Perpetually confused about Op-Amps

Discussion in 'General Electronics Chat' started by gerases, Mar 19, 2013.

  1. gerases

    Thread Starter Member

    Oct 29, 2012
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    I don't get this rule about Op-Amps:

    But how can an Op-Amp amplify anything if the inputs are at the same potential? Isn't it all about amplifying the difference? I've read this 20 times on different web sites and no one seems to notice what confusion this can create within the mind of a nub.

    Can anybody explain it to me?
     
    Last edited: Mar 19, 2013
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    This kind of explanation expects the opamp to have infinite gain, which is never the truth. Beacuse the real opamp has finite gain, the difference is only very close to zero, but not truly zero, so the almost zero times open loop gain equals to the output voltage you see.

    I would appreciate if someone could show why this holds when the gain really is infinite :)
     
  3. gerases

    Thread Starter Member

    Oct 29, 2012
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    Let's say I have:

    1. 5V signal on the V+;
    2. 0V on the V_
    3. The power supplied to the Op Amp is 15V

    How do we get close to zero on the V+ if we start with 5V?
     
  4. Sue_AF6LJ

    Member

    Mar 16, 2013
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    When you look at a simple inverting amplifier using an OP-Amp the feedback resistance sets the gain, where the input resistor remains constant.

    If the feedback resistance is zero the gain is zero.
    If the feedback resistance is infinity the gain of an ideal OP-Amp would be infinity. Putting it simply the gain of an OP-Amp is the feedback resistor over the input resistor for an inverting amplifier.
    If I make the input resistor and the feedback resistor equal I have zero gain. If I make the input resistor larger than the feedback resistor I have negative gain. To have positive gain the feedback resistor must be larger than the input resistor.

    That is the simplified explanation.....
     
  5. #12

    Expert

    Nov 30, 2010
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    I'm a pragmatic person. I work with practical models where the difference in input voltage is so close to zero that you can assume it is zero and the math will work properly. I believe crutschow is the guy that can do the math for a gain of hundreds of thousands of volts per volt times a microscopic voltage difference at the input pins of an op amp. Perhaps he will grace us with the proper explanation of how a few microvolts is amplified to create a useful output voltage.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    If you just apply 5V to the V+ input and 0V to the V- input then the output of the opamp will saturate and will be the most positive voltage it can output.

    When you have a feedback network, the voltage at one (or possibly both) of the input pins is a function of the output voltage. If you have negative feedback, then if the difference between the input pins is some value the output will change and the feedback network will have the effect of reducing the magnitude of the difference.

    For practical opamps the open loop gain is typically 100k to 1M. If it is 1M, then if the opamp is outputing 10V and it is not saturated, then the differential input voltage will only be 10μV. There are some other non-ideal properties that might make the actual difference somewhat more, but this is the general order of differences you are looking at. For a run-of-the-mill opamp you would generally see differences in the mV range or less.
     
  7. Sue_AF6LJ

    Member

    Mar 16, 2013
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    This is the circuit we are talking about....

    [​IMG]

    Nobody ever..
    Well almost never connects the inputs directly to voltage sources unless this is being used as a comparator or a voltage follower..
    Regardless of the gain whatever you put at the input; the OP-Amp equalizes or attempts to equalize the voltages at the inverting and non inverting inputs.
    In this circuit the ratio of the feedback resistor to the input resistor sets the gain.
     
  8. gerases

    Thread Starter Member

    Oct 29, 2012
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    This is the part I'm interested in -- how does it equalize them?
     
  9. Sue_AF6LJ

    Member

    Mar 16, 2013
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    The Op-Amp has at it's input a circuit called a Differential Amplifier
    See below...
    [​IMG]
    Sorry for the large image...
    There isn't much on the net to choose from..
    In this circuit when both inputs are at rest the output voltage at each collector are equal. Keep in mind this circuit is almost oversimplified.
    In a real OP-Amp the emitter resistor the two transistors have in common is really a current source made up of a transistor or two and one or more diodes and resistors.

    If I put a voltage on the base of Q1, Q1 draws current which raises the emitter voltage of both transistors causing the collector voltage on Q2 to rise. If I do the same to Q2 the emitter voltage on both transistors will rise and the collector voltage of Q1 will rise. The collectors of Q1,2 are the outputs of the differential amplifier. In an OP-Amp these outputs continue on to a push-pull output stage that has a single ended output.

