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Why Opamp with potentiometer?

Discussion in 'General Electronics Chat' started by sagh, May 5, 2014.

  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes but if we got gain up to the output of the current mirrors with higher slew rate then that means we saw an advantage, at least up to that point. What we do after that will depend on what we need after that for the application. We might find a use for that.

    For a simple example, perhaps we can use a cheaper op amp as the first op amp and a more expensive one as the output stage, assuming we need that kind of 'buffering' for the application.
    Or maybe build our own op amp for the output stage, or maybe just some transistor buffering.
     
  2. sagh

    sagh Member

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    I read a paper about application of circuits kind of similar to this one a few months ago. I try to find and post it here. Meanwhile I look for the article can you tell me the correct definition of gain bandwidth product. I checked definition of Gain Bandwidth Product in op amp design. but they were slightly different from each other.
     
  3. alec_t

    alec_t Well-Known Member Most Helpful Member

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  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    The gain bandwidth product GBP is a very simple concept. An amplifier has limitations, and the GBP is one that involves the gain you can get from the amp vs the frequency you intend to use it at.

    As the frequency rises, the amplifier gain decreases naturally due to internal structure. The gain goes down inversely proportional to frequency. So if you get a gain of 10 at 100kHz, then you get a gain of 100 at 10kHz, and a gain of 1000 at 1kHz. Going up, at 1MHz you only get a gain of 1.
    This means you can look up the GBP of the op amp and divide by the frequency in order to get the maximum gain you can get at that frequency.

    For example, for a GBP of 1000000 (one million) if we want to operate at 1Khz then we divide this by 1000, and we get 1000 as the result:
    1000000/1000=1000
    This means the max gain we can get is 1000

    It should be carefully noted that this is the "max" gain that we can get, but there is another constraint called the slew rate that also limits the frequency we can use the op amp at without distortion. The slew rate limits the rate of rise of the output of the op amp, regardless what the gain is. The only way we can really get the max gain is if we keep the output level below a certain peak level.
    We can talk more about this once you have had time to think about the GBP.

    More examples for an op amp with a GBP of 1000000:
    F=10, max gain=100000
    F=20, max gain=50000
    F=500kHz, max gain=2
     
  6. sagh

    sagh Member

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    Actually you say limiting the frequency response limits the rate of change that can occur at the output, right?

    The slew rate of each op amp is specified in datasheet. Is it possible to calculate the overall slew rate of a circuit and guess after what frequency of input signal the op amps will be even less able to keep up and therefore the amplitude of the output waveform will decrease?
     
  7. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Just as the strength of a chain is limited by the weakest link, won't the overall slew rate be that of the slowest opamp?
     
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    sagh:
    Yes. The frequency response is important because we cant pass a frequency higher without seeing unwanted attenuation or distortion.

    And yes it is possible to calculate the overall slew rate. It's based on the op amps and other circuitry.


    Alec:
    In most cases it is true that the overall slew rate will be limited by the slowest op amp, assuming no special filtering is being done, but with this circuit we've seen that transistors can be used to increase the slew rate by quite a bit. Transistors are fast compared to most lower end op amps, so we see an advantage there. Whether this feature can be used or not depends on the cleverness of the designer as to how it gets incorporated into the total design.
     
  9. sagh

    sagh Member

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    Here is the article that I read a few months ago. I would prefer not to assemble such complicated circuits like the third one for testing in real world. So I decided to try a simpler version of this model. Even though I assembled part of the third circuit on solderless breadboard it started oscillating and never has stability. After MrAl suggested adding 2 voltage buffers and changing resistors at the output of current mirrors quite a bit Circuit started working in a normal way.
     

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  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    That looks interesting. I'll have to read it over a little later. It appears they are after good CMRR.
     
  11. sagh

    sagh Member

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    Yes, That is why I would want to get CMRR of the circuit that we have because in real world its performance is better than their circuits. besides their circuits are more complex. In real life Using 8 transistors for current mirrors and even more for improvement of CMRR creates many problems.
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yeah i looks like maybe they were looking at this with the goal of creating a new op amp die, so that all the new parts would be put on the same chip as the original op amp. There's no way i would build a circuit with 8 transistors like that, it's just too much and without matching it might be totally worthless.

    LATER:
    Couple things...

    I just did a CMRR test on the LM358 using a different simulator and got a result at low frequency (like 100Hz) of 85db. This is very close to the data sheet.

    With 1v peak input, the max frequency for testing will be 80kHz with a gain of 1. After that the slew rate will have too much of an effect.

    A simplified calculation of the maximum frequency for an op amp is:
    F=0.16*sr/A
    where
    F is the frequency in Hertz,
    sr is the slew rate in volts per second (not volts per microsecond),
    A is the peak sine output amplitude (not rms).
    So for example an op amp that has 0.5v/us slew and GBP of 1MHz, the max frequency is:
    F=0.16*500000/1=80kHz
    for a peak sine output of 1 volt. If the gain is 1 then the peak sine input is 1 volt also. This is valid because 80000 times the gain of 1 is less than 1Mhz:
    80000*1<1000000
     
    Last edited: Jun 27, 2014
  13. sagh

    sagh Member

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    would you please tell me why did you do this test? It's obvious that you did this test on a simple amplifier because you compared the result with the result in datasheet. but what was your objective in doing this test?

    In your post you said " the max frequency for testing will be 80kHz with a gain of 1. After that the slew rate will have too much of an effect". what do you mean?
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    I did the test on the LM358 because you had said before that you got a result of 45db didnt you? Or did i remember that wrong? I wanted to verify that because i never got around to doing this until just today. Since the result came out to be a reasonable value, maybe the model needs updating.

    With the LM358, the max frequency will be 80kHz with a gain of 1 and input voltage of 1v. The slew rate messes up the distortion too much after that.
    This kind of thing is true of any op amp, but the frequency and output voltage will be different if the slew rate is something other than 0.5v/us which is what the LM358 has. For an op amp with 5v/us we could use as high as 800kHz, but after that it seems pointless to test the CMRR because the amp wont do a good job with the signal regardless of what the CMRR is.
    Make sense?
     
  15. sagh

    sagh Member

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    I did that test on solderless breadboard and in LTspice for 741 op amp. in simulator I had a correct result but in real world I didn't have very precise equipment like digital voltmeter and couldn't have a right measurement for common mode gain so I had a wrong result.

    Yeah it makes sense. :)

    I have a question How does change in resistor values of output first buffers have impact on performance of transistors in current mirrors?
     
  16. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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  17. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well, when the resistor is lowered in value the current goes up, so the current in the current mirrors goes up. That may be good if the transistors work better at higher current. You cant go too high though, because the op amp can not drive with too much current on the output.
     
  18. sagh

    sagh Member

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    Below picture is the CMRR of circuit in LTspice. The dotted line shows something. what's that?
     

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    Last edited: Jun 28, 2014
  19. MrAl

    MrAl Well-Known Member Most Helpful Member

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    (ignore)
     
  20. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Phase :)
     
  21. sagh

    sagh Member

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    In below picture the voltage nodes of 009 and 007 are 0.5v instead of 1v. I simulated this circuit with LM358 and LT1001 op amps. why are they 0.5v?
     

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