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Why Opamp with potentiometer?

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

  1. sagh

    sagh Member

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    For adjusting the scale of signal I set the volts per division knob of oscilloscope on 5 v per div for output voltage and 2 v per div for input voltage.
    I don't understand why did you mention "minor"?:(
     
    Last edited: Jul 10, 2014
  2. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Normally the big squares are called "divisions", while the little small tic marks within the squares are called "minor divisions".
    If you are saying that you set the input to 2v per division, then you must mean 2v per minor division. That means the little tiny tic marks are each 2v.

    But that doesnt work either. The signal is higher than 2v. So that means the scope is out of calibration or else you turned the adjustment knob as well. So what you will have to do is measure the input or output with a meter.

    You can calibrate the scope using the following method:
    Using a generator or even the line frequency divided down with resistors, generate a sine wave and measure it with the meter. The frequency range of the signal must match the capabilities of the meter, so the meter should be accurate at that frequency. With the input signal connected, set the scope to 2v per minor division (that's probably 10v per division).
    Note the mark where the signal goes up to. If it does not go up to one minor division, then raise or lower the signal level until it does. When it does, read the meter. The meter now reads the true volts per minor division.
    For the 20v settings, on the other channel set the scope to 10v per division, not 10v per minor division. Check that the same way only use a higher signal level. If you have it set at 10v peak, then you should read 10v peak on the meter and on the scope too.

    My guess at this point is that the scope is off so much that you are getting very fictitious readings. The readings are higher than what is really present in the circuit. If you multiply all the peak readings by the error factor (less than 1) you would get the true measurements. Not all scopes are accurate on the vertical scales, some are off quite a bit.
     
  3. sagh

    sagh Member

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    I think I don't know something important about scope. Please correct me if I am wrong.
    The input voltage applied by function generator was about 0.5v. voltage per divisions knob of scope for input voltage was set on 2v per division I thought that this knob showed the unit of big squares. I mean when I set this knob on 2v then each big square is 2v and every little tick mark is 0.4v. In picture attached to a few posts ago peak of input voltage on the scope was 0.4v. Am I right?
     
  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Maybe you where not showing the pic you wanted to show?

    If you have the knob on "2v per division" then yes it means usually that the large squares are 2v each in amplitude. So two divisions means 4v, three means 6v, etc. If there are 4 tic marks between each square, then yes each one is 2/5=0.4v each. One tic would mean 0.4v. You are right 100 percent.

    But you showed that picture and described inputing 2v didnt you? And then the output doesnt look right either.
    Let me take another look at that pic....brb
     
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Back. That is what you said about the input and output voltages on the scope, so yes the input is 0.4v. But the output is then only 5v peak, approximately, so it appears that your output clips at LESS that 5v (5v is one square high).
    It also appears that your input may go slightly lower than it goes high, so maybe the input sine zero is also off. That's a secondary point however maybe not that important right now.
     
  7. sagh

    sagh Member

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    Positive peak of voltage clipped before 5v while negative peak of voltage went down to -5v. I thought if I increase value of supply voltage to 20v then there wouldn't be no problem but I have to first find out the reason of clipping the positive peak of output.
     
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    Well if that was more apparent from the beginning i could have told you right away, or at least have a good guess anyway :)

    Many op amps clip the positive output at less than the supply, while they sometimes go all the way to or near to the negative rail. So that is probably what you are seeing. It is important to know however that this changes with the op amp, and it can change quite a bit. For example, the LM358 would do just what you are seeing, clipping at near to -5v but as low as +3.5v, which is 1.5v below the +5v supply rail. That's quite a bit really. Other op amps could be very different because it depends highly on the output stage used. You really have to consult the data sheet on this one for the particular op amp.

    So going higher on the supply voltages will help the clipping a lot.
     
  9. sagh

    sagh Member

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    (ignore)
     
    Last edited: Jul 13, 2014
  10. sagh

    sagh Member

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    Above passage is part of what you said in many posts ago (post #139). I have one question about that. you said with 10k on the output and 1k at the output of input buffer of first stage the slew rate is increased. you meant 10k on the output of current mirror, right?

    one more thing for increasing the usable bandwidth I think we better set the gain of output stage to 1 and get it increased from output of current mirrors but it seems a little impossible with this circuit configuration.
     
    Last edited: Jul 13, 2014
  11. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes i meant the output of the current mirrors.

    If it wont work right with the gain of 1 then that's life i guess :)
     
  12. sagh

    sagh Member

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    I have a basic question about slew rate. All kind of voltage amplifiers better have high slew rate or it depends on the appliction of amplifier? As far as I know high slew rate is necessary for high frequency which means if we need the application to work in higher frequency we need to have higher slew rate. So the amplifier is supposed to work in low frequencies doesn't need high slew rate. Please correct me if I'm wrong.
     
  13. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes that is correct. You can calculate the slew rate needed for a given application.
     
  14. sagh

    sagh Member

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    So high slew rate is not necessary for all amplifiers. it depends on the application of amplifier. how can I calculate slew rate of the circuit that I have?
     
  15. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Find the slew rate of the buffer amp (data sheet).
    Find the slew rate of the output of the current mirrors (data sheet and resistor values).
    Find the slew rate of the output stage (data sheet).
    The slew rate that is the slowest will be the overall slew rate.
     
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  16. sagh

    sagh Member

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    For measuring CMRR in LTspice, I want to input 2 AC voltage sources (1v and -1v) to differential circuit. I have a question. For -1v, I have to set both SIN voltage amplitude and AC amplitude on 1 also set the AC phase on 180, right?
     
  17. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    If you need a sine that is 180 degrees out of phase then yes you do. But you may have to use pi instead.
    Why do you have to invert the sine? Is that because you want to try it that way?
     
  18. sagh

    sagh Member

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    Yes, I just want to try.

    I want to know how mismatch between parameters of transistors used in circuit affects the CMRR measurement in LTspice. I don't know how can I apply mismatch in transistor parameters, I did change resistor values and saw their effect on simulation but I can't see mismatch of transistors result in CMRR.
     
  19. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well for example you can put a battery in series with the base to see what happens with a Vbe mismatch. If the normal voltage is around 0.65v, then a battery of voltage 0.0065v will mean a forced change of 1 percent. You could then see the effect by doing another test for CMRR. If that is not enough, increase to 2 percent.
     

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