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Analyse this amplifier for me please :)

  1. Jun 28, 2015 #1
    Hi folks, I've designed this amplifier, it uses an Ne5532p op-amp for the voltage amplification and feedback in the non-inverting configuration. The output stage is handled by two Darlington transistors, TIP 121 and TIP 126, these are hopefully biased by TIP 117 and TIP 112.

    I was hoping that some more experienced engineers could take a look at this circuit and give me feedback on its performance, and any tips to improve it would be greatly appreciated.

    I have managed to build this circuit and it sounds okay to me, I can't hear any distortion, even at low volumes, but are my ears deceiving me? It operates off +- 18v, and I was also wondering about the output stage, is it operating in class AB, or is it purely a class B stage? I don't think it's class B because surely the crossover distortion would be horrible but If you guys could clear up my confusion I would greatly appreciate it, many thanks,

    snipd.PNG
     
  2. jcsd
  3. Jun 28, 2015 #2
    The way to find out would be to take a digital sample of the input and output, then do some kind of statistical analysis of the comparison instead of using your ears.
     
  4. Jun 28, 2015 #3
    assuming you did all the math on the biasing right you'll need a scope to check the transistor inputs as well as running a function generator with sine wave and doing a sweep, one thing I do notice that your missing is a low pf cap across the 56k resistor to prevent high frequency oscillation.
     
  5. Jun 28, 2015 #4
    on second thought I notice another problem, your output bias, you don't have any positive feedback for the output transistors thermal regulation and you may have a problem with the input and/or output transistors where you'll need .7 volts lifted at the gates, this can be done with a couple of small signal diodes in the right place.

    /me waits for a more experienced member to comment.
     
  6. Jun 29, 2015 #5
    What output power do you want from this amplifier ??
    Well I see you try to use a diamond buffer as a output stage. But why with two Darlingtons ?
    Also your output stage quiescent current is not well defined. Because now your quiescent current depends on vbe mismatch.
    Also connect your power supply directly to output stage and use RC for supply the OpAmp.
     
  7. Jun 29, 2015 #6

    Baluncore

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    If you download a copy of the free LTspice from the Linear Technology site you will be able to simulate it yourself.
    http://www.linear.com/designtools/software/
    If you have problems getting LTspice or your circuit to work you can ask or attach .asc files as text files to a PF post.
     
  8. Jun 29, 2015 #7
    Firstly, many thanks for the replies folks,
    - I was under the impression that this voltage was provided by the driver transistors TIP 117 and TIP 112, is this incorrect?

    - I was hoping to get around 25w into 4Ω @ ±18v. Is this obtainable, my circuit draws 30mA with no input connected.

    - I ordered them by accident, so I tried to make them work I realise its not ideal. The downside is my circuit seems to clip @ ±14.5v with power rails @ ±18v. I think this might be because the darlingtons require around 1.2v to turn on, is this correct? Or could it be the op-amp not capable of reaching the rails.

    - Should I make the TIP 117 and TIP 126 drivers say both TIP 126 instead so they are the same part. Then do the same with the NPN transistors.
    Leaving me with matching pairs? Would this help?

    Many thanks :)
     
  9. Jun 29, 2015 #8

    Averagesupernova

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    How can you make both outputs the same part when one needs to be PNP and the other NPN? I would say you are correct in saying that the way the driver transistors are configured should bias the outputs correctly. However, this assumes the drivers base-emitter voltage is similar to the outputs base-emitter voltage. If the outputs are darlingtons then of course this won't be true.
     
  10. Jun 29, 2015 #9
    Thanks for your input Averagesupernova,

    What I meant is make all the NPN transistors the same model and all PNP transistors the same, so the above schematic would have (2x) TIP 121 NPN and (2x) TIP 126 PNP.
     
  11. Jun 29, 2015 #10
    Last edited: Jun 29, 2015
  12. Jun 29, 2015 #11

    Averagesupernova

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    I would not make the drivers the same as the outputs. The base currents will be much larger in the outputs compared to the drivers so the base-emitter voltage drop will also be different and you want the to be as close to the same as possible.
     
  13. Jun 29, 2015 #12
    I've not used a diamond buffer.

    It looks like it would have a low output impedance, but the input impedance seems high due to the ß > 1000 of the darlingtons. Since your emitter resistors are in series with the load, the input resistance to the darlinton emitter follower stage will vary with your speaker impedance (which dominates the input resistance equation). The speaker impedance varies with frequency. This might cause an impedance mismatch between your biasing transistor stage and your darlington stage that will vary with frequency. It should be a small effect and act somewhat like a graphic equalizer, so it's no big deal unless you're an audiophile. (The mismatch seems to be varying from about the right value to several times too high. Too high is better than too low.)

    Or I could be wrong.
     
  14. Jun 29, 2015 #13

    Averagesupernova

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    The negative feedback in this amplifier should help a lot with frequency response.
     
  15. Jun 29, 2015 #14

    meBigGuy

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    Last edited: Jun 29, 2015
  16. Jun 29, 2015 #15

    Averagesupernova

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    I can see where this output stage cannot be expected to drive very close to power supply rails as the only thing pulling on the base of the output transistors is a resistor. A somewhat clever biasing scheme though. I had never seen this til now but it jumped out at me and made sense.
     
  17. Jun 29, 2015 #16

    meBigGuy

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    So what sets the quiescent current at 0 volts out to make it class AB.
     