    What you see above is the actual heart of most all bi-polar Op-Amps, the FET input version is more or less configured the same way and does the same thing, the only difference is they use FETs.

    The first OP-Amps used vacuum tubes, same circuit, different era....

    I hope this helps.
    Once you understand how Op-Amps work you can do whole lot....
     
  10. ScottWang

    Moderator

    Aug 23, 2012
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    Your question is ideal situation as we wish the forward voltage of diode is zero, but it is around 0.7V, and we also wish the vce of transistor is zero, but it is about 0.2V, so just take it easy.
     
  11. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    Here's an analysis of simple op amp circuits. The simplified equation for a non-inverting op amp circuit is shown on page 5. Op Amp Eq.gif
    Obviously for high values of Av the denominator is very close to 1. To answer kubeek's query, you can see that if the gain is infinity, then the denominator becomes (1+R2/R1)/∞ + 1 = 1 (since (anything)/∞ is 0). The result then becomes the familiar ideal gain equation, Vo/Vin = -R2/R1, that we use for an inverting op amp circuit.
     
    Last edited: Mar 19, 2013
  12. WBahn

    Moderator

    Mar 31, 2012
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    Do you mean how an opamp achieves such high gain values, or just how you can analyze a circuit using the high gain value instead of the assumption of the virtual short.

    The quick answer to the former is that it is a carefully designed multistage differential amplifier.

    For a lot of the ICs that I've designed, we used a highly simplified one-stage amplifier that had an open loop gain of only a few hundred. Sometimes it was so simple that it only had a gain of 10 to 20. Even so, our basic design analyses would still use the virtual short method, but then we always backed up the design analyses with simulation results. Occasionally, we would need higher gain in which case we would tweak the design or, space/power permitting, add a second stage.


    For the latter, let's consider the classic inverting amplifier stage where you have the V+ input tied to GND and the V- input is at the junction of an input resistor, Ro, and the feedback resistor, Rf.

    The output of the amplifier is Vout and the input is Vin. The voltage gain of the amplifier is Av.

    So the output is

    Vout = Av(V+ - V-)

    V+ = 0V

    V- = Vin + [(Vout-Vin)/(Ro+Rf)]Ro

    (or you could write it as V- = Vout - [(Vout-Vin)/(Ro+Rf)]Rf )

    Plugging these in we have

    Vout = Av(0V -{Vin + [(Vout-Vin)/(Ro+Rf)]Ro})

    Vout = -Av {Vin + (Vout-Vin)/(Ro+Rf)]Ro }

    Vout = -Av {Vin + (Vout-Vin)[Ro/(Ro+Rf)]}

    Vout = -Av(Vout)[Ro/(Ro+Rf)] + Av(Vin)([Ro/(Ro+Rf)]-1)

    Vout + Av(Vout)[Ro/(Ro+Rf)] = + Av(Vin)([Ro/(Ro+Rf)]-1)

    Vout {1 + Av[Ro/(Ro+Rf)]} = + Av(Vin)([Ro/(Ro+Rf)]-1)

    Vout/Vin = Av ([Ro/(Ro+Rf)]-1) / {1 + Av[Ro/(Ro+Rf)]}

    Multiply top and bottom by (Ro+Rf)/Ro and you get

    Vout/Vin = Av ( 1 - (Ro+Rf)/Ro ) / {(Ro+Rf)/Ro + Av}

    Vout/Vin = Av (-Rf/Ro) / {(1 + Rf/Ro) + Av}

    Now divide top and bottom by Av and you get

    Vout/Vin = (-Rf/Ro) / {1 + (1 + Rf/Ro)/Av}

    So let's say that you wanted a gain of -4 and chose Rf = 4Ro. Your actual gain would be

    Vout/Vin = -4 / {1 + 5/Av}

    If you had a really crappy opamp, such as is not uncommon for an opamp in a pixel of a SPrITE array (Signal Processing in the Element), you might have an open loop gain of 10 and so your actual gain would be

    Vout/Vin = -4 (2/3) = 2.67

    But if A was even 100, your actual gain would be

    Vout/Vin = -4 (95.2%) = 3.81

    If you had wanted a gain of -100 instead of -4, then even with an open loop gain of 100 your closed loop gain would only be about half of the desired gain.

