Audio Amplifier Clipping Below Rail

Discussion in 'The Projects Forum' started by aws505, Mar 11, 2013.

  1. aws505

    Thread Starter Member

    Mar 11, 2013
    59
    7
    Hello All,

    I'm building a small audio amplifier circuit, very, very similar to the amplifier shown on the final step of the Class B Audio Amplifier page. I have uploaded a schematic that reflects my changes. It seems to work well below my power rails, but when my output signal becomes close to my power rails, the output clips before I expect it to. My rails are ±12V, but my signal seems to clip when it comes close to 5V. Am I doing something here that would make my circuit clip prematurely?

    I have one last question out of academic curiosity. In my circuit, the feedback portion of the second op-amp does not connect directly to the load. This connection is added without explanation in the link above. What purpose does the connection from the load to the feedback of the second op-amp provide?

    Thanks for your thoughts!
    - Andrew
     
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  2. wmodavis

    Well-Known Member

    Oct 23, 2010
    737
    150
    Perhaps an earlier stage is clipping. Check each stage's output with an o'scope.
     
  3. patricktoday

    Member

    Feb 12, 2013
    157
    42
    A couple places where clipping might occur would be the second op amp and the output stage. You could lose as much as 2 volts off each rail from the op amps and higher current output will bring it lower. Then, your output transistors will drop perhaps another volt since they must follow the op amp.

    Also, consider the case where the output voltage is, say, +9V. You circuit must provide 9V/8 Ohms at that point into the load and your top transistor must provide that much current. The op amp must provide that much current divided by the beta of the transistor which can be pretty low at high currents, perhaps 50 or less. If the beta is low it may be maxing out your op amp's current capabilities. If you're simulating this, try upping the load to a higher value and see how much swing you get then.

    And connecting the final output of the amplifier back to the op amp is probably meant to keep the voltage at zero volts when there is zero volts input and to compensate for the slight differences in the transistors.
     
  4. aws505

    Thread Starter Member

    Mar 11, 2013
    59
    7
    Hey All,

    I've checked further back in my circuit and it doesn't look like it's really being clipped by the first amplification stage. The second amplification stage, though, does begin to clip the signal when it gets around 7V amplitude. That's a full 5V off my rail! When I was running this test, I had the transistor networked grouned and pushed my op-amp directly into my oscope. Is it possible that I simply have a bad op-amp? Is a full 5V off a rail acceptable? After the op-amps begin to saturate around 7V, I can turn up the gain more and the saturated value will move towards 12V, but the phase at which the saturation begins remains unchanged...

    Hmm...
    - Andrew
     
  5. patricktoday

    Member

    Feb 12, 2013
    157
    42
    I don't think 5V off the rail is acceptable :) That's not very easy to work with. When you did your last test did you actually remove the connection from the second op amp output to the transistors and just leave the feedback wire in place? Are your supply rails sagging under the load?
     
  6. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    What is your gain set to when you get the 5v clipping? Or, what is the value of the feedback resistor when that occurs?

    Large current draws from the amplifier will also cause the output voltage to become limited. In the original schematic from the ebook, the opamp did not have to drive the feedback resistor, but only the transistors and transistor bias circuit. There is a figure in the datasheet that shows the effect, figure 9 I believe.
     
  7. aws505

    Thread Starter Member

    Mar 11, 2013
    59
    7
    Hey Patrick,

    During my last tests, I did remove the connection from the second op-amp to the transistor network. I left the feedback resistor in place. Instead of the transistor network, my second op-amp was pushing directly into an oscope.

    I seem to have found my problem, though. Your comments about the current limits of op-amps got me thinking. I began to worry about my first amplifier (the voltage follower) trying too hard to push into the second amplifier (the gain amplifier). I replaced the 68 Ohms with 150 Ohms and have got my clipping to move out to approximately 10.5V. That seems much more acceptable!! I'm going to increase that resistor even more -- maybe to 300 Ohms or so. I have a 10 kOhm feedback, which should provide more than sufficient gain.

    As a matter of curiosity, I tried to look up the output current limit on the TL082's, but can't find a clear answer. Am I totally missing this parameter listed on the datasheet? Or are these kinds of things something a designer needs to guess at?
    EDIT: Ahh! As StayatHomeElectronics has pointed out, there does seem to be an indication of this effect in the datasheet -- specifically Figure 9. A serious question arises in this case, though: If the op-amp was limited by the amount of current it was pushing through the 68 Ohms, then why would the circuit work at all if the 68 Ohms is, instead, a short (as in the link in OP)? I would imagine that the same mechanism that protects a shorted voltage-follower from supplying infinite current would also protect my voltage follower from pushing too much current into my second op-amp. Any ideas?

    Thanks again!
    - Andrew
     
    Last edited: Mar 11, 2013
  8. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    In the original diagram the follower was driving into a variable resistor that was used as both the input and the feedback resistor. The total resistance of which is 10kohms. So, depending on the gain setting the output load of the first stage can be multiple kiloohms.
     
  9. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    The 68 Ohm resistor makes no sense at all. The non inverting input of the second op amp is a virtual ground so the first op amp is driving a 68 Ohm load which is too low. Op amps can only source maybe 5 mA typical.

    FYI: I would never use the circuit shown in the schem. The output stage will have severe notch distortion.

    Here's a much better design with selectable gain. It will run very well on +/- 12V.
     
    Last edited: Mar 11, 2013
  10. patricktoday

    Member

    Feb 12, 2013
    157
    42
    Well, it's not exactly a short circuit it's just pulling a lot more current than what the device is intended to provide. So you get degraded results. Since your original source voltage is probably quite small it can still put out a reasonable output even at a 68 ohm load. But, yes, best practices: make sure the input to op amp 2 is over 1k; the math is the same as far as how much the feedback resistor will cause the signal to be amplified (Rf just needs to be proportionally larger) and the only downside is in increased noise (as you make the resistors around op amp 2 larger, noise will increase).

    Whatever the current value of op amp 2's feedback resistance is at a given moment, if you divide that by the resistor that goes to op amp 2's base, that's your gain. You could increase the value of the 68 ohm resistor to, say, 4.7k and adjust the feedback resistor accordingly and get the same gain results. All this is just to say: yes, jack up the 68 ohm resistor a lot and recalculate :) Once you get your desired voltage gain at the output of op amp 2, you can deal with the output stage.
     
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