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How to clasify if it's an inverting/non-inverting op amp

  1. Feb 2, 2015 #1
    Hello,

    Like the title said, how can you tell if it's an inverting or non-inverting op amp just by looking the the schematic like this one.

    upload_2015-2-2_0-53-17.png
    textbook said this is an inverting op amp. I thought whichever input node (+/-) the signal go through dictate the classification of an op amp if it's inverting or non-inverting.

    Questions I still have regarding op amp:
    • Op amp needed feedback (through '-' non-inverting input) to "stabilize" it. What does it means? I'm thinking that since the gains is most often in the magnitude of >100k, the feedback is there to take away some of the amplification so that the amplified voltage could be in the range of the supplied voltage of the op amp.
    • what is it mean when the op amp is "saturated"?

    Thanks ahead.
    X
     
    Last edited: Feb 2, 2015
  2. jcsd
  3. Feb 2, 2015 #2

    Svein

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    There is no such thing as "an inverting or non-inverting op amp". An operational amplifier has two inputs, one inverting and one non-inverting. The circuit decides how you use it. The schematic shows it used in the non-inverting mode, which again means that it is not used as an amplifier (analyze the circuit yourself - assume that Vout max = 14V and Vout min = -14V).
     
  4. Feb 2, 2015 #3

    davenn

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    hi there

    I have seen that configuration before on here
    The interesting things is, when you google images for inverting or non-inverting opamps
    you never see that particular configuration for either mode

    will wait for out resident opamp experts to chime in

    Dave
     
  5. Feb 2, 2015 #4

    LvW

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    The shown circuit cannot work as an amplifier at all (neither inverting or non-inverting) because it contains POSITIVE feedback. As a consequence, it will immediately latch/saturate at one of the supply rails. Each working opamp based circuit needs negative feedback for stabilizing a suitable operating point.
    After exchanging both opamp input nodes the circuit resembles an inverting amplifier with a gain of -9/3=-3.
     
  6. Feb 2, 2015 #5
    Okay, so conceptually. This doesn't work in practice, but the textbook stating that this configuration is "inverting" is wrong, correct?
     
  7. Feb 2, 2015 #6

    LvW

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    Yes - it is wrong (just an unintended mixing of both input terminals).
     
  8. Feb 2, 2015 #7
    So my initial thought was correct, whichever input is active will classify the op amp type? That is to say:
    signals going through non-inverting input will make that op amp "non-inverting" and vice versa. What if they both active?
     
  9. Feb 2, 2015 #8

    Svein

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    Just to show you a quick simulation of the shown circuit (caveat: I was using an op-amp with only ±6V supply voltage and an input signal of 100Hz AC). The green signal is input, the red is output.
    upload_2015-2-2_10-37-49.png
     
  10. Feb 2, 2015 #9
    That data is purrty data man. If you don't mind, can you point out what is not right about it?
     
  11. Feb 2, 2015 #10

    NascentOxygen

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    Providing it really is a proper amplifier, yes.
    One will dominate. But usually the circuit will involve frequency-dependent impedances (capacitors, etc), so the sign of the gain may change from + to - over some part of the usable range, and have a phase somewhere in between for most of the intermediate range.
     
  12. Feb 2, 2015 #11

    NascentOxygen

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    The output waveform doesn't resemble the input sinusoid! So not an "amplifier" in the ordinary sense of the word. The output is not linearly related to the input.
     
  13. Feb 2, 2015 #12

    davenn

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    and that was my whole point about never seeing this configuration, I couldn't see how it could be correct :)
     
  14. Feb 2, 2015 #13

    NascentOxygen

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    The configuration is okay. It's a Schmitt comparator, and common enough. It's just not a linear amplifier.
     
  15. Feb 2, 2015 #14

    davenn

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    Thankyou !!
    which is why it didn't show up in the inverting or non-inverting ccts
    I really have to remember that one for next time I see it :)

    Dave
     
  16. Feb 2, 2015 #15

    jim hardy

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    oops i see many others posted while i was typing sorry if this is redundant... old jim


    "Operational Amplifier" unfortunately has dual meaning.
    It is the name given to an amplifier that could perform some math operation provided it has been surrounded by circuitry that lets it make its two inputs equal. That's how it "operates".
    It has fallen into common usage as the name given to a circuit containing both such an amplifier and its surrounding circuitry when used with negative feedback to make its inputs equal. It performs a math operation, usually but not always a linear one.

    The way to tell whether the latter inverts is to work it in your head. i talk my way through them.
    Looking at yours
    First question is can this operate ? That is, can it hold its + and - input terminals equal?
    upload_2015-2-2_0-53-17-png.78539.png
    Let's see what we know:
    we know input- is held firmly at 0.
    we know input+ will be at some voltage that lies in between Vin and Vout, because the 3k and 9k resistors make a voltage divider.
    we know that making input+ more positive makes Vout more positive because that's what the + means.
    So at first glance it appears
    applying a tiny positive voltage to Vin will make Vout go positive
    so input+ will lie between Vin which is positive and Vout which is also positive

    Now - the condition for operation was it must make its inputs equal, that is make input+ equal zero.
    Can zero lie between two positive numbers? I think not.

