Transistor will not shut off

Discussion in 'The Projects Forum' started by hensle, Apr 23, 2012.

  1. hensle

    Thread Starter Member

    Dec 31, 2011
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    I am having trouble with a project and I was wondering if anyone here could help. It is pretty straight forward so I would assume that someone has run into this problem before. I have attached a schematic of the circuit with related info. As an overview, the object of the project is to drive an inductive load of about 10 Ohms at a frequency around 900 kHz with a peak current of 1.8 amps. The schematic has the two parts finished so far. To the left is an armstrong oscillator which provides the correct frequency. It provides the necessary voltage as measured and displayed at voltage point A. This part works. To the right, the voltage is sent to two complimentary transistors to create a class-b (push-pull) amplifier. The voltage is sufficient, but I need to amplify the current. The load is not shown. The two 10k resistors are placed between the transistors while I test to make sure that the transistors are switching properly (before I actually start letting sizeable current flow through). Now here is the problem: The transistors are not shutting off. It seems to defy everything that a transistor should do. The oscilloscope readings are shown at points B and C, the expected reading and the actual reading.

    The maximum peak voltage at point A is around 40V. Now I thought that this might be the problem, but by messing around with the oscillator I was able to reduce this to a mere 2V peak to test this and still the transistors did not shut off.

    The frequency is fairly high and I thought that maybe capacitance was an issue. The transition frequency of the transistors is 10 MHz which should be sufficient, but to test this idea I reduced the freqency of the oscillator to 60 kHz and there was no change.

    I thought that maybe the transistors were bad, so I replaced them and also with other types of transistors, some that didn't even have the proper frequency response. While the voltage at B and C was significantly reduced with the wrong transistors, the transistors still did not shut off.

    Then I thought that I would put some diodes after point A to rectify the current before it even got to the transistors. Because of the frequency and voltage I used high power shottky diodes. The diodes DID NOT rectify the current at all!

    This is not my first project. In the past I have been able to shut off a transistor and have been able to rectify with a diode. What is it about this circuit that causes all the components to defy the laws of physics?
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    No laws are being broken. You have no path to ground from the bases. Transistors require base current. The capacitor blocks it. For your posted circuit, you can simply add 10k to ground from the bases. For a 1.8A load, your oscillator will not run due to loading. You will need a high-power buffer between the oscillator and the emitter followers.
    What is the vallue of your inductor? To get a peak voltage of 40V, and peak current of 1.8A into an inductor with 10 ohms series resistance, your inductor has to be less than 3.5uH.
     
    Last edited: Apr 23, 2012
  3. crutschow

    Expert

    Mar 14, 2008
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    As noted, you have no DC bias path for the bases of the output transistors. You cannot drive a transistor with only a capacitor.

    Try connecting a resistor between point A and the output (1kΩ to start). You also may have to increase the value of C4 (say to 0.1μF).

    But I'm don't understand why you are getting transistor current with no input. :confused: Are you sure the transistors are wired correctly?
     
  4. Ron H

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    Apr 14, 2005
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    I hate to say this, but I think we are wrong - sorta. In ideal transistors, with equal betas, the half-cycle of one transistor's base current charges the cap, and the other discharges it. Given a large enough cap, the voltage change across the cap is negligible. With real-world transistors, having unequal betas, the charge on the cap will be adjusted until the average base currents are equal.
    The problem in the real world is Vbe breakdown. The above scenario only works for input voltages<≈6V peak.
     
    Last edited: Apr 23, 2012
  5. mcasale

    Member

    Jul 18, 2011
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    I haven't done anything like this in a long time. My guess is that BASE leakage current might be keeping the transistors on - especially if they have a large BETA. Try sticking a big (10K or so) resistor across the Base-Emitter. This should help them turn off.
     
  6. Ron H

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    I edited post #4.
     
  7. hensle

    Thread Starter Member

    Dec 31, 2011
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    Thanks for the suggestions.

    I tried the resistor in any case and nothing changed. I did not expect anything because in this set up you do not have to bias the resistors. Typically designers will bias it a little bit to get rid of cross-over distortion, but this is not required here. In this case I definitely want to avoid biasing so that the transistors cut off properly, otherwise there will be a short.

    If this were the case no emitter follower could ever have an output pulse greater than 6 volts because the output pulse voltage is typically the same as the base pulse voltage (usually a teeny bit less). I have had them work up to 50V. That is because the vbe breakdown depends on the voltage *difference* between the emitter and base.

    BUT just in case, I did try the circuit with a 2V pulse and the transistors still did not shut off.

