating.Which is a universal lesson about device properties: for the BJT, the continuous and pulsed ratings are also conveniently close to the real physical ratings. You can't draw much more current than that, and it's irritating to achieve (see how low the hFE is at this current?) and may be destructive (because the base contact isn't designed to handle that much current, and eventually fails).The fT peak is also usually before the hFE peak (in this case, slightly after, at 20mA, whereas hFE starts dropping off around 10mA), which is around where you can expect the best switching times as well. (BJTs tend to slow down at high currents; most notably, at high forward base current, it takes a longstorage timeto turn off again. And, like MOSFETs, that time is determined by how much current youremovefrom the base during the turn-off time.)So what DO you need to make this circuit work? If you're trying to draw a peak current of about 2A, you need a transistor rated more than 2A. A low-Vce(sat) type, like PBSS303NX/PX, or any number of related PBSS and ZTX parts, will do the job excellently. In that case, you'll only need a few mA from Qlvl1, which will also set the speed of the circuit, because (now with enough extra current capacity) the emitter follower only goes as fast as it's driven.It's still not obvious what you're trying to build, or if it'll be any good, based on what you've just shown here -- but this should at least give you some ideas about this section.Tim

author: rfeecs
author_rank: Frequent Contributor
timestamp: July 24, 2016, 10:56:51 pm
content: How about just reading the datasheet that you linked to?https://www.fairchildsemi.com/datasheets/MM/MMBT5551.pdfLook at figure 1 and figure 2 and see what happens when you try to put 1A through it.This part is intended to operate at around 10ma or so, not 1A.

author: acolomitchi
author_rank: Contributor
timestamp: July 24, 2016, 11:11:05 pm
content: Quote from: T3sl4co1l on July 24, 2016, 09:26:06 pmJust to confirm my understanding: you are saying that IC(max)is not the Ic over which a BJT will vaporize if you drive through it, but the maximum current that can ever be provided by the transistor, no matter how hard you try (meaning: it won't be anover-ICcurrentthat will destroy your BJT - just forget it, it won't happen, you can't reach it - but your attempts to convince the ICto go higher - i.e. the IBovercurrent).This corrects a serious misrepresentation in my mind about BJT operation, special thanks for that.And thanks for the extra references.Quote from: T3sl4co1l on July 24, 2016, 09:26:06 pmIn regards with "what I'm trying to build" - not necessarily something, rather to answer the question "how can one charge a capacitor the fastest way possible using BJT?".The answer seems to be "take a BJT that's able to deliver the amps you need in the mid of their spec - their max may be unobtainable" (or, if masochistic enough and with spare PCB area, use parallel BJTes with emitter ballasts).

author: acolomitchi
author_rank: Contributor
timestamp: July 24, 2016, 11:26:16 pm
content: Quote from: rfeecs on July 24, 2016, 10:56:51 pmFigure 2 is the most enlightening, thanks.(Vce(sat)going well up with increased Ic - the external sign of BJT trying to cope with increased current. It's obvious,once pointed out, thanks for that)In regards with the "intended to operate", the thread title already contains "abuse" - so I was accepting (looking for) scary operation modes.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 25, 2016, 10:20:56 pm
content: Quote from: acolomitchi on July 24, 2016, 11:11:05 pmYes, that's more-or-less the most you can get out of it, other limitations notwithstanding.MOSFETs are somewhat similar, but Id(pk) is usually only possible at Vds(max) -- at lower voltages, Rds(on) dominates, and the current flow is proportional to voltage. (In other words, current flow is resistive until a high enough voltage, at which point it levels off. Transistors are ultimately constant-current devices, it just might take an inconveniently high voltage or current to reach that region of operation.)Thyristors (SCRs and TRIACs), and diodes, have the curious property of being able to deliver almost unlimited current. Like BJTs, they are minority carrier devices*, but the BJT can only source as much current as the base-emitter junction can handle; eventually it saturates. Diodes can keep going above that level (but die of other mechanisms, like heating and electromigration), and honestly I forget why SCRs can too, but it probably has something to do with the junctions all being driven by internal current flows, whereas the BJT is only driven by external input current.*Meaning, the source of electrons and holes, within the semiconductor itself, comes from semiconductor physics (thermal motion and doping). Whereas in schottky diodes and MOSFETs, the electrons come directly from a metal, or follow similar physics at least.QuoteYup! Or use MOSFETs, but you still need enough capacity (low enough Rds(on), high enough Vds and Id(max)), and you still need to drivetheirgates fast enough. It's turtles all the way down...Tim

author: Marco
author_rank: Super Contributor
timestamp: July 25, 2016, 10:29:01 pm
content: Why can IGBTs handle higher current densities than BJTs though? They can be made short circuit proof for significant time, while BJTs would already be in secondary breakdown. Is it because the IGBT never goes into saturation?

