BJT transistor all terminals shorted - Page 1
Forum section: electronics / beginners

author: piguy101
author_rank: Regular Contributor
timestamp: August 13, 2017, 01:53:46 pm
content: Hi. I am making a CC/CV power supply, using a BJT NPN transistor as the main power element. The base is controlled by an op amp as shown in the picture. The power supply seemed to be working well in both CC/CV modes. I would set it to something like 5 volts and short the output and it would limit current to my setpoint (which was around 0.5 A). I tried increasing the voltage to 12 V and then shorted the output again, and BANG the fuse blew.I then tested the pass transistor and all of the terminals are totally shorted together (<1 ohm between any two pins). So I guess when I shorted the output, the transistor must have failed short and the fuse blew. Have any of you seen this behavior before with all pins shorting? Do you have any idea what caused the failure and how to fix it?Would adding a large capacitor in parallel with the load, such as 100-1000 uF reduce the stresses on the transistor when the output is shorted?Thanks

author: Ghydda
author_rank: Regular Contributor
timestamp: August 13, 2017, 02:16:20 pm
content: Quote from: piguy101 on August 13, 2017, 01:53:46 pmA number of different things could have caused your pass transistor to go belly up.We need more information: A schematic would be good. With part numbers on.Cheers!

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 13, 2017, 02:17:44 pm
content: you dont have limiting resistor between opamp output to bjt's base. that probably will create chain reaction when BE diode is damaged due to excessive current running pass through it..

author: Yansi
author_rank: Super Contributor
timestamp: August 13, 2017, 02:26:00 pm
content: Post the full schematic of what you are building, otherwise we can only start polishing our oracle dodecahedrons...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 13, 2017, 02:30:11 pm
content: Quote from: Yansi on August 13, 2017, 02:26:00 pm(I know, it's not quite a dodecahedron, but it's the most commonly large, ball-like crystal I could think of.)Tim

author: piguy101
author_rank: Regular Contributor
timestamp: August 13, 2017, 02:34:35 pm
content: Sorry, I should have given more details. The schematic doesn't actually exist yet. So I'll post it soon. My apologies

author: Audioguru
author_rank: Super Contributor
timestamp: August 13, 2017, 02:57:16 pm
content: Your circuit controls only the output voltage. A second opamp is needed to sense and control the current by reducing the output voltage until the current is at the current setting you want.By the way, if you use an opamp with an input phase inversion problem like a TL081 then when its input voltage becomes within a few volts from its negative supply voltage (which is probably ground in your circuit) then the output goes as high as it can.

author: piguy101
author_rank: Regular Contributor
timestamp: August 13, 2017, 03:34:26 pm
content: Thanks for the replies and sorry for being unclear. Attached is part of the schematic that should be relevant. There are two setpoints, Vref and Iref, which are both controlled with potentiometers. There are two feedback lines, Vmeas and Imeas. Op amps C and D act as comparators to see if the supply should be in CC or CV mode, depending on which one is more limiting, just like how a real bench power supply operates. Op amps A and B are precision rectifiers, effectively choosing the lower of the two outputs from C and D and put that voltage on the base of Q2. The LEDs are used both as circuit diodes for the precision rectifier circuit and as indicators showing whether the supply is operating in CC or CV mode. The transistor configuration is darlington, and Q1 is the transistor that shorted on me. Q2 is still healthy. The circuit did seem to work at low power (voltage, current?) in CC and CV modes, so I am not sure what killed it.Audioguru, I don't think that I am having that problem because I am feeding a slightly negative voltage to the op amps in order to get the output down to 0 V, but I may be totally wrong.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 13, 2017, 03:46:29 pm
content: The inputs are unspecified, and there are no feedback/measurement points. Assuming the inputs are arbitrary signals, you've got two wired-OR comparators and an emitter follower blasting a resistor.It's no surprise that it shorted, but it's also nowhere near a CC/CV supply!Tim

author: MK14
author_rank: Super Contributor
timestamp: August 13, 2017, 04:21:29 pm
content: Ignoring all the other obvious circuit issues, including what others are saying.The LEDs (which you don't seem to have specified, so I can't check the datasheet), probably have a reverse voltage limit of around 5 volts. Anyway, I doubt it is anywhere near 27.6 (24 + 3.6) volts (minus somewhat small voltage drops, still leaving it above about 24 volts or so).

author: Zero999
author_rank: Super Contributor
timestamp: August 13, 2017, 04:24:02 pm
content: Quote from: Mechatrommer on August 13, 2017, 02:17:44 pmA base resistor isn't necessary to limit the current. The transistor is configured as an emitter follower, so the base current should be self-limiting. An increase in the base voltage, causes an increase in emitter current, resulting in the emitter voltage rising, thus reducing VBEand the base current. If the circuit oscillates, then a base resistor might be added to reduce the gain, but that's a different matter.

