Circuit Problem, keep smoking resistors.

Discussion in 'General Electronics Chat' started by ihateelectronics, May 29, 2011.

  1. ihateelectronics

    Thread Starter New Member

    May 29, 2011
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    Hello everyone, i'm new to the forum. How is everybody?

    Anyways, i'm teaching myself how to design amplifiers, and I built an output stage that was in a textbook of mine that I tweaked to fit the stages before it. In LT Spice everything seems fine, within range and what not, but when I built it and powered it up, the emitter resistor of Q10 automatically smokes and the current output on my PSU goes between 1-.5A. Any help would be greatly appreciated. Thanks everyone.
     
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  2. #12

    Expert

    Nov 30, 2010
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    I think it's an assembly error because the circuit checks out, as well as my feeble mind can see, and it's no match for LT Spice. Try installing 4.7k for R34. That will keep the smoke in long enough to do some measurements.
     
  3. newbies_hobbyist

    Member

    Jun 4, 2010
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    what is the wattage of your resistor? try to increase the wattage of R34 first.
     
  4. #12

    Expert

    Nov 30, 2010
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    The design puts .0135 watts on R34 if all the parts are good and the assembly was correct. One seventy-fourth of a watt. Not likely the resistor is too small to survive the design wattage. Excess voltage is arriving at R34. 4.7k would dissipate an eighth of a watt if the whole 24 volts tried to get through it.

    Maybe one of the diodes is open or they are not grounded. If the base of Q10 goes high, so does the base of Q9, and that applies the whole power supply to R34.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    One or both of the diodes connected backward could do it.
     
  6. #12

    Expert

    Nov 30, 2010
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    Try measuring and thinking. That's what this site is about, not replacing random parts without a clue in your head.
     
  7. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Hi,

    By simulating different scenarios, what I found so far using computer simulation only, is that 3 things would need to be in place simultaneously for your 15 ohm resistor to go up in smoke.

    1) a diode is not conducting either disconnected or in the circuit backwards.
    2) and the 100K resistor is a 10K resistor by mistake.
    3) and Q9 is shorted out so full supply is getting to Q10's collector.

    Under all 3 of these conditions together did the simulator show around 1 watt consumed in the R34 resistor.

    Another way would be if Q9 base was mistakenly connected to R27 and Q10 base connected to R33 that gives close to 190mW. thats with no diodes connected. according to simulatoer.
     
    Last edited: May 30, 2011
  8. #12

    Expert

    Nov 30, 2010
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    If a diode is not conducting, for whatever reason, the base of Q10 receives some current and Q10 starts conducting. The collector voltage of Q10 moves toward ground which lowers the voltage on the base of Q8. As Q8 shuts off, its collector goes high and that voltage feeds the base of Q9, which allows the 24 volt supply to flow into Q10. Because this is a DC coupled amplifier, it amplifies the DC errors and immediately goes into saturation, thus, the smoke.

    It is my opinion that thinking beats simulating.
     
  9. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Simulated Q8 out of cirucit entirely, also diodes removed, power in R34 = 49 mW.
     
  10. #12

    Expert

    Nov 30, 2010
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    I guess that proves 34 milliwatts is enough to burn up the resistor.
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    BC184 has the collector and emitter swapped relative to some other transistors, e.g. 2N3904. If the OP were in the habit of using EBC transistors, he might install BC184s upside down.
    I have a homemade model of 2N3904 which includes Vbe breakdown. If all the transistors were installed "upside down", Q9 and Q10 (in simulation) draw about 600mA each, and dissipate (briefly) about 6W each. They also dump about half an amp through the two diodes. If one of the diodes were to fail open before a transistor failed, then the dissipation of R34 would jump from about 100mW to a little over 2W. However, the transistors are now still at 3.5W apiece. Now, if R34 fails next, then Q9 drops to about 50mW, and Q10 Pdiss is basically zero.
    Yeah, this is a very questionable scenario. The only thing that seems to be a certainty is that at least one component would fail, maybe more, if Q9 and Q10 were installed upside down. Q8 doesn't really matter in this case.
     
  12. Laird Scooby

    New Member

    May 31, 2011
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    Was your power supply a self build job? Try connecting the earth rail of the amplifier to earth on the chassis and also make sure the earth of the power supply is connected to earth on the chassis. I had the same problem many years ago doing the same thing and that was the answer.

    Hope that helps!
     
  13. Adjuster

    Well-Known Member

    Dec 26, 2010
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    You must be joking! (Indeed, perhaps you are joking.) The rapid failure the OP is observing suggests that the resistor is running well above its rating - so what would that be, 25mW?