author: rfeecs
author_rank: Frequent Contributor
timestamp: July 25, 2016, 11:44:27 pm
content: BJTs have a thin base layer, and the base current is fed from the edge of the base and flows in the direction parallel to the base layer. So the thin layer has a high resistance that causes a voltage drop across the base for higher base currents. Higher base current results in current crowding near the contact edge of the base which reduces the effective area of the transistor and eventually limits the current or causes hot spots, secondary breakdown, etc.SCRs, once they turn on, don't need the external current supply in the parallel direction through the gate layer. The current flows in the direction perpendicular to the layers through a wide area.IGBT similarly don't have that thin base layer.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 26, 2016, 01:44:48 am
content: IGBTs handle high currents, sort of by a similar current-multiplying mechanism as SCRs; except in SCRs, it's the counter-flow of electrons and holes, which feeds back on itself (hence the latching behavior). In IGBTs, it's majority carriers (from the MOSFET part) feeding into electrons and holes (the BJT part).Coincidentally, or perhaps unsurprisingly, IGBTs do have a four-layer (SCR) design, which can lead to latch-up.(Modern power switching devices are designed to avoid latchup for any collector and gate voltage within ratings. Special-purpose types may still be prone to it, like this curious device:http://www.onsemi.com/pub_link/Collateral/TIG058E8-D.PDFwhich carries a maximum dV/dt rating. They don't talk about it, but it's a dynamic latchup effect. Like SCRs, IGBTs can turn on again if the terminal voltage rises too quickly after turn-off. Think of it as capacitance from collector to internal-latching base.)I think it's popularly believed that IGBTs are somehow more rugged, but they aren't. That should be apparent from the headline claim: greater current density! They're subject to all the same limitations as BJTs and MOSFETs, thermally speaking (including the 150/175C Tjmax rating), but ramped up several-fold because of the higher current density AND the lingering threat of latchup. Indeed, this is the motivation for providing short-circuit ratings at all -- to show that a particular deviceisn'tas fragile as glass, but can withstand at least a little abuse!In other words, BJTs and MOSFETs rarely (if ever) carry a short-circuit rating, because it's in the 100s or 1000s of microseconds range, and fits with standard (linear) operation anyway, so you simply plot your operating point on the SOA curve, and check it out.So, SOA curves. IGBTs aren't intended for linear operation, so their SOA curves are often rather spartan. For example,http://rohmfs.rohm.com/en/products/databook/datasheet/discrete/igbt/rgt8ns65d.pdfFig.3 shows only two curves, 10us and 100us. And the 100us curve has a suspicious dip at high voltages (2nd breakdown).IXYS covers it in more detail, Fig. 11http://ixapps.ixys.com/DataSheet/DS100541B(IXYP10N65C3).pdfCuriously, most manufacturers show constant-power curves, which is more than a little disturbing to me:ST Fig. 14-15DM00149621.pdfAOS Fig. 9http://aosmd.com/res/data_sheets/AOD5B65M1.pdfFairchild Fig. 10https://www.fairchildsemi.com/datasheets/FG/FGD3N60UNDF.pdfOthers (including IR and Microsemi, I guess) provide the RBSOA, which is a lot less straightforward than the FBSOA: points are plotted based on the currentbeforeswitch-off (or after switch-on), and the voltageafterswitch-off (or before switch-on). I would say, in these cases, they don't provide an FBSOA because they don't intend them to be used in the linear range.For the ones that do provide normal-looking FBSOAs, I would be very interested in seeing actual test results, at high voltages. Even with test data, I don't think I'd put such a thing in a real design!Tim

author: acolomitchi
author_rank: Contributor
timestamp: July 27, 2016, 01:44:17 pm
content: Quote from: T3sl4co1l on July 25, 2016, 10:20:56 pmStumbled uponthis sm74611- an IC using a(n internal) FET masquerading a bypass diode:IF(AVG)8A typical, 15A maxVF(AVG)@ 8A - 26mVIR- 0.3uANo reverse recovery time spec - given the intended use and built, I wouldn't be surprised to find it in the usec range.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 27, 2016, 05:54:21 pm
content: Heh, neat.Check out that scope shot they did... note the 20ms/div time scale... try 2ms, not 2us!There had been a faster one, from On Semi I think, that was good enough for, say, mains rectification (but not really switching supplies, I think). Just not popular enough, I guess. LT also makes a FWB controller chip (add 4 external NMOS) for high efficiency mains rectification.Tim