author: Ian.M
author_rank: Super Contributor
timestamp: August 13, 2017, 04:57:00 pm
content: *ASSUMING* there is actually a current control loop, Its quite possible it isn't fast enough and shorting the output transiently took the transistor outside its S.O.A.If you examine the S.O.A graph in theD44H8 datasheet, the DC limit is a little over 4A @12V Vce and around 1.5A @25V, and for a 1ms pulse, 20A @12V, dropping to about 7A@25V.If one added a 0.47R emitter resistor to Q1, with the B-E junction of another NPN transistor across it, and its collector connected to the base of Q2, it would shunt the Darlington pair's base drive at about 1.5A load current making it essentially bulletproof as long as the supply is 24V or less.If the current control loop is adequate to keep the transistor within its DC S.O.A line, the emitter resistor could be decreased to 0.1R, limiting the peak current to between 6A and 7A which will prevent the output transistor exceeding its 1ms pulse S.O.A. line.I'd also add a fast reverse biassed diode across the load so the output cannot be taken negative by current flowing in the load inductance when the C.C. limit trips.

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 13, 2017, 06:07:43 pm
content: Quote from: Hero999 on August 13, 2017, 04:24:02 pmthe OP was saying about shorting the output, i assume shorting the Rload, so there must be tremendous current going through the BE, shorted diode as well, no?

author: Ian.M
author_rank: Super Contributor
timestamp: August 13, 2017, 06:48:53 pm
content: Quote from: Mechatrommer on August 13, 2017, 06:07:43 pmNo. The LEDs prevent the OPAMPs sourcing current into the base of theKSC1009driver transistor.(Although their nominal reverse voltage will be significantly exceeded if the load is shorted and the OPAMP outputs rail high, LED reverse voltage is usually specified to prevent junction defect growth that excessively degrades their luminous efficiency over their rated lifespan, and is not an avalanche breakdown limit)Therefore R7 will limit the driver base current to about 10mA. The KSC1009 current gain drops rapidly above 200mA Ic, and its unlikely for there to ever be more than 0.5A base drive to the output transistor. As the KSC1009 survived, and its rated for 800mW dissipation, that would suggest it cant have supplied more than 33mA for an extended period. An x10 overload (330mA) for one second would almost certainly have blown it, which provides another strong indication that the D44H8 base current was limited.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 13, 2017, 07:11:31 pm
content: Quote from: MK14 on August 13, 2017, 04:21:29 pmGood observation, from a design standpoint.In practice, red LEDs avalanche around 30V. But they're only tested at 5V. Which one are you going to trust?Tim

author: Mechatrommer
author_rank: Super Contributor
timestamp: August 13, 2017, 07:15:46 pm
content: Quote from: Ian.M on August 13, 2017, 06:48:53 pmright, possibly that. my answer was posted before the OP showed the actual circuit (which still fail to indicate feedback loop). even though D44H8's base was limited, i guess still enough to make it work beyond its SOA rating during the short, hence the magic smoke...

author: MK14
author_rank: Super Contributor
timestamp: August 13, 2017, 07:25:01 pm
content: Quote from: T3sl4co1l on August 13, 2017, 07:11:31 pmEDIT N.B. To be clear, I agree with you. But there is a small possibility that the red LEDs, may have badly reacted to voltages less than 30V, hence the rest of this post.Well looking at this thread (which ironically we both participated in):https://www.eevblog.com/forum/beginners/what-is-the-reverse-breakdown-voltage-of-a-red-or-green-5-mm-led/msg1114021/#msg1114021It seems that potentially, at around the 20V .. 23V, it fails (open, closed or avalanche), forSOMEof the RED LEDs, others at higher than 23V or so. I.e. it depends on what red LEDs they used and luck.tl;drThe LEDsMAYhave caused the output transistor failure.Without seeing theFULLschematic, I'd prefer to NOT try to over-analyse this circuit.E.g. the actual current measurement circuit, may have played a role in all of this.Also perhaps the common-mode input range of those op-amps is being exceeded, which can cause problems.

author: Zero999
author_rank: Super Contributor
timestamp: August 13, 2017, 07:47:27 pm
content: Quote from: Ian.M on August 13, 2017, 06:48:53 pmTo be fair, in the post Mechatrommer was responding to, there was only an op-amp and transistor in the feedback loop, with nothing else. In that case, I still doubt it would be the base-emitter to blow, even if the output is short circuited. Op-amps limit their output current to around 20mA to 60mA, which won't hurt a power BJT and it would be the enormous collector current and power dissipation which would have destroyed the transistor, rather than excessive base current.

author: piguy101
author_rank: Regular Contributor
timestamp: August 13, 2017, 08:39:52 pm
content: Quote from: MK14 on August 13, 2017, 04:21:29 pmThat is a legitimate concern, and they are junk bin 5 mm red LEDs. I just tested them now. I put 32 V across the LED in reverse and measured <0.1 uA flow for each LED. So at least in steady state, they seem fine. Is this test sufficient?Quote from: T3sl4co1l on August 13, 2017, 03:46:29 pmThe feedback lines are for Vmeas, a voltage divider scaling the variable max 24 V output to 5 V max into the comparator. The current feedback, Imeas is a 0.1 ohm shunt on the low side before ground. Vref and Iref come from potentiometers.Please do criticize this circuit, as it is very helpful for my learning! Thanks

author: MK14
author_rank: Super Contributor
timestamp: August 13, 2017, 08:45:16 pm
content: Quote from: piguy101 on August 13, 2017, 08:39:52 pmThat does seem to be sufficient. That test helps, because it narrows down what has caused your problems.So other things caused your problem(s). Obviously, the LEDs should not really have such large reverse voltages.