    These SMD resistors are rated at 50mW, and I don't think I could even see one, let alone solder it! http://www.resistor.com/assets/pdf/0402std.pdf
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I'm pretty sure #12's comment was sarcasm.
     
  15. #12

    Expert

    Nov 30, 2010
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    Yes, I was sarcasing. I recently learned the word, "hubris" (on this site) and I think it applies to people who act like, "I can Sim so I must have thought of everything (without measuring anything) and this could only happen with these three unlikely but simultaneous errors".

    My philosophy says that, with the information provided, the next step is measuring, thinking, and maybe even using a calculator. Replacing parts without measuring anything does nothing to consider assembly errors, and an assembly error is likely because the design seems to be good.

    I stand by the statement that a 4.7 k (1/2 watt) in the R34 position will stabilize the circuit so it resists smoking long enough for you to measure where the voltage and current went wrong. Five minutes with a voltmeter should clear this right up.

    ps, My philosophy also objects to teaching beginners bad habits, like not thinking or measuring, but just replacing parts willy-nilly.
     
    Last edited: Jun 1, 2011
  16. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Hi,

    I rather not use a simulator when designing circuits, If I can not explain and have a working knowledge of why my circuits working the way it is, then I would rather study up on it ,then just stick components in and out until something gives.

    When I first started out I would do a lot of experimenting that way, but I was more interested in having a working knowledge of why a circuit worked the way it did, not just becuase the Radioshack manual said to do it that way.

    My first 160in one kit I got from RS. was taken apart pretty quickly when I would start burning components left and right, trying to experiment with a lot of (what if scenarios), I wanted to kjow WHY to use that component, and the only way was to change values and KEEP a LOGBOOK close by.

    I Have my 2 original binders full of circuit configurations, and the why's of using components I used.

    Well after taking my electronics courses from NRI, and CIE, I began to get a better appreciation for circuit design. I loved learning how to design circuits from scratch.

    I have my electronics simulators, that I used quite extensively when I forst got them, but as the prices kept going up anf my simulators are from the 90's now, I had realized, am I going to rely on simulators for this hobby. NOT ANY MORE,

    Soa couple years ago, I decided I'm going to use calculator, data books, and pencil and paper, because they are affordable, and I'm going to no longer rely on rules of thumb to design circuits, but NOW I'm going to learn how to design circuits the way proffesionals do,
    that is they have a specific parameters to work under, so there designs are implenmted from different areas of the circuit, sometimes the input sometimes the output, circuit design is not a cut and paste methos, each design is unique on how it is approached depending on what it is being used for.

    So that's when I began to design circuits from the output, then from the inputs, then arbitrarily choose a outrageous parameter, and see how close can I come to a working circuit, or what chenges do I need to make to accomodate this parameter, and so the ENJOYMENT of circuit design has taken a very well turn for the bwetter for me as a hobby in this field.

    Now this is what I was in the process of writing to the OP when I got stumped by realizing the diodes could not be the only problem and decided to just simulate and see what could cause such catrostrophic failuere.

    Here is what I originally was goiing to post that evening.

    Hi,

    Here is how I would trouble shoot your circuit.

    1.) Take out all the AC components, (capacitors) so as to check bias levels only.

    2.) Remove Q10 from the circuit entirely check to make sure diodes are installed properly, then check for voltage across the diodes, if the diodes are conducting properly, yiou should have around 1.3V to 1.6V from ground.
    If you are getting a value close to the supply, then your diodes if installed properly, one or both may be opened. From there reinstall new diodes.
    DO NOT install new diodes UNTIL this measurement test proves outr, so as to eliminate one possibility of failure.

    3.) Now doing a current calculation, your R34 cannot be burning out through the base terminal.
    according to ( doing a first order approximation) ...[(VCC - Vbe) / 100K ohms] = 233uA.

    So even if your diodes were backwards and the collector was disconnected and max current flow through the base diode would still only be in the uA range.

    So the base current is not a deciding factor in this. Due to a high value for R27.

    My conclusion is R27 is a value much lower than 100K for the base current to cause high wattage at R34.

    So lets determine what value of resistor would cause this effect if the diodes were not operating.

    (I^2 x R)n = P so using 250mW as our base, lets rearange the equastion to solve for current needed. would be around 130mA to flow directly through R34 to max it out.

