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Inut and output of an Op.Amp.

  1. Jul 4, 2009 #1
    The question is about the concept of the operational amplifier in electronics.
    For an ideal op. amp., it is said that the input impedance is infiinite and the current allowed by the op.amp. through its input terminal zero or in the case of a practical op.amp., it is nearly zero or negligibly small. It means all the current arriving at the input terminal passes through the outer path containing the feedback resistor R_f. The normal questions a student asks is -"When that is the case, what is the need of the circuit elements at all since all the current flows through only a path outside the op.amp. device? What is the exact role played by several resistors and diodes that are making up the circuit diagram of the op.amp.? How does an enlarged output appear without taking any input?"
    Is it possible to give a simple, understandable and convincing explanation to a student of higher secondary classes without iindulging in any jargons of advanced and brain teasing explanations using higher electronic terminolgy, for the above questions? I myself find difficult to comprehend the above phenomenon. No reference books in electronics deal with such basic notions and doubts arising in the minds of an average reader as they straight away start analysing everything mathematically.
  2. jcsd
  3. Jul 4, 2009 #2


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    http://en.wikipedia.org/wiki/Operational_amplifier" [Broken]
    No math here. But if
    then use math.
    Last edited by a moderator: May 4, 2017
  4. Jul 4, 2009 #3

    The Electrician

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    You seem to be assuming that some input current must be taken in order to produce an output. There are two things I can say in answer to this.

    First is to say that (in theory) there are devices that are "voltage-controlled", such as vacuum tubes and FETs. In practice, they draw some current; leakage currents, for example, or if the input signal is AC, they draw capacitive currents. The ideal versions of these devices don't need to "take" a current at their input in order to operate.

    Second, in practice, the currents that such devices draw are very small, often in the picoamp range (10-12 amps). For engineering purposes, such small currents can be assumed to be zero, and these devices are "voltage-controlled".

    Some opamps will take an input current in the microamp range (which is still a rather small current), and this is not a problem if these opamps are used in an application where this current is taken into account.

    So, the answer to your query is that opamp input currents are so small that they can be ignored in most first-order calculations, which I think you already know.

    Perhaps the reason you are confused is that the concept of a "voltage-controlled" device hasn't been explained to you.

    The opamp can be treated as a "voltage-controlled" device. That is what you are doing when you ignore the input current of an opamp.

    It's impossible to describe the quantitative role of the passive components without using simple algebra. All that can be said without algebra is that they determine the transfer function. This is the key concept of feedback; throw away gain to make the transfer function depend on the passive components only. To describe the details requires math.

    There's a reason that mathematics is a prerequisite for most courses of study in electronics. Without it, true understanding is elusive.

    I see math there; simple algebra, but math nonetheless.
    Last edited by a moderator: May 4, 2017
  5. Jul 4, 2009 #4


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    I was referring to the quote. I thought this was a good explanation without math.
  6. Jul 5, 2009 #5
    Thank you all very much for the kind efforts you have taken in explaining the concept.
  7. Jul 5, 2009 #6
    The opamp amplifies the input voltage and not current. The total power of the input signal is less than the power of the output signal (as P=V*I, I is constant, but V increases). This additional power comes from the d.c. supply of the opamp.

    The feedback circuit (resistors, diodes etc) help in modifying the signal to our needs. This may be amplification, rectification (analog to digital converters), or a number of other applications. It all depends on what you want the circuit to do. The feedback elements mold the signal so it fits whatever application we need it for.
  8. Jul 5, 2009 #7


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    Hi ananthu. I hope you don't mind me asking but I was wondering if you are teaching this at secondary school level and if so what subject is it. We teach it as part of an elective component to year 12 physics here in Australia.
  9. Jul 6, 2009 #8
    Dear uart, It is incorporated in the curriculam of Higher secondary physics (not secondary),
    ie.+2(12th standard class), in Tamil nadu,India. It is part of the electronics lessons and the op.amp.is there as inverting amp., non inverting amp., adder and subtractor with very brief details. Though I didn't find any extraordinary difficulties in explaining the points mentioned in the text to a minimum understanding level of an average student so that he can easily deliver them in the examination, I find always some kind of internal vagueness while dealing with some of these concepts in electronics. In one of the recent orientation programmes held for hihger secondary teachers, a college professor who has been all these years teaching only classical physics, jokularly remarked to his collegue who is an electronics professor:" I am unable to digest most of the concepts in your subject..I am afraid your elecronics is not at all a branch of our physics..It should be some thiing entirely a new entity!"
  10. Jul 6, 2009 #9


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    This is actually a very insightful question, students should be commended for asking it.

    Others explained that practical opamps do take in some very small amount power when they measure the input signal, but they did not explain where the energy for the amplified signal comes from. In the real (as in not ideal) opamp device it comes from a second set of inputs to the device (the inputs are labeled vs+ and vs- in the picture below) which is connected to a second power source. The ideal opamp schematic symbol does not draw these inputs to keep the pictures cleaner (tricky huh?).

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