author: MK14
author_rank: Super Contributor
timestamp: August 13, 2017, 08:57:20 pm
content: Quote from: piguy101 on August 13, 2017, 08:39:52 pmI think you may be violating the common mode input voltage, on the second set of op-amps (the ones on the right), when it is brought up towards 24 volts (whenever there is NO load, or very little), via the 2K2 resistor. This can badly upset the op-amps (get them to misbehave). Such an effect can cause latch-up, depending on the type of op-amps used.Since you have not supplied a complete schematic, I've had to visualize it (from the partial schematics and text descriptions), and I could be wrong.When I checked it earlier, I used the following datasheet, which seems to show that the common mode input range extends to about 2 volts below the positive supply rail.http://www.onsemi.com/pub/Collateral/MC34071-D.PDF

author: Zero999
author_rank: Super Contributor
timestamp: August 13, 2017, 10:17:16 pm
content: Quote from: piguy101 on August 13, 2017, 03:34:26 pmNo the op-amps aren't comparators but error amplifiers. A comparator's output is either on or off, depending on which input voltage is greater. An error amplifier, is put inside the feedback loop of a more complex circuit/system. It compares the required voltage on one of its inputs, with the voltage at the feedback node on its output input and alters its output to try to make the voltages on its inputs equal one another.QuoteOne disadvantage about using a Darlington pair, is it will have a voltage loss of 1.2V at tiny currents and approaching 3V at very high currents. Another issue is, as the output voltage is increased, the available base drive current decreases. When the output voltage is near zero, the base current will be around 10mA but when it's set to 20V, the base current will drop to 364µA. R7 could be replaced with a constant current source with a low drop-out voltage, to keep the base current constant, irrespective of the output voltage.By the way, how are you sensing the voltage and current? That makes a difference, as it might be oscillating. Have you connected something which draws a pulsed current waveform, such as a MOSFET and resistor with the gate driven by a signal generator and looked at the output on an oscilloscope? It wouldn't surprise me if it oscillates, when a load is suddenly connected/disconnected.Quote from: MK14 on August 13, 2017, 08:57:20 pmI agree. I think the current limiting is failing when the input voltage is taken outside the CMMR, causing the op-amp's output to jam on and the output transistor to overheat.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 14, 2017, 12:01:47 am
content: Quote from: Hero999 on August 13, 2017, 10:17:16 pmNote that the distinction is not very meaningful here, as the circuit is not compensated: it will most likely oscillate, regardless of whether, say, LM358 or LM393 were used.To add compensation, you need a series input resistance to the -in pin, and an R+C from out (the op-amp output) to -in. The series resistor will probably be 1-10k or so, and the R+C will depend on the behavior of the feedback loop around the op-amp. A typical RC time constant of 100s of us would be a good starting point, and R ~= Rs. Adjust values until the step response is as fast as it can be without ringing (you need an oscilloscope for this).You have the added trick of CC to CV operation. When one op-amp is closing the loop, the other is saturated to +V. When a threshold is crossed, the op-amp's output must fall all the way from +V to the voltage where the LED pulls on. Op-amps have limited slew rate, so this takes a lot of time: microseconds. During this time, if you've suddenly applied a short circuit, for instance, the transistors are simply delivering whatever voltage they can -- the short-circuit current is unlimited for those precious microseconds!Microseconds aren't enough to destroy transistors (usually), but if the recovery time constant is milliseconds instead of microseconds, you've got dead silicon on the bench!It's often better to design a power supply with CC operation in mind first, then to put a second control loop around that, to enforce CV operation. This way, the CC loop reacts instantly, limiting surge current. (To go with that: the transistor would be PNP, common emitter, so its own output has a constant-current characteristic, even before the op-amp responds.) The CC loop is "programmed" with a setpoint; that setpoint has a limited range, so even if the voltage controller (that's generating the "program" signal) goes nuts, it simply can't call for any more than maximum current!What's best, is this architecture flows seamlessly into SMPS design: tack on a PWM modulator, switch and inductor (so that you are controlling inductor current), and you have the most robust SMPS controller!Tim

author: Zero999
author_rank: Super Contributor
timestamp: August 14, 2017, 08:27:57 am
content: I agree, it's oscillating. The trouble is, it might be stable under certain conditions and only oscillated when the load current is stepped or it changes from CV to CC or vice versa.One of the problems, is there are too many op-amps in the feedback path. IC4A and UC4B are not required and can be removed. The ORing diodes will still work with IC4C and IC4D connected to Q2.