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Homework Help: Op-amp question

  1. Oct 28, 2016 #1
    1. The problem statement, all variables and given/known data
    So we just covered op-amps in class and when going through the lecture notes and reading the topic about ideal op amps, I came across this. It says that if the gain A is infinite(which it is in ideal op amps), the only way the output voltage v0 can remain bounded is if the input voltage vd between the input terminals is 0.

    That makes very little sense to me. I understand that between the inputs of the op-amp there is an infinite input resistance so no current can flow. However I thought that the limiting factor of the output voltage was the V+ and V- power supply inputs to the op amp.

    Can someone explain to me how vd=0 bounds the output voltage v0. I mean if input voltage vd is 0, how can you amplify 0? Confused.


    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 28, 2016 #2


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    Staff: Mentor

    A *really* ideal op-amp requires no power supply and can produce any voltage and current at its output. You don't want to use an ideal op-amp without some external constraint (a feedback network) to limit the device, unless of course you happen to enjoy lightning bolts and boiling metal on your workbench :wideeyed: :smile:

    When you're dealing with (theoretical) infinities, you also end up dealing with mathematical limits and differential quantities. So it's more accurate to say that the potential difference between the ideal op-amp's inputs tends to zero in the limit.

    Real-life op-amps have constraints like power supply limits for their voltage swings as you note. But they really do have some impressive open-loop gains on the order of 105 or even 106. That's enough to pin the output to one of the power rails for even very small potential differences between the inputs. For normal operation where the op-amp is producing an output between the bounding limits, the differential input must be very small. Not zero precisely, but very small indeed. So small that for almost all practical purposes it can be taken to be zero when you're designing or analyzing a circuit.

    The small potential difference between the inputs is enforced by negative feedback provided by the surrounding circuit. You should notice that you almost always find a circuit path that leads from the op-amp output back to its input. That's the feedback loop. Sometimes the loop can be hard to spot if it includes another circuit stage or two. But it will be there.
  4. Oct 29, 2016 #3


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    We must realize that in electronics - in particular, in transistor-based active electronics - no expression/formula is correct by 100%. It is common practice to neglect effects which do not contribute considerably to the result (gain, input/output resistances,...). This is good engineering practice because it makes no sense to consider effects which would improve the accuracy by 1% if we have to face main parameter tolereances of 5% or more.
    Therefore, during calculations we always neglect also the finite differential input voltage (µV range) of an opamp with feedback if compared with all other voltages within the circuit (in particular, input and output voltage). This is equivalent to assuming an opamp open-loop gain of infinity (remember: mathematically, the product zero times infinity may give a finite value!).
  5. Oct 29, 2016 #4
    Ah. So basically the input isn't really 0. its very close to 0 and so due to the high gain of the op-amp, you get a large output voltage. But in ideal terms we consider it to be 0. So in calculations there will always be something else connected to the op-amp to limit it? And whatever that something else is(v source/power supply) that will determine the output voltage.
  6. Oct 29, 2016 #5


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    Staff: Mentor

    I'd say in practical terms we consider it to be zero. It's certainly possible to mathematically analyze the circuit (ideal or otherwise) to find expressions for the "actual" potential difference. The ideal case will involve dealing with infinities and limits. The non-ideal case involves employing a realistic model of the real op-amp, including actual open-loop gain, input impedances, output impedance, offset currents, and so on.

    The behavior of an op-amp in a circuit is defined by the surrounding circuit. This is what makes op-amps so very useful as a general purpose part. They are practically ideal :smile:
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