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Characteristics of Ideal Operational Amplifiers

  1. Apr 15, 2012 #1
    One of the characteristics of an ideal op amp is that it have infinite input impedance/resistance.

    Now, I want you to tell me if I am correct :

    The voltage source is likely to have some internal resistance.

    By having an infinite input resistance, no current is drawn from the voltage source. Hence, if there is no current there is potential drop across the internal resistance of this voltage source.

    Could you also explain the importance and requirement of an infinite open loop gain ?
     
  2. jcsd
  3. Apr 15, 2012 #2
    Also, I'm having a little trouble with this :

    IMG_0235.JPG

    I fully understand the above derivation, but my textbook provides the following formula for the open loop gain....

    A0/(1+A0)

    Which is correct ?
     
  4. Apr 15, 2012 #3
    Only with infinite open loop gain it is possible to bring non-inverting input equal to inverting input.

    As for the op-amp

    The real, physical op-amp (without any feedback loop) is nothing more them just a simply differential amplifier.
    The output voltage is the difference between the + and - inputs multiplied by the open-loop gain:

    Vout = (V"+" − V"−") * G_open_loop
    .

    attachment.php?attachmentid=46244&stc=1&d=1334484368.png

    So when the voltage at the input "+" (non-inverting) rise,output voltage is also increases.
    Increase voltage at "-" input (inverting input) causes, decrease the output voltage.
    Decrease voltage at input "-" increases the output voltage.

    attachment.php?attachmentid=46245&stc=1&d=1334484384.png

    This circuit has negative feedback apply by R2

    attachment.php?attachmentid=46246&stc=1&d=1334484487.png

    We apply 2V at inverting input so the op amp see the potential difference between his inputs. So the output voltage start to decrease. And it will be decrease until V"+" = V"-". As you can see nothing magical is happening here.
    When we apply positive voltage at inverting input, the op amp output voltage start to decrease.
    But thanks to R2 some part of the output voltage is feedback to inverting input. And when V"+" = V"-" = VD = 0.
    The op amp see 0V difference between his inputs terminals. And output voltage stop changing.
    From the observer perspective op amp "magically" brought his input terminal differential voltages to 0V (virtual short).

    The more accurate explanation:

    Now let us understand how the negative feedback returned through R2 affects the amplifier operation.
    To begin our discussion, let us momentarily freeze the input signal as it passes through 0 volts.
    At this instant, the op amp has no input voltage (i.e., VD = 0 = voltage between the (+) and (-) input terminals (VD)).
    It is this differential input voltage that is amplified by the gain of the op amp to become the output voltage.
    In this case, the output voltage will be 0. Now suppose the output voltage tried to drift in a positive direction. Can you see that this positive change would be felt through R2 and would cause the inverting pin (-) of the op amp to become slightly positive Since essentially no current flows in or out of the op amp input and the (+) input of the op amp is at ground potential. This causes VD to be greater than 0 with the (-) terminal being the most positive. When VD is amplified by the op amp it appears in the output as a negative voltage (inverting amplifier action). This forces the output, which had initially tried to drift in a positive direction, to return to its 0 state. A similar, but opposite, action would occur if the output tried to drift in the negative direction. Thus, as long as the input is held at 0 volts, the output is forced to stay at 0 volts.

    Now suppose we allow the input signal to rise to at +2 volt instantaneous level and freeze it for purposes of the following discussion.
    With +2 volts applied to R1 and 0V at the output of the op amp, the voltage divider made up of R2 and R1 will have two volts across it. Since the (-) terminal of the op amp does not draw any significant current, the voltage divider is essentially unloaded. We can see, even without calculating values, that the (-) input will now be positive. Its value will be somewhat less than 2 volts because of the voltage divider action, but it will definitely be positive.
    The op amp will now amplify this voltage (VD) to produce a negative-going output.
    As the output starts increasing in the negative direction, the voltage divider now has a positive voltage (+2 volts) on one end and a negative voltage (increasing output) on the other end.
    Therefore the (-) input may still be positive, but it will be decreasing as the output gets more negative.
    If the output goes sufficiently negative, then the (-) pin (VD) will become negative. If, however, this pin ever becomes negative then the voltage would be amplified and appear at the output as a positive going signal.
    So, you see, for a given instantaneous voltage at the input, the output will quickly ramp up or down until the output voltage is large enough to cause VD to return to its near-0 state.
    All of this action happens nearly instantaneously so that the output appears to be immediately affected by changes at the input.

    Now let as change the type of the feedback from negative feedback to positive feedback and see what will happen.

    16_1334484554.png

    We apply positive input voltage and see what happen.
    As the input voltage increases, the voltage on the (+) input also increases.
    Once the (+) input goes above the voltage on the (-) input, even momentarily, the output of the op amp will go toward positive direction. This rising potential, through R2, further increases the potential on the (+) input pin, in our case, the output will drive all the way to its positive saturation limit (positive supply voltage).
    I hope that now you see difference between positive and negative feedback.
     

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  5. Apr 15, 2012 #4
    Your textbook show a positive feedback amplifier.

    For the negative feedback the equation look like this

    attachment.php?attachmentid=46247&stc=1&d=1334485223.png
    Based on diagram we get three equations

    (1) V1= Vin - A
    (2) Vout = V1*Ao
    (3) A = Vout*β

    Ao - is a open-loop gain of a opamp ( forward gain ) = 500[V/V]
    β- feedback factor ( feedback gain ), sometime we use letter "K" for feedback gain.

    And now we can calculate the close loop gain, substitute 1 to 2

    Vout =V1*Ao = (Vin - A)*Ao

    And 3 for A

    Vout = (Vin - Vout*β)*Ao

    Vout = Vin*Ao - Vout*Ao*β

    Vout + Vout*Ao*β = Vin*Ao

    Vout = Vin*Ao/(1 + Ao*β)

    Vout/vin = Ao/( 1 + Ao*β)

    Or dividing all by Ao we get:

    Vout/vin = 1 / [ (1/Ao) + β ]
     

    Attached Files:

  6. Apr 15, 2012 #5
    So the equation in the ATTACHMENT is wrong for the closed loop gain of an inverting op amp with negative feedback ?
     
  7. Apr 15, 2012 #6
    Yes, the equation in ATTACHMENT is wrong.
    They analyses the positive feedback amplifier.

    The correct equation for closed loop gain of an non- inverting op amp (only) with negative feedback look like this
    Vout/vin = Ao/( 1 + Ao*β)
     
  8. Apr 15, 2012 #7
    What if Beta was negative in the equation in the ATTACHMENT ?
     
  9. Apr 15, 2012 #8
    Then we have the same result.
    But how can a simply voltage divider create negative Beta?

    And in feedback theory

    Vout/vin = Ao/( 1 + Ao*β) ---> negative feedback

    Vout/vin = Ao/( 1 - Ao*β) ---> positive feedback
     
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