    What series resistor would be needed for R27 to provide this current.
    roughly 165 ohms in series with 15 ohms of R34.
    Is it possible you have a 100 ohm resistor mistakenly in there, and the diodes are not connected properly?

    I'll simulate this and see what could be.
    ---------------------------------------------------------
    Well by the time I got done writing all this, I decided to just simulate it just for curiousity and see what could make that resistor get so hot so quickly.

    Because the diodes being backwards still stumped me, unless the R27 was mistakenly put in too low.

    So I simulated it and tried to come up with all the scenarios I could think of at the moment.

     
  17. hobbyist

    Distinguished Member

    Aug 10, 2008
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    56
    Hi,

    I am not being sarcastic in this reply, but mean this in good intentions,

    I don't think putting another value resistor in its place of greater magnitude in value and wattage is going to serve a good purpose in locating the fault.

    Because the design is for the 15 ohm resistor, so now by putting a much higher value in there will upset any voltage reqirements for the circuit to work.
    It is being taken out of its designed parametes.

    The circuit needs to be functioning in its proper biasing values.
    My suggestion is to remove the Q10 altogether, and in its place put a resistor of a value that will drop 12v. across it so the output has a 12v. quiescent value.

    See if any other components are failing, if the output remains stable then take confirming measurements in key places around the citrcuit, and then begin to lower this test rersitor, a little at a time and check voltage drops in key places, until you find where excessive currents are flowing.

    Then you have a base to work off, of where the component values cause problems.

    A way I would do it is on a sepeate breadboard construct each stage individually, check for proper voltages, then begin to connect each one at a time, until you find excessive currents.

    It may mean to use temporary, resistors instead of Q8 for the base voltage resistor for Q9, temporarily, just to check individual component tolerances, once every stage works properly, then begin to recionstruct it constantly checking measurements until a failure arises.

    Another thing that happens often on breadboards, the OP may have components leads shorting out somewhere.

    That's just one of many ways to approach it.
     
    Last edited: Jun 1, 2011
  18. Laird Scooby

    New Member

    May 31, 2011
    11
    0
    There's only one proper way to do this, assuming the amplifier earth/ground is connected to the power supply earth/ground and that the power supply itself isn't faulty and causing the problem. The way to do it is start at the beginning - ensure all components are as specified and in the correct way round. If this is ok, go through the entire circuit with a multimeter to check that resistors haven't burnt out, verify the diodes and transistors etc.

    If it's all looking good at this point, connect the multimeter across the supply rails on voltage and switch the PSU on. Observe the voltage and make a note of it. It should be steady anyway. Now start probing round the circuit measuring voltages WRT earth and make a note of them.

    Having just had my first proper look at the circuit, i'm a little confused why you have 2 1N4148 diodes in series across the base of Q10. This will put 1.2-1.4V on the base of it, causing it to conduct hard all the time. Perhaps this is the problem? One final question - did you adapt it from having a "matched pair" output stage, in other words a PNP and an NPN transistor instead of the two BC184s? If so, it's highly likely the biassing is all wrong and will cause either the problem you're getting or absolutely nothing at all!

    For test purposes, you could temporarily disconnect R27 so Q10 would then be off and less likely to fry!

    Hope this helps,

    Dave
     
  19. Laird Scooby

    New Member

    May 31, 2011
    11
    0
    Just noticed something else - you appear to be running on 24V and the emitter resistor on Q10 is 15Ω. Under DC conditions, this means it will pass just over 1.5A. Quick mental arithmetic says it will be dissipating something in the region of 34W - almost as much as a light bulb! Put another way,

    P= I x I x R so 1.5 x 1.5 x 15 = 2.25 x 15 ≈ 34W.

    Granted it shouldn't be running dc conditions most of the time but it's still going to be dissipating a serious amount of power!

    Have you got the original circuit as it appeared in whatever you copied it from? I'm guessing LT Spice is an author but not one i'm familiar with, sorry!
     
  20. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Hi,

    Again I mean this in good intentions,
    and I say this in all due respect,

    Am I missing something then?

    The R27 is in series with the base of Q10, it drops the voltage, and thus the current
    by [(VCC - Vbe) / (R27+R34)]. Not including HFE parameter which would raise the input impedance higher.
    This is with collector opened.

    The 2 diodes keeps a good base bias voltage on Q10 to help put the amp into Class AB mode.

    Needs to have some base current to keep from cross over distortion.

    For fun I breadboarded the circuit last night using different resistor values to design it for an output current of 1mA, with no load, and got a 16vP-P with no crossover distortion.
     
    Last edited: Jun 1, 2011
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