    Soooo, unfortunately the circuit still disobeys the laws of physics.
     
  8. hensle

    Thread Starter Member

    Dec 31, 2011
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    Thanks for the suggestion. I tried it and nothing changed.

    I tell you. This problem is driving me crazy, because it really should work as it is shown in the schematic. Another thing is how pristine the output wave is. Not only do the transistors not shut off, but there is not the slightest indication of the slightest effort by the transistors to do so.

    And this has been tested from 2-40V peaks and 60-900 kHz frequency,
    and with 4 different sets of transistors.

    I couldn't even get the diodes to rectify the current.

    I am at a complete loss.
     
  9. Ron H

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    Apr 14, 2005
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    With all due respect, you are wrong. You have the emitter resistors grounded, so the NPN emitter can never go below 0V. Meanwhile, the base is going to -40V. You better believe the BE junction will break down. The same is true for the PNP.
     
    Last edited: Apr 23, 2012
  10. CDRIVE

    Senior Member

    Jul 1, 2008
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    Judging by your remarks I think your confusing VEBO with forward bias voltage. Ron is referring to Maximum permitted reverse emitter base voltage (VEBO), which is only 5V for both your NPN & PNP. Once VEBO is exceeded and break-over occurs, the transistor is toast unless the reverse current was greatly limited. By the way.. exceeding VEBO is very similar to exceeding the PRV rating of a Diode. It kills them.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    Sounds like the transistors are toast and that's why they aren't working. Your connection of the individual output resistors to ground is the problem. Your attempt to protect the transistors may have caused their demise. ;)
     
  12. CDRIVE

    Senior Member

    Jul 1, 2008
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    If you don't want to fry any more transistors I suggest trying either one of these schemes. They will both work.

    Edit: After further analysis the circuit on the right will not be as effective as the one on the left. Since rectifier diodes can exhibit substantial parallel capacitance the circuit on the right may let a spike sneak through. This may also be a problem as frequency increases and the diode's junction capacitance progressively passes more reverse biased current.
     
    Last edited: Apr 24, 2012
  13. hensle

    Thread Starter Member

    Dec 31, 2011
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    Then no class b amplifier over 6V could ever work. Why? because at some point the reverse current would exceed 6V and it would break down. So how do such amplifiers work? I could be wrong...but I don't see how they would then work.
     
  14. crutschow

    Expert

    Mar 14, 2008
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    The problem is that you isolated the two emitter outputs with the two emitter resistors to ground. That is not a normal Class B amplifier connection.

    If there were one resistor between the emitters or one resistor from the emitters to ground, then the two emitters would always be at nearly the same voltage and the high reverse base-emitter voltage condition would not occur.
     
  15. CDRIVE

    Senior Member

    Jul 1, 2008
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    I posted two schematics to prevent exceeding Vebo but if you want to keep popping transistors you can just ignore that post.
     
  16. hensle

    Thread Starter Member

    Dec 31, 2011
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    Thank you for the suggestion.

    However, if you read the original post, I mentioned that I placed diodes into the circuit as they are placed in your schematic to the right. The idea was to rectify the current before it got to the transistors. This way there would be no high reverse voltage at the transistors. The problem was the diodes failed to rectify the current, even though they were high voltage rectifiers and should have been able to handle a reverse voltage of over a 100V.

    In fact, they had no effect on the voltage at all. Of course, this makes no sense and I am at a loss to explain it. Did I fry the diodes as well?
     
  17. CDRIVE

    Senior Member

    Jul 1, 2008
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    I assumed that this was not his finals but rather a driver to his finals.
     
  18. CDRIVE

    Senior Member

    Jul 1, 2008
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    Yes, it makes no sense at all. Did you try this on the existing transistors? I mean... they're shot!
     
  19. hensle

    Thread Starter Member

    Dec 31, 2011
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    This is interesting, but I do have one question. If I did connect the emitters and have the load to ground, how would this reduce the reverse voltage on the transistor bases? If the two emitters are at the same voltage, does that reduce the reverse voltage at the base?
     
  20. hensle

    Thread Starter Member

    Dec 31, 2011
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    OK. I replaced the transistors with brand new unused transistors of the same type. I then applied a sine wave to the base (from the oscillator) and maintained a small 2.5V peak at all times. There is no way that this small signal should fry the transistors.

    What happened?

    Exactly the same thing as before. The transistors did not shut off. The voltage at B and C where full waves identical to the wave coming to the base.

    The answer is not shot transistors.

    My best guess at this time is that there is a tear in the fabric of the space in my lab and the laws of physics no longer apply.
     
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