Op-amp

Discussion in 'General Electronics Chat' started by Hurdy, Feb 28, 2006.

  1. Hurdy

    Thread Starter Senior Member

    Feb 27, 2006
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    Ok I have read up on op-amps at let me know if i'm getting this right.

    Firstly if there is a more positive voltage applied to the + (noninverting) input the output tends to drive high.

    If there more positive voltage applied to the - (inverting) input the output tend to drive low.

    And the reverse. If a more negative voltage is applied to the + (noninverting) input the output drives low.

    If there is a more negative voltage applied to the - (inverting) input the output drives high.


    Now to work out the output I take the difference of the two input voltages and multiply it by the gain? How can I tell what the gain of a LM348 is?

    Do you think I am missing any more important information concerning op-amps?
     
  2. windoze killa

    AAC Fanatic!

    Feb 23, 2006
    605
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    Firstly, you should have added this to your previous post.

    Secondly, the data sheet for the LM348 should give you the formulas to calculate this. Basically it is the ratio between the bias and feedback resistors. Do a search for the OP AMP cookbook, this will tell you all you will ever know.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I think you are missing the concepts of negative feedback and virtual ground. I'm so old that op amps were not taught in any of my college classes. I studied on my own, and had an epiphany one day when all was revealed to me.
    Take a look at the schematics below. Keep in mind that the differential gain (for the voltage between the inputs) is nearly infinite, while the common mode gain (for voltages common to both inputs) is zero. Real op amps, of course, are not ideal, but most are very good.
     
  4. CoulombMagician

    Active Member

    Jan 10, 2006
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    Don't forget that the ideal opamp also has infinite bandwidth and infinite power supply rejection.

    The next approximation to the opamp is that the open loop differential input voltage is given by

    (a*s+b )*Vout

    where b is 1/(finite open loop gain at DC) and b/a is the -3dB(1/2) gain angular frequency of the open loop gain.
    Input and output impedances are finite and nonzero.
    Common mode rejection ratio and power supply rejection ratio are finite.
    Output voltage range is finite and equal to the supply voltage difference.

    As the approximation becomes better the gain becomes a higher order polynomial in s( say 3 poles for instance).

    The voltage controlled voltage source in SPICE is a good starting point for modeling an opamp. You can build in these characteristics with verious passive components.
     
  5. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    On a more basic level, my rules of opamps are:

    1. The output will always be between the positive supply(+9V in your function generator) and the negative supply(GND in your function generator). For now don't worry about circuits where this is not true.

    2. The amplifier uses it's output to make the voltage DIFFERENCE at its inputs equal to zero. In other word it tries to make the voltages at the two input EQUAL. BTW it can only do this in the presence of FEEDBACK.

    3. The output will be subjected to a non-linear limiting function if it gets close to either supply rail. This means it will no longer increase or decrease with a change in input voltage.

    Now using these three rules, go back to your function generator, and convince yourself that LM348:D is configured as a "follower". That is the "output" will follow the "input"
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Errmm, if we're talking about the function generator schematic, I think you meant that op amp A is configured as a follower. Op amp D is used as a comparator in a Schmitt trigger.
    EDIT: Has the schematic changed? I just read aac's response, and it also has some of the op amp designations switched around. Either way, it's gonna confuse the hell out of Hurdy.
     
  7. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    Ron H
    You are quite correct. I was trying to answer from memory at my office whilst the saved orignial is at home. I was able to locate the repost of the drawing. Thanks.
     
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