Help with burning component

Discussion in 'General Electronics Chat' started by Veracohr, Feb 26, 2014.

  1. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    I'm hoping someone can help me figure out why a component keeps burning on the circuit I'm testing.

    In the attached schematic, the part of the circuit in the upper right is a current source that is supplying bias currents to the OTA's. It's designed to supply up to 600μA to each OTA, the upper end of which is limited by R98. In simulation it is limited to 4.84mA total out of Q4. I simulated each of the eight OTA's as two diodes in series, because the Iabc input to the OTA (LM13700) is two diode drops above the negative supply. Even simulating a short from the collector of Q4 to the negative supply doesn't cause the output current to rise above the limit set by R98.

    There's a little more to the circuit than what you see in the schematic, but I cut it out for simplicity. I don't think anything I cut out would be part of the problem.

    The problem is that unless I put a resistor between the collector of Q4 and the OTA's, Q4 burns. Not always immediately, sometimes it takes a few seconds after turning on power, but it's always Q4, and I'm pretty sure it's always the emitter of Q4 that has a visible burn mark on the component. On my breadboard test circuit Q3 and Q4 come from a FPQ2907, which is four 2N2907's in a PDIP package. I've burned up a bunch of these things now. I have used a resistor down to 470Ω without anything burning up.

    - The transistor is rated for 150mA, I'm only asking it to supply 5mA.
    - The transistor is rated for Vce of 40V, it should only be getting Vce of 22-23V.
    - The OTA's aren't getting damaged, only Q4. I know from past experience how easy it is to burn up a LM13700 by supplying too much current to the Iabc input, but that's not happening here.
    - I've used the same current source, with different resistor values to provide different current limits, in OTA circuits using two and four OTA's, without adding another resistor and without Q4 burning. Those circuits worked as designed and simulated.
    - I've tested the output of Q4 by disconnecting it from the OTA's and putting the current into a resistor to ground. Testing this way, it's pretty darned close to what it should be.

    I've also attached a picture of my breadboard. The PDIP containing Q3 and Q4 is the first black one on the left (not counting the SOIC on the green adapter board); the four to the right of it are the OTA's. It's kind of hard to see but there's a resistor from the upper right pin of the transistor array to a point between the two DIPs, and from there it goes to the four OTA packages through a short red wire, two long loopy green ones and a long loopy black one. Each of those four wires goes to one Iabc input of the OTA chip, and then a short jumper wire goes to the second Iabc input in the package (two OTA's per package). Without that resistor, I had those four wires going directly to the upper right pin of the transistor array (collector of Q4).

    Sorry for the length of this post, I just wanted to give you all the info. Any idea why Q4 is burning without a resistor?
     
    Last edited: Feb 26, 2014
  2. crutschow

    Expert

    Mar 14, 2008
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    You circuit schematic appears to be in error. There is no path for the base current for Q3 and Q4 so they could not turn on. I assume the bases are connected to the collector of Q3 to make a current mirror(?).
     
  3. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    Oh yeah. Boy that could have caused me some completely different problems later on, thanks.

    You're correct. The schematic is updated now.
     
  4. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    It looks to me like your schematic does not match the breadboard layout in the area of the current source. The schematic shows a pin 16 and a pin 15, but the chip looks like a 14 pin DIP. I can't find a pinout or a datasheet for an FPQ2907, but I did find one for an MPQ2907, which is a 14 pin DIP.

    In the breadboard photo, it looks as if there is no connection at all to pins 4-11, which would mean that Q2 and Q3 are not being used. The emitters of Q1 and Q4 are connected together and then to a power rail, the bases of Q1 and Q4 are connected together, the input lead from the TLA074 is connected to the collector of Q1, and the collector of Q4 is connected to a resistor, which feeds the OTAs.

    I am confused.
     
  5. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    No, the schematic and board don't match exactly. The final project (it's a school project) will use mostly surface mount IC's, whereas I used DIP where I could for the prototype testing for ease on the breadboard. In the final project the four transistors will come from MMPQ6700, which contains 2 NPN and 2 PNP. I didn't have any DIP equivalent so I used a MMPQ2222 with a SOIC-DIP adapter for the NPN's, and a FPQ2907 for the PNP's (pinout matches MPQ2907). Each package has 4 transistors, of which I'm using 2. For the PNP's, I'm using the top 2 in the package, Q1 and Q4 in the datasheet:

    Datasheet --> Schematic
    Q1 --> Q3
    Q4 --> Q4

    The 2 opamps labeled TL074 in the schematic come from a TL072 on the board which is in the lower left by the blue trimpot.

    I'm positive everything is hooked up right. As I said, it works just fine when I put a resistor in there, and it works just fine without a resistor when using 4 or 2 OTA's. Only when I'm using 8 OTA's as in this circuit does it burn up.

    Although you can't see it in the picture, the base of Q1 (schematic Q3) is also connected with a tiny jumper to the collector of Q1. I'm not at home at the moment to verify but it must be hidden behind the red and black wires. The flat green wire connected to Q1's collector is going to the NPN array that is on the green SOIC adapter.
     
  6. tracecom

    AAC Fanatic!

    Apr 16, 2010
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    Then, why would you post a picture of the breadboard?

    I have no idea what your circuit is supposed to do, but i have built lots of circuits on solderless breadboards. Trying to assemble a circuit of this complexity without a schematic that exactly defines what you are building is begging for assembly errors.

    But, assuming for a moment that there are no assembly errors, then there are only two possibilities that I can think of: erroneous design or defective components.

    For my edification, please post a link to the datasheet for the FPQ2907.

    ETA: Now that I understand that an MMPQ6700 is simply an IC containing two 2N3904 and two 2N3906 transistors, I wonder why you assembled the solderless breadboard with two ICs and an SOIC adapter (a total of over 40 leads) instead of four (very common) discrete transistors (a total of 12 leads)? Not only would discrete transistors have made assembly easier and troubleshooting simpler, but also would have reduced the cost of "burning" Q4 to less than ten cents.
     
    Last edited: Feb 26, 2014
  7. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Q3 and Q4 must be identical and biased identically for a true current mirror. Although they share a common package, plugging them into a breadboard could introduce contact resistance such that they don't pass identical currents. Hence Q4 current could perhaps have a considerably higher current than Q3?
     
  8. Veracohr

    Thread Starter Well-Known Member

    Jan 3, 2011
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    In case there's some quirk of the layout or connection scheme that may have had some unforeseen consequence that someone with more experience than I could spot.

    FPQ2907 Datasheet (doesn't contain pinout)
    The pinout diagram I've been using I think comes from the MPQ2907 datasheet. I verified with testing that the pinout matches.

    I assumed that two transistors in one package, even if they're not explicitly "matched", would likely be more closely matched than two discrete transistors which may not be from the same manufacturing batch at all.

    "Considerably" higher? Even if that were the case, that current would have to go somewhere. If it were high enough to burn up the transistor (rated for 150mA), it would be more than high enough to burn up the Iabc input on at least one of the OTA's (absolute max 2mA). And there's also the fact that if I disconnect the OTA's from the current source and use a resistor as the load, I measure the expected current.
     
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