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Homework Help: Schmitt Trigger question

  1. Feb 24, 2016 #1
    1. The problem statement, all variables and given/known data
    FIGURE 2 shows an op-amp circuit with a constant input voltage V connected to the inverting input. If the output
    voltage is +2.4 V, calculate the value of V.

    Schmitt trigger values:
    R1=4.8 ohms
    R2=2 ohms
    V OUT= +2.4V

    2. Relevant equations
    V+=(R1*V SAT)/(R1+R2)

    3. The attempt at a solution

    Now this is the Reference Voltage going to the non inverting terminal but I'm lost as to where to go from here, this would indicate that V is less than 1.69?
    Any help to point me in the right direction would be much appreciated. This is the first time I've studied the electronics side of things and it's safe to say I'm struggling a little.
    Last edited: Feb 24, 2016
  2. jcsd
  3. Feb 24, 2016 #2


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    Hi StripesUK, Welcome to Physics Forums.

    Where is "FIGURE 2"? We'll need to see the circuit if we're to check your attempt or offer advice.
  4. Feb 24, 2016 #3
    Sorry about that. Please see attached.

    Attached Files:

  5. Feb 24, 2016 #4


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    Why don't you try assuming some finite gain "A" for the op-amp and writing an expression for the output voltage for a given input voltage? Then you can see what happens when the gain is allowed to be very large. Don't assume any non-linear (Schmitt or comparator) qualities at the outset.
  6. Feb 25, 2016 #5
    I think I have it now...

    So if,



  7. Feb 25, 2016 #6


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    You haven't defined what V1 and V2 are. Can you show more of your work in detail?
  8. Feb 25, 2016 #7
    R1=4.8 ohms
    R2=2 ohms

    [itex]V= \frac{R1V2}{-R2} [/itex]

    [itex]V= \frac{4.8*2.4}{-2}=-5.76V [/itex]

    I derived all this from the equation, (V1 being the voltage(V) that I am looking for);

    [itex]\frac{V2}{V1}= \frac{-R2}{R1}[/itex]

    Both sides can be used to find the gain.
  9. Feb 25, 2016 #8


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    Those should be in kilo ohms (although it won't change the voltage results).
    I don't see where this comes from. Why the negative sign on R2?

    In this circuit feedback is provided from the output to the input via a voltage divider (R1 & R2). The result is to "lift" the potential of the + input of the op amp by a fraction of the output voltage. If the output happens to be positive, then this "lift" is also positive. The very large gain of the op amp will ensure that this "lift" will be driven towards matching the potential at the - input of the op amp (enforcing a zero potential difference between the inputs).

    You should be able to derive this result by first assuming that the op amp has some finite gain A and writing the circuit equations to solve for ##V_{out}## in terms of ##V_{in}##, then taking the limit as A gets very large.
  10. Feb 25, 2016 #9

    jim hardy

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    uhhh,, mr stripes

    as drawn
    that opamp has positive feedback , so it can't drive its inputs equal unless its gain A were to be negative.

    mr gniel has hinted at that

    your calculation for V+ in post 1 is correct, to the digits you showed
  11. Feb 26, 2016 #10
    Hi guys, I too am struggling to get my head around this question and was wondering if you could see if I'm on the right path to my final answer and give me a bit of guidance if I am way off.

    My attempt so far:

    Vout = +2.4V
    Rf = 2k ohms
    R1 = 4.8k ohms
    Vin = ?

    First I calculated V+ by using:

    V+ = (R1*Vout)/(R1+Rf)
    V+ = 11520/6800
    V+ = 1.694V

    I then calculated the gain, as stated above because it has positive feedback the gain should be negative therefore I have used:

    A = -Rf/R1
    A = -0.416

    I then used the following and have transposed to try and calculate Vin (or V-):

    Vout = A(V+ - V-)

    Transposing for Vin I came out with:

    Vin = V+ - (Vout/A)
    Vin = 1.694 - (2.4/-0.416)
    Vin = 1.694 - (-5.76)
    Vin = 7.454V

    So I believe this is my final answer however I am not totally confident as I was expecting Vin < V+ due to Vout being a positive value, my only explanation is that my initial assumption was wrong and Vout is actually its lower limit but again I am not 100% sure and this is where my confusion lies. Any help or guidance would be greatly appreciated as I would love to get my head around this.
  12. Feb 26, 2016 #11


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    Your answer is not correct. You've made an assumption about the gain being negative which is not necessarily true, and you've used a formula for the circuit gain that is not correct.

    Don't calculate the actual gain of the circuit. Just assume that the op amp has an open loop gain of some value A (where A is a positive number). Then analyze the circuit and determine what happens when A is very large.
  13. Feb 26, 2016 #12
    So if I just give the gain a practical value such as:

    A = 10^5

    Then use this value in:

    Vin = 1.694118 - (2.4/10^5)
    Vin = 1.694094

    Vin is now less than V+ but only by a very small amount but this is still enough for Vout to be at its high limit. Is this more accurate or am I still going in the wrong direction? Apologies if I'm missing the obvious but I'm finding this topic a little difficult to wrap my head around.
  14. Feb 26, 2016 #13


