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Low-Frequency Differentiator (need help with the calculus)

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data
    This is a low-frequency differentiator circuit.
    Vo(t) = -0.001*[dVi(t)/dt]

    Vi=0.5Vpk
    Frequency at 1kHz

    I am confused on how to apply calculus to this. My calculus is a little rusty but I am not use to seeing it in this applied forum. I only did calculus when it was in simple mathematic forum and I guess I never connected the dots.

    I am hoping someone could walk me through this problem.


    I have to build this in Multi-sim and show the graph but I am getting a saturated voltage out of 0.5Vpk. Which is fine if the value ends up being over 14Volts. *Note this is another problem and not the one shown but I wanted to change the values*

    The concept is this circuit takes a waveform and differentiates it. I can fine with the circuit itself, just proving my voltage level using the above equation is what is throwing me off.
     
  2. jcsd
  3. Jul 18, 2015 #2

    gneill

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    Can you post the circuit diagram? I am guessing that it is a simple RC arrangement with an op-amp at its heart, but it's best to be sure.

    Are you familiar with the Laplace transform approach to circuit analysis? It's a direct path to discovering the differential equation associated with a transfer function. If not, you can also use the basic current-voltage relationships of reactive components to write the circuit equations.
     
  4. Jul 18, 2015 #3
    circuit.png Yes here is the circuit diagram for the problem. The circuit portion itself is not an issue I believe. Is it possible to solve that above equation using Vi=0.5Vpk with just calculus? Ignoring the circuit itself.
     
  5. Jul 18, 2015 #4

    gneill

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    Thanks for the diagram.

    I'm not sure what you mean by solving the equation using Vi=0.5Vpk, but I imagine that the exercise is to determine the relationship between Vo and Vin given the above circuit arrangement. This will involve writing equations that pertain to the circuit. That in turn means using KVL and/or KVL and knowledge of the VI characteristics of the components. You'll end up deriving an expression for Vo in terms of Vin that will have the form of Vo(t) = -0.001*[dVi(t)/dt], with an expression involving actual component values in place of the "0.001" constant. Start by considering how for a capacitor the current is related to the change in voltage.
     
  6. Jul 18, 2015 #5

    LvW

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    gneill - it must be your goal to calculate the transfer function Vout(s)/Vin(s)=T(s) in the frequency domain for finding a corresponding circuit.
    In principle, the circuit as shown by you is correct - however, the parts values are NOT.
    What you need is the corresponding RC time constant of the differentiating circuit.
    This is easily found by applying the LAPLACE transform to the expression dV/dt.
    This is a very simple task and can be found in each book dealing with LAPLACE (or, of course, in the web).
     
  7. Jul 18, 2015 #6

    gneill

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    Hi LvW. I'm not the OP. I'm trying to find out exactly what the OP is trying to work out. If it's just the magnitude of Vo for a given sinusoidal Vin, then he has options including Laplace, complex impedance, or direct differential equation approaches.
     
  8. Jul 18, 2015 #7
    Yes I am trying to gain the magnitude of Vo for a given sinusoidal Vin. This is where I am stuck. I will do some research on Laplace.
     
  9. Jul 18, 2015 #8

    gneill

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    If it's just the magnitude that you need then an impedance approach would be expedient. You know the input frequency so you can find the impedance of the capacitor. Then you can determine the gain of the simple op-amp layout.
     
  10. Jul 19, 2015 #9

    LvW

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    Sorry for this mistake.
     
  11. Jul 19, 2015 #10

    gneill

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    No worries :smile:
     
  12. Jul 19, 2015 #11

    rude man

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    For sinusoidal inputs you don't need Laplace, in fact it's overkill (using the transform for a sine input gives you the transient as well as the steady-state response. You want the steady-state only, most likely).

    So just use complex algebra: Z of a capacitor = -j/wC, Z of a resistor = R. And BTW the relation Vout = -k dVin/dt does not hold for high frequencies (f > 500 Hz approx., due to the finite value of Ri).
     
  13. Jul 20, 2015 #12

    LvW

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    I cannot agree to the above contribution. It is no problem to realize an active differentiator for frequencies>500 Hz. By the way: What is meant with the symbol "Ri"?

    For finding the required transfer function of the differentiator (as a first step to realize an electronic circuit) it is necessary to know how the process dV/dt (time domain) looks in the frequency domain. For this purpose you need the LAPLACE transform as a tool which connects both domains. However, it is a very simple relationship that can be found at the first place of each correspondence table (books, wikipedia,...)
     
  14. Jul 20, 2015 #13

    rude man

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    Look again at the OP's last statement: "Yes I am trying to gain the magnitude of Vo for a given sinusoidal Vin. This is where I am stuck."

    For that, laplace is not called for. In fact, it's misleading as I explained.

    Look again at the schematic the OP furnished. Ri is shown as a 159 ohm resistor in series with C1. The presence of Ri prevents the circuit from being a differentiator at high frequencies. At high frequencies ( ~1000 Hz or above) the circuit is a simple gain of -20K/159 ohms. Since the input is stated as 1000 Hz the circuit is definitely not differentiating the input.

    (In "real life" Ri is needed to prevent excessive high-frequency noise. In the old days of analog computers, if you wanted to simulate a differentiator, you would not use this circuit (minus the Ri). You would wrap an integrator around itself instead.)

    If the OP really wanted to "use calculus" he could go with capacitor i = C dV/dt and resistor i = V/R of course, then KVL or KCL or whatever. He could also form the Laplace transfer function Vout/Vin and rearrange in the form Vout*F1(s) = Vin*F2(s), then substitute s = d/dt, i.e. interpreting s as the differentiating operator.
     
    Last edited: Jul 20, 2015
  15. Jul 20, 2015 #14

    LvW

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    Oh yes - I agree. I didn`t look at the second post of the questioner. Sorry for that.
    Of course, if the value is not changed, the circuit does not work as desired. Therefore my answer in post'5 that - in principle - the circuit is OK, however the values must be adapted to the required time constant. And yes, I know about the common problems in realizing such a task with real opamps (tendency to oscillations at the point where the loop gain is unity). Not only because of excessive noise! Hence, a suitable Ri is necessary.
    There was a misunderstanding between us - my fault.
     
  16. Jul 20, 2015 #15

    LvW

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    Sorry - but this comment is, for my opinion, misleading.
    In contrary, for designing a workable differentiating circuit it is necessary to know the corresponding transfer function in the s-domain.
    This means that I must be able to transform the operation dV/dT into the s-domain (as you know, this simply gives "s")
    Hence, the corresponding transfer function is H(s)=-0.001*s.
    This knowledge - together with the frequency response of the used real opamp - is required for finding a suitable value for the stabilizing resistor Ri .

    A rough calculation - assuming a general-purpose opamp with GBW=1E6 Hz - shows that a design with Rf=1kOhm, C=1µF and Ri=25 Ohms allows a stable differentiating process at 1kHz with a phase margin of app. 45 deg.
     
    Last edited: Jul 20, 2015
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