    So i think your textbook has thrown you a trick question - that circuit isn't a linear operational amplifier circuit, even though it has an operational amplifier in its heart.
    Now there do exist the math operators for comparison, > for greater than and < for less than.
    Your circuit can do that function.

    Let's apply some arithmetic to your circuit.
    Voltage between Vin and Vout is of course Vin-Vout
    so voltage across the 3k is ¼(Vin-Vout)
    and voltage across the 9k is ¾(Vin-Vout)
    Write KVL around this closed loop::: Vin to common to input+ through 3k back to Vin
    and you get
    Vin - (input+) - ¼(Vin-Vout) = 0
    re-arrange
    input+ = Vin - ¼(Vin-Vout)

    input+ = ¾Vin + ¼Vout

    so in order for input+ to be zero, Vin and Vout must have opposite polarity. That's what is meant by inverting opamp circuit.
    In the circuit above they don't, but if you swap the amplifier's input pins they will.
    So as is, that circuit doesn't invert.

    If this was a trick question, the circuit indeed doesn't invert so it must be noninverting. But i wouldn't call it an operational amplifier circuit

    Since the opamp has supply of +/- 15 volts that's the highest its Vout can go.

    What if Vout were at that limit, let's just say postive 15 volts?
    could we drive input+ to zero by applying negative Vin ? If we went just a little further would we tip the balance and make Vout go to negative 15? Of course we could.

    So let Vout be +15
    what Vin will drive input+ to zero?
    input+ = ¾Vin + ¼Vout
    0 = ¾Vin + ¼(+15)
    ¾Vin = - ¼(+15)
    Vin = -5

    So that circuit flips its output state whenever input edit: reaches -Vout/3, as shown in svein's delightful post #8 scope trace . crosses 5 volts with polarity opposite Vout

    which would be described as a comparator with hysteresis
    also known as a "Schmitt Trigger"
    and it does a logical math operation, the inequality "greater than" with memory of last state

    and since output polarity follows input polarity
    i'd call it a "non-inverting comparator with 5 volt hysteresis"

    KVL is great for figuring out op-amps.

    Ask teacher if that was a trick question.

    Have fun. Talk to your circuits, it'll amuse your roommates no end.

    old jim
     
    Last edited: Feb 2, 2015
  17. Feb 2, 2015 #16

    jim hardy

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    Svein's 'scope trace in post 8 shows Schmitt Trigger action quite nicely.

    NIce Job !
     
  18. Feb 2, 2015 #17

    LvW

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    Yes - it is a so called "Schmitt trigger". Watch the hysteresis effect (output goes high at another level as it goes down again).

    Warning: One should know that some simulation runs do not reveal the unability of the circuit to amplify. For example, doing an ac analysis you get a result which looks quite "normal" (as expected).
    However, you should not blame the program in this case. The result is quite correct.
    Why? Because this analysis assumes (a) that the circuit was powered since infinite times (without switching transients) and (b) that there are absolutely no external influences (noise, temp. variations) which would disturb the quiescent conditions.
    You must do a TRAN analysis (time domain) to show that the circuit cannot work in a linear mode.
    This case is similar to the mechanical model of two balls - one riding upon another one. Under ideal conditions this arrangement could also be stable.
     
    Last edited: Feb 2, 2015
  19. Feb 2, 2015 #18

    Svein

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    Just to show the basics:
    upload_2015-2-2_12-34-27.png
    Non-inverting circuit:
    • Gain = 1 + R3/R2
    • Input impedance: R1
    Inverting circuit:
    • Gain = -R5/R4
    • Input impedance: R4
     
  20. Feb 4, 2015 #19

    jim hardy

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    @xconwing


    haven't heard from you in a while.

    Maybe in our enthusiasm we missed your original question:

    ......
    That's exactly right.
    We don't call it "taking away the amplification" though
    we call it "negative feedback" because the output is "fed back" in such direction to subtract from the input signal. So it more takes away signal than amplification.
    So, (input - feedback) X (gain of >100K) < supply. Maybe supply is 15 volts? If so, (input-feedback) must be less than 15/gain, around 150 microvolts. We usually round that off to zero for analysis.
    Your 100K gain is called "open loop gain" and is designated on datasheets by Avol
    When surrounded by a circuit including feedback, the gain of the entire circuit becomes called "closed loop gain"
    To use almost your exact words, it means the amplified voltage ISN'T within the range of the supplied voltage of the op amp.
    that is,
    try as it may the poor little amplifier just can't make its output voltage go all the way to (gain X input) because there's not that much voltage available to it.
    Most amplifiers can drive output to about a volt shy of the supply voltage
    so called "rail to rail" opamps can drive output to within millivolts of the supply voltage.


    Your questions suggest that you have an intuitive feel for these circuits.
    Be rigorous in your thinking and vocabulary and you will do very well.

    Good luck in your studies !!!!

    hmm where'd all those extra dots come from? see my signature...

    old jim



     
  21. Feb 4, 2015 #20
    @ jim hardy
    I was just letting the thread chill for a bit so that I could soak up (or recalling) some of the information mentioned.

    THANK YOU all for the replies.
     
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