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    No, it is not. You have assumed a finite gain for the op amp, so you won't get infinitely magnified differences. What you've discovered is that the input voltage will be very close in value to the potential at the op amp - input, even with a finite gain of 105.
    You should analyze the circuit purely symbolically first, without plugging in numbers or making assumptions about the op amp open loop gain or expressions for the circuit's closed loop gain. Give the op amp an open loop gain of A and find the relationship between Vin and Vout. Then see what happens when A becomes very large.
  15. Feb 26, 2016 #14
    Again, this may just be my stupidity but I thought I had already seen what happens when A becomes large by using the value 10^5? I can see that the closer my value of V- is to V+ then the higher the value of gain A is? I just don't understand how I am supposed to calculate Vin without an actual gain value?
  16. Feb 26, 2016 #15


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    The actual circuit gain comes from analyzing the circuit (which I'm hoping to coax you into doing), but your approximation using a fixed value for the gain of the op amp should give you a clue that you can expect the difference between the V+ and V- potentials to tend to zero as the op amp open loop gain grows very large.
  17. Feb 26, 2016 #16

    jim hardy

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    yes, make yourself a table
    columns for A and Vin

    "A" being gain of the amplifier itself whic you might vary from 0.1 to 10^6 in steps of ten, ie 0.1, 1, 10, 100 1000 etc
    you'll see gneill's point that if the circuit is "operating", ie output not saturated, amplifier's inputs must be different by Vout/A .

    you mentioned A of Rf/R1
    make yourself aware that there's two different gains
    one is the gain of the raw amplifier by itself, usually quite high for op-amps, like 10^5

    the other is the gain of a circuit encompassing an op-amp which the circuit designer makes whatever he wants by selecting resistors for feedback and input. That one we call "closed loop gain" .
    So long as the amplifier is able to push its inputs equal, it can be said to be "operating" that's why we call it an operational amplifier circuit.

    In the circuit you presented
    the amplifier is unable to push its inputs equal
    look - raising V will lower Vout driving the inputs farther apart
    so that question is either badly phrased or a trick question,
    i'd say an unfair one to give to a beginner
    we must rigorously do the arithmetic to get the answers
    but if you build that circuit and apply input that you can adjust with a knob
    you will be unable to hold Vout steady. It's a math dream only.

    That said:
    Swap the + and - inputs \ and it'll work beautifully, grab a LM324 and try it.

    old jim
  18. Feb 27, 2016 #17


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    Then it wouldn't be a schmitt trigger.

    I'm pretty sure your original answer of <1.69V is correct. Reading the other posts, it seems at least some assume an inverting amplifier rather than a schmitt trigger.
  19. Feb 28, 2016 #18

    jim hardy

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    Right !

    what i shoulda said

  20. Feb 29, 2016 #19
    The only other equation I can find involving Gain, Vin, and Vout is,



    [itex]V-=V+ - \frac {Vo}{G}[/itex]

    This would mean that V- is proportional to the Gain. The larger the Gain the larger V- becomes, this would also be the same for Vout.
    Am I now just looking for the correct calculation of the circuit gain G?
    Last edited: Feb 29, 2016
  21. Feb 29, 2016 #20

    jim hardy

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    Do you see now that circuit is counter-intuitive?
    They have fixed Vout at 2.4 volts and ask you to calculate the Vin that would cause that Vout .

    That's working the circuit backward. Adding insult to injury, the circuit is unstable .
    Very confusing and unfair to treat a beginner that way.


    did you mean "The larger the Gain the more nearly equal V- and V+ must become ? " That's required if the circuit is to not saturate.

    The term " gain " for a schmitt trigger circuit is meaningless because its output is not linear but bistable.

  22. Feb 29, 2016 #21
    Yes, sorry I can see that is correct from my different calculations of the Gain.

    So now because the feedback is positive the calculation of the Gain from the circuit would be [itex]G=1+ \frac {R2}{R1}[/itex]?
  23. Feb 29, 2016 #22

    jim hardy

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    If gain =## \frac {Δoutput}{Δinput}##

    what is the Δin required to cause a one volt change at output ?


    G X ( Vout X (4.8/6.8) - Vin ) = Vout

    Vin = Vout/G + Vout X (4.8/6.8)
    Vin = Vout X (1/G + 4.8/6.8)

    so the circuit can exist precariously balanced at any input Vin = Vout X (1/G + 4.8/6.8)
    but if Vin moves one single nanovolt away from that value
    positive feedback will move the + input AWAY from that balance , causing more imbalance
    by the time Δoutput reaches our 1 volt, imbalance at input has grown from our nanovolt to (1/G + 4.8/6.8 +1X10-9) volts
    that's what positive feedback does, instead of balancing things it further unbalances them

    so our gain of ##\frac {Δoutput}{Δinput}## has a denominator that's made infinitesimal by positive feedback
    that's what positive feedback does, raises gain.

    Gain infers linear operation
    and a schmitt is not a linear circuit.

    I still think this was an unfair problem to give to beginners.

    How did you make those nice looking fractions ?
    Last edited by a moderator: Mar 1, 2016
  24. Feb 29, 2016 #23
  25. Feb 29, 2016 #24

    jim hardy

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    i dont see a schmitt on that page

    TI doesn't even mention "gain" and i think it's a mistake to apply the term to a schmitt, gain is infinite... sum of a divergent series.

    Best way to learn them is get a 555, tie pins 2&6 together as input and you have a handy schmitt that'll interface with most any logic...

    wiki has a decent page
    What's gain at those vertical transitions ?
  26. Feb 29, 2016 #25
    Did you click on it before I edited the link? I originally posted an incorrect link.
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