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Op-amp: Current and bandwidth question

  1. Mar 17, 2015 #1
    300px-Op-Amp_Inverting_Amplifier.svg.png

    Does current flow through the feedback resistor without going through the op-amp at all? After all, the op-amp is supposed to have very high input impedance right, so if the current just flows through the feedback resistor, wouldn't there be nothing going through the op-amp?

    Also, how does the negative feedback increase the effective bandwidth? I get that the gain is smaller, so the signal won't be saturated and thus effective bandwidth is increased. But why bother with the op-amp at all then? Can't we just use the raw voltage is the gain is low anyway? Is it physically impossible to produce an op-amp with a more reasonable open-loop gain, making feedbacks necessary?

    Op-amps confuse me :frown:

    P.S this is my first time encountering op-amps so I think they are pretty much introductory questions??
     
  2. jcsd
  3. Mar 17, 2015 #2

    analogdesign

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    Op amps are tricky, that's for sure.

    First, yes you're right the current flows only through the resistors if the op amp is ideal. There is plenty of current going through the opamp but it going through a path not shown in your diagram, namely from the opamp power supply input to ground.

    Negative feedback increase closed-loop bandwidth, not bandwidth. The opamp will always respond fastest if you don't have feedback but it will not be very useful in that case.

    An opamp is design such that its closed-loop gain multiplied by its bandwidth are constant. This is called Gain-Bandwidth Product. If you have learned about frequency response yet this is because it has a single dominant pole. Anyway, since GBW is constant, RF = 0 and you have a unity follower, the bandwidth is highest and if RF/Rin is large, the bandwidth is lower.

    You bother with an opamp because it gives you stable, repeatable gain. If RF and Rin are accurate to 0.1%, for example, then your amplifier is accurate to 0.1%. This is really, really hard to do any other way. Basically negative feedback desensities the amplifier from issues with the opamp. If you need a really high-frequency amplifier you can't use an opamp anyway.

    If you don't want an opamp and your circuit doesn't need to be precise you can use a common-emitter amplifier (look it up). It won't be as useful as the opamp in general because it will have a much smaller gain. It turns out lots of circuits besides amplifiers benefit greatly from having a very large open-loop gain.
     
  4. Mar 17, 2015 #3
    An op amp requires very little signal current. The feedback current is small and basiclly equal to the current drawn from the signal source.

    Bandwidth gain product is about constant for a given type of op amp Expensive op amps have a greater BW gain product and are move complex devices. In any event if you want higher bandwidth you will have to settle for less gain for a given op amp. The open loop gain of an op amp is usually too high 105 and greater for most applications without feedback. When you use it for small signals, noise becomes a problem as noise in the range of 100's of uV. If you signal is too large then you can drive the op amp into saturation basically overloading it. So feed back gives you a way of controlling the gain for your particular use. High gain and feedback are the characteristics that makes them so valuable and interesting. Op amps are fascinating devices. You can often build a circuit with op amps that are much easier and cheaper to design and build compared to circuits built with transistor, resistors etc.
     
  5. Mar 17, 2015 #4

    LvW

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    Does current flow through the feedback resistor without going through the op-amp at all? After all, the op-amp is supposed to have very high input impedance right, so if the current just flows through the feedback resistor, wouldn't there be nothing going through the op-amp?

    The opamp is a very high-gain device reacting upon the VOLTAGE between both input terminals. Therefore, as long as the output voltage is within the limits set by the supply rails this differential input voltage is so small that it can be neglected during calculations. This is the so-called "virtual ground principle". As a consequence, there is the same current through the series connection Rin-Rf is driven by TWO voltages: The small input voltage and the amplified output voltage (with a negative sign!). There is only a very small current going into the opamp which normally is neglected (nano amperes). And the output voltage automatically adjust itself to a value that allows the "virtual ground" at the inverting opamp terminal. This is the result and effect of negative feedback

    Also, how does the negative feedback increase the effective bandwidth?
    Most opamps are unity-gain compensated (for stability reasons) which means: The open-loop gain is rather large (100...120 dB) but with a rather small open-loop bandwidth (20...100 Hz). The open-loop gain decreases with a slope of -20dB/dec. Based on the principle "Gain*Bandwidth=const". we note that the bandwidth increases for decreasing the closed-loop gain.

    Is it physically impossible to produce an op-amp with a more reasonable open-loop gain, making feedbacks necessary?

    Yes - of course, we can design amplifiers with lower open-loop gains without the necessity of negative feedback. But, in this case, we loose all the advantages connected with negative feedback:
    *Gain determined by external resistors only,
    *Improved linearity
    *Reduced output impedance
    *Increased bandwidth
    *Reduced sensitivity to supply voltage variations

    Op-amps confuse me :frown:

    Why? They are the most versatile and interesting active components. They are the "heart" of active filters.

    LvW
     
  6. Mar 17, 2015 #5

    jim hardy

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    I well remember my profound consternation at my first introduction to them. "Gain of a million? Infinite Zin? Yeah right Professor, you're getting daffy in your old age."
    For me the mental hangup was that the word "operational" refers to how they're used not to how they're built. They are just a plain old amplifier used to perform some mathematical operation.


    So, to get a handle on opamps it is necessary to train our mind to walk on a leash, as follows:

    To use your circuit for an example, ask yourself :
    1. Will Vout be something reasonable, like between +-15 volts? Of course it will, that's all the opamp can make.
    2. Having accepted (1) , what constraint does that place on voltage at opamp's - input?
    Well,
    IF the amplifier has gain of a million, and its output is 15 volts or fewer, and it's not being overdriven,
    THEN the voltage at its -input must be 15 microvolts or fewer. That rounds off to zero.....and that's important.
    Do you see that the circuit designer has to surround the amplifier with a circuit that allows it to hold its inputs equal?
    (well, actually within Vsupply/(open loop gain) of equal which is a very small difference, in the range of microvolts that we round off to equal)
    SO: We can say "It is the duty of the designer to surround the amplifier with a circuit that lets it hold its inputs equal".
    Then and only then can it perform a nice linear math operation for us.

    To continue the example
    Let us set Rf and Rin in your circuit equal to one another.
    Then by voltage divider action the voltage at -input will be halfway between Vin and Vout. Halfway between those voltages would be their average.
    May i call voltage at -input simply Vminus, just to present a silly example?
    If so, we might write
    Vminus = (Vin +Vout )/2 ,, their average
    and since Vminus rounds off to zero
    0 = (Vin +Vout )/2
    which resolves to
    Vout = -Vin
    and that is the mathematical operation performed by your circuit with equal Rin and Rf . That is, so long as the amplifier can hold Vminus at zero, equal to Vplus.




    300px-Op-Amp_Inverting_Amplifier.svg.png

    If you repeat the above exercise with unequal Rin and Rf you'll begin to grasp how handy opamp circuits are for developing mathematical transfer functions. Rf could become Zf with complex arithmetic....

    In summary , to grasp opamp basics it is necessary to kinda reverse our thinking about gain. Gain constrains the input, not determines the output.
    I work opamps by writing KVL for both inputs, Vplus and and Vminus, then equating them.

    Try it. It soon becomes familiar.

    I hope above helps you over this common stumbling block
     
  7. Mar 17, 2015 #6

    jim hardy

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    Last edited: Mar 17, 2015
  8. Mar 17, 2015 #7

    davenn

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  9. Mar 22, 2015 #8
    Thanks everyone! The Dave Jones video was really helpful :)
     
  10. Mar 22, 2015 #9

    davenn

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    He has a bit of an irritating voice .... but dang, I have learnt so much from his many videos :smile:
     
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