  18. Jun 30, 2015 #17
    The output stage is push-pull. Any quiescent current flowing through the speaker will cause a small voltage drop which will feed back to the negative input of the op-amp, changing the bias points of the push-pull transistors to limit the current. It's a small effect, but it only needs to overcome the mismatch in the darlinton betas.

    The quiescent bias current through the darlingtons seems a little wonky since they are being biased by the voltage drop across a regular BJT. (They "should" have twice that.) I think that would limit their RF response, but I don't think it matters much at audio frequencies. It does mean a low quiescent current through the darlingtons, but I don't see a problem there; more of a feature really. (There's a lot of non-linearity as well, but with a push-pull pair I don't think that's a problem either.)

    A very clever design.
     
  19. Jun 30, 2015 #18

    Svein

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    The circuit is not bad, but there are some areas that could stand an analysis and a redesign. Off the top of my head:
    • The power dissipation in the driver transistors will be a bit on the high side
    • Since there is almost no local feedback, only an overall feedback, I see a possible problem with TIM (Transient Intermodulation Distortion)
    I suggest that you take a look at http://www.w5dor.com/AppNotes/AN485-Hi-Power-Audio-Amps.pdf [Broken] for a thorough discussion of the stages in a Hi-Fi amplifier.
     
    Last edited by a moderator: May 7, 2017
  20. Jun 30, 2015 #19
    Why will the power dissipation be high? The current and voltage drop need to be dissipated somewhere. Admittedly there are switched amplifiers with lower dissipation, but power transistors are built for this sort of job. Make sure the heat sink is up to the task. (He said he built the circuit around the darlingtons, so I assume they are expensive/hefty.)

    It's my understanding TIM is when non-linear amplifiers unintentionally act as mixers. Push pull amps are tend to be pretty linear. That applies to the op-amp as well as the darlington stages. So the bias transistors would be the problem children. They look to be emitter followers with a linear gain ≅ their beta, but perhaps I'm missing something.
     
  21. Jun 30, 2015 #20

    Svein

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    No. The Ottala paper states that TIM is due to clipping and saturation in the intermediate stages when little or no local feedback is used. See http://www.shabad.ru/IEEE/otala.pdf.
    Yes, but in a power amplifier class AB the current in the output stages is kept low when the signal is small. The drivers in this case are constantly driving a current through the emitter resistor (about 6mA) with a voltage across them of about 18V (18V*6mA = 108mW). Again, see my reference for a very good design.
     
  22. Jun 30, 2015 #21
    I'm not sure I understood all that. It's been a while since control theory.

    But what I think it's saying is that the maximum frequency is limited by something akin to the gain/bandwidth product. When higher frequency signals come in they drive the system to the rails which limits low frequency gain to 1 which is non-linear (at least for a small bit) and mixes. Darlingtons have lots of parasitic capacitance, so that amplifies the problem. But the voltage gain is only 28 (as opposed to 100 or more for typical applications) so that limits the problem. These transistors seem to be made for about 100 kHz operation. (It was hard to tell from the data sheet, but they used that value in a test.) It gives a collector-base capacitance (~150 pF at 18V) but I think the important value would be internal. (The collector-base capacitance on the first transistor inside the pair is important, I think.) Anyway, I have no idea how to figure what the cutoff frequency is. 100 kHZ is clearly problematic, but I'm sure the transistor can manage well above that. Just how far?...

    I can see where limiting the high frequency end of the input could help eliminate this problem. Maybe toss in a filter set for 30 kHz or so.
     
  23. Jun 30, 2015 #22

    Averagesupernova

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    I am not sure power dissipation in the driver stage is that big of a deal compared to other schemes. It's going to take so much drive on the bases of the outputs to get so much power out of them. After a certain point this probably requires an emitter follower (which is exactly what this design uses) and this often means some power dissipated in an emitter resistor. This brings up another point. Does an emitter follower technically have feedback built in with its emitter resistors? When we think of a common collector transistor stage we think of the emitter resistors being part of a feedback path. I would say that the point of the overall feedback in this circuit is to flatten the frequency response and keep the output at zero volts DC. The voltage gain from the output of the op-amp in the first stage to the speaker output is slightly less than one. Pretty hard to apply more 'local' feedback here. I think for what it is in components it is a nice little circuit. I don't think anyone should expect it to outperform the output circuits available in IC form. That would be silly.
     
  24. Jun 30, 2015 #23

    meBigGuy

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    In a class AB amplifier, current should flow through the output transistors when the output voltage is exactly zero. That is, a reliable quiescent bias current through the final PNP and the npn NPN when the speaker voltage is exactly zero. How is that accomplished here?
     
  25. Jun 30, 2015 #24
    Hi folks thank you for all your replies, I have been reading through them carefully. Svein I will take a look at that link, thank you. And thank you thankz (lol) for the book suggestion - looks good.

    I have not changed the circuit yet, but I have taken some measurements @ ±18v of the circuit which I have here on the breadboard.
    The new measurements are in pink.

    These measurements are taken with NO input signal, and with a 6Ω speaker connected.

    snipe.PNG
     
  26. Jun 30, 2015 #25
    The darlington pair is somewhat under-biased, but there is some bias voltage/current. Thus the quiescent current is near, but not at zero. In high frequency applications this would probably slow the response. But I doubt it's a problem in audio applications.
     
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