resistor at the output of an op amp?

Discussion in 'Homework Help' started by mikelearns, Jan 12, 2007.

  1. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    Let's say we're using an opamp ua741 to function as a comparator.
    An ac wave 4 Vpeak, 50Hz is fed into positive input of the opamp. The negative
    input is tied to ground. The output of the op amp will be connected to a 10k resistor (R load) and then be shorted to ground
    (output->resistor->ground). I expect we'll get a square wave at the output.

    Then here's my doubt.
    How does the resistance value of output resistor (R load) affect the output voltage waveform??

    Thanks.
     
  2. Distort10n

    Active Member

    Dec 25, 2006
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    You will have to see how much current the uA741 can source to the load. If you have a heavier load; i.e., lower resistance, then the load will demand more current from the op-amp in order to maintain a given voltage swing.
    When you reach the current limit of the output stage, you will start to see a drop in voltage. Ohm's Law and nothing sinister beyond that.
    10k is usually good enough. +/-15V with a 10k load is +/-1.5mA of current. I am sure the uA741 "All Father" can source that amount of current.
     
  3. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    Thanks for reply knight :)

    I wonder how i can find that piece of information. In the data sheet, i suppose? What does it call? output short circuit current? How to interpret the data? The opamp i'm using is manufactured by ST. I have attached the data sheet for your convenience.

    V=IR, lower resistance, higher current, but why is there a drop in voltage at the current limit?

    Thanks.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Op amps do have output resistance. When inside a feedback loop, the output resistance is reduced by the loop gain, usually making the output resistance insignificant, at least at low frequencies (assuming the load current is within the maximum current range).
    When used as a comparator, the op amp is running open loop, so the output resistance is what it is. The TI uA741 datasheet says the typical open loop output resistance is 75 ohms, so the output voltage will be somewhat affected by the load resistance.
    Be aware that, due to the common mode range limits, your comparator won't work with the inputs at zero volts unless you have at least -3V on the negative supply pin.
     
  5. Distort10n

    Active Member

    Dec 25, 2006
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    You would not look for short circuit output current. That is when the output is directly shorted to ground. You would look for output current (Io) which appears to be absent from the ST datasheet and as well as the TI datasheet.
    So I appears to me, the uA741 can source ~2.5 mA with an 8k ohm load.


    Think of it this way. Let's say you want a +/-10V output swing with a 1 ohm load. What would be the current through that load? +/-10A. Meaning the op-amp would have to source +/-10A to maintain a +/-10V swing.
    Most precision amplifiers will not even begin to come close. Most will source +/-50 mA. So, you still have a 1 ohm load but the op-amp can only give you +/-50mA. What would the voltage swing be? +/-50 mV.

    With everything else being ideal of course...
     
  6. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    that's truly informative. thank you. :)

    another question.

    i'm still using an opamp ua741 to function as a comparator but now instead of an AC, i'm feeding a ripple voltage which fluctuates from -8V to -9.6V. I'm to produce a square wave. So after simulation, i found the reference voltage should be -8.8 V. However, when I construct the circuit, the reference voltage is -8.88 V. What could be the reasons leading this discrepancy?

    Thanks.
     
  7. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    ----------
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Mike,

    Lady luck must shine on you.

    According to page three of your datasheet:


    IOS Output short Circuit Current
    minimum 10 mA
    typical 25 mA
    max 40 mA
     
  9. Distort10n

    Active Member

    Dec 25, 2006
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    Post your schematic and we will take a look.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    If you're using a resistor divider to create the reference, there are 3 possible sources of error:
    1. The power supply voltage
    2. Resistor tolerances (most likely)
    3. Op amp input current
     
  11. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    how does the opamp input current causes the error? care to explain?
     
  12. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    what do these figures represent?
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    In the schematic below, note that Vref is a function of vcc, R1, R2, and Ibias.
     
  14. Distort10n

    Active Member

    Dec 25, 2006
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    These are meaningless for what you need. Short circuit current is the current the op-amp will source to ground when directly shorted. You would consider this for fault conditions:
    1) Can the op-amp withstand a short to ground; i.e., brief periods of time or indefinite.
    2) When it is shorted to ground, how much current will it pull from your supply. Would suck if your supply were batteries.
     
  15. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    Thanks for your reply.

    Are you trying to say that actually there is a current going into the op amp in real world which is usually neglected by simulation software?

    I haven't really got it yet, how does this Ibias causes the difference?

    And also, referring back to your previous post, which power supply voltage were you referring to? How does it causes the difference?
     
  16. mikelearns

    Thread Starter Member

    Jan 12, 2007
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    attached is my cct.
     
  17. beenthere

    Retired Moderator

    Apr 20, 2004
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    Hi,

    I don't think your circuit is going to output a square wave. The negative terminal is at -8.8 volts, which tries to force the output to + infinite volts (no feeback). The positive terminal is going to be at about -10 volts, which will try to force the output to - infinite volts (also no feedback). Your output is going to be about 13 volts negative, or as close to the negative rail as the 741 can swing.

    Your resistor network is probably not giving the voltage you calculate because your resistors are 5% accurate, and so there will be some variance away from the stated value.
     
  18. Distort10n

    Active Member

    Dec 25, 2006
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    This is possible depending on the sophistication of the simulation models being used.

    If you had an ideal voltage source, resistors of equal value, and op-amp, Vref would be 1/2 VCC dead nuts. Since practical op-amps have a minute input bias current that will cause a small voltage drop across R1. This voltage drop will cause unsymetrical readings in your divider circuit. I would imagine they would be quite small since input bias current is in the nA to fA range (FET inputs). I did not look at the 741's input bias spec so it could be worse;i.e., uA range.

    If you use KCL on RonH's circuit, you will develop the equation:

    0 = (Vcc-Vref)/R1 - Vref/R2 - Ibias

    Solving for Vref, you will find:

    Vref = (Vcc - IbiasR1)(R2/(R1+R2))

    A poor voltage source. Noisy and unregulated possibly. Stay away from ripple. Use batteries.
     
  19. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You neglected to account for the ripple. The output should be close to a square wave.
     
  20. beenthere

    Retired Moderator

    Apr 20, 2004
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    Hi,

    I'm still not counting on a 1 mill bleed current off a 100 mike cap causing appreciable ripple, certainly not as much as 1.2 volts. With 10 mikes, maybe. But I think the 100 mike cap will charge right up to peak.
     
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