    Looking back at the full equation

    Vout/Vin = (-Rf/Ro) / {1 + (1 + Rf/Ro)/Av}


    And defining the nominal closed loop gain to be Ao=-Rf/Ro, we have

    Vout/Vin = Ao / {1 + (1-Ao)/Av}

    Vout/Vin = Ao / {1 - (Ao/Av) + (1/Av)}

    As Av gets larger, particularly in comparison to Ao, then the second two terms in the denominator get smaller and smaller leaving you with the leading 1 and result that the actual gain is very close to the nominal gain. In the limit that Av is infinite, then you are left with

    Vout/Vin = Ao = -Rf/Ro

    Let's say you had an opamp with an open loop gain of 100,000 and you chose resistors to give a nominal gain of 1000. You actual gain would then turn out to be

    Vout/Vin = Ao / {1 - (1/100) + (1/100000)}

    Vout/Vin = Ao (99.01%)

    or a gain error of only 1%, which would be comparable to what you would expect from using 1% resistors in the circuit.

    If you were only shooting for a gain of -100, then your gain error due to finite opamp gain would only be 0.1% and would likely be negligible compared to other error sources.

    If you trule could make an opamp with infinite gain, the you wouldn't have any gain errors, but the higher the opamp gain the harder it is to keep it stable and prevent oscillation. There are opamps out there with gains in excess of ten million, but stability is a real concern in designs that use them. With infinite gain, this task would be all but impossible.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    Boy, did this thread moved fast! I started working on my last response (Post #12) immediately after posting the prior one (Post #6).

    Sometimes fast moving threads result in disjointed flows of thought, so hopefully that didn't happen too bad.
     
  14. patricktoday

    Member

    Feb 12, 2013
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    Does it make it any easier to picture it that the op amp will push its output with all its mortal might until the inverting input is brought back to the (almost) exact voltage of the non-inverting input? If you have a 1k resistor to the (-) input and you're doing an inverting amplifier configuration, no matter _what_ value you choose for the feedback resistor, the output will push downward until the current across it satisfies Ohms Law to ensure its life's mission: namely that (-) becomes equal to (+). Vout minus V(+) divided by the resistor will be made equal to the current flowing through the 1k resistor at the (-) input, and this is done by pushing down the voltage at the output as much as necessary. To me, the problem is more intuitive viewed in reverse. (And I'm not trying to explain the input offset voltage.) HTH!
     
  15. gerases

    Thread Starter Member

    Oct 29, 2012
    177
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    Thanks, all, but I need it dumbed down a bit.

    The best explanation I read was in Electronics for Dummies, but it still left me wondering. It said that because of the negative feedback network, the output voltage fed back is "divided" according to the ratio of the resistors (which form a voltage divider).

    If the resistors R1 and R2 are equal, it will be halved in a loop many, many times until it's almost zero.

    But then why doesn't it keep being divided forever and how can it happen so quickly with an alternating voltage source?
     
  16. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The voltage divider/amplifier is a single pass ordeal, not a infinite repetition.

    If you think of the output being the voltage required to make the two inputs equal, it makes more sense.

    If inverting input is at 5V and the non-inverting input is at 0V, with a 1k resistor at the inverting input, and 10k from output to the inverting input, picture those two as a divider. The output going through the 10k resistor will need to be higher to match the 5V input (with 1k input impedance), hence, gain.
     
  17. bobweb1234

    New Member

    Mar 18, 2013
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    Think of the input going to zero as balancing. The negative feedback simply increases until it balances at the inputs. Then the input difference is zero, and nothing further happens.

    bob
     
  18. WBahn

    Moderator

    Mar 31, 2012
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    Your confusion is coming as a result of not having done the prelimary worrk of understanding basic circuit analysis involving bothing but voltages and resistors.

    If you have two resistors in series, Ro on the left and Rf on the right, what will the voltage at the junction between them be if a voltage Vin is applied to the left end of the string and a voltage of Vout is applied to the right end of the string. Forget about opamaps. Just focus on answering this question involving two resistors and two voltage sources.
     
  19. gerases

    Thread Starter Member

    Oct 29, 2012
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    Very likely.

    Is Vout an independent voltage source or is it a result of Vin? Do they common ground? Sorry, I don't really need to be challenged more than I already am.
     
  20. gerases

    Thread Starter Member

    Oct 29, 2012
    177
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    I've seen the math that refers to an ideal amplifier, but I want to understand it intuitively. Perhaps through an analogy?
     
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