Nonconstant Resistance (newb) Question

Discussion in 'The Projects Forum' started by Ippocrate, Jun 5, 2007.

  1. Ippocrate

    Thread Starter New Member

    Jun 5, 2007
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    Hi everyone,

    I feel almost ashamed to ask this - but I am mechanical engineer so hey :p

    We are trying to calculate the resistance of a mechanical device (microcantilever).
    We actually want to have the whole impedance model of the cantilever, but having
    at least the resistance would be a good start.

    Here is the problem. Following the circuit which I have attached, the cantilever
    resistance would be given by:
    Rc = (Vout/Vin - 1)*Rb

    meh.jpg

    The tricky part is that we are supplying a voltage which has both AC and DC part,
    therefore when carrying out that division a very strange curve appears for the
    resistance ( [A*sin(wt) + B]/[C*sin(wt) +D] ).

    Is it possible to just divide like we did, or should we separate AC and DC
    calculations? The two approaches give out different results for the resistance
    (one gives it as a constant - if we separate ac and dc calculations,
    the other results in a quite strange curve as mentioned above).

    I would say we should separate ac and dc, but something in my head also says to me that since
    we have got only resistances in there ... dividing mixed ac and dc signals
    should actually also work fine...

    Btw, we expect to see a non-constant resistance in there, therefore I am not
    happy with the solution we find by separating ac and dc, because if we separate,
    and just look at the dc part, we have constants Vout and Vin, therefore constant
    Rc.

    Which one is the right path ? o_O
     
  2. Ippocrate

    Thread Starter New Member

    Jun 5, 2007
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    Argh I see now i forgot, Vin is of course the voltage taken right
    before the cantilever Resistance (Rc) and the Vout
    is the voltage taken right after Rc.

    Also, sry if I am posting in the wrong section :p
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Why are you using AC+DC? Also, I think your equation is wrong. It should read:

    Rc = (Vin/Vout - 1)*Rb
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    290
    If I were going to come up with a resistance figure, I would use DC (or an ohmmeter) for the test.

    Impedance involves changing resistance as a function of frequency, so that test would need not only AC, but a couple of frequencies applied. A plot of the response to a frequency sweep would be quite valuable.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    If the Rc varies, then How can Vout be constant? With Vin being DC, varying Rc will cause Vout to vary. The same is true if Vin is AC, but this is more difficult to measure.
     
  6. Ippocrate

    Thread Starter New Member

    Jun 5, 2007
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    You are right, I had that right even before but I typed the
    equation from the top of my head - we had at least that
    correct ;)

    I understand that if we had a simple DC, our calculations
    would have been very easy.

    However we want to send in a sinusoidal input to measure
    the change in resistance due to the change in temperature
    caused by heating fluctuations (very small).

    Also we need a sinusoidal input so that in a successive step
    we will be able to figure out the whole impedance of the
    cantilever (not just resistance) by doing a frequency sweep
    as has been suggested.
     
  7. Ippocrate

    Thread Starter New Member

    Jun 5, 2007
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    You are right !
    The problem is that we couldnt see any change in the resistance
    because we would take as DC value for the Vout the
    (constant) value which comes out of a fitting.

    What do you suggest in order to get, from a messy output
    (which contains DC plus AC first harmonic plus a bunch of
    other harmonics) only the "time dependent" DC value?
    I was thinking whether a filter that cuts everything and only
    lets in compontents at around 0Hz could do the job..
    0Hz is DC right?

    But I guess that with this I am just doing the total
    average of the whole signal, therefore I have again only a
    constant value.

    Also, you mention this measurement of resistance could also be
    done in AC. And that is what we want to do. How would you do that?
    Because that is where the dilemma with [A*sin(wt)+B]/ [C*sin(wt)+D]
    comes in, if we just take the above equation and blindly substitute
    the curves which contain both AC and DC
     
  8. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
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    The DC is simply an offset of the AC. If you had a +5V battery and applied AC of 2Vp-p, you would have an AC signal from +4V to +6V.

    To measure, connect an oscilloscope using "AC coupling" mode (which will remove the DC component) and measure the AC voltage drop across what should be the purely resistive and known value of Rb that you add in series with your device Rc. The main issue to watch for, if your device is inductive, is to be sure that you are NOT over-driving it into saturation.
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    How big is this microcantilever? Do you really think it has a reactive (non-resistive) impedance component (at less than microwave frequencies)? Is it acting basically as a strain guage?
     
  10. Ippocrate

    Thread Starter New Member

    Jun 5, 2007
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    You see, also it seems to us that if you do not apply a DC
    value, when you do the division from the equation above
    in order to get the value for Rc, then you are going to
    get a division by 0.

    Basically, when you do Vin/Vout and Vin and Vout are pure AC
    signals (with same freq) you will have all the points at
    which Vin and Vout are 0 that will give you problems.
    Or how can we get around this?
     
  11. Ippocrate

    Thread Starter New Member

    Jun 5, 2007
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    I don't have the specifics for the cantilever (but it is really tiny :p)

    We are goin to drive it up until 50kHz, and what we want
    to measure is the amplitude of fluctuation of the resistance
    at each frequency. Maybe you have some advice on how
    to calculate those fluctuations of resistance in an other
    way? :|
     
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I wouldn't expect resistance of a tiny element to be a function of frequency, unless it is piezoelectric.
     
  13. nomurphy

    AAC Fanatic!

    Aug 8, 2005
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    By using an oscilloscope to measure only the AC component across Rb, you can determine the value of Rc.

    As an example: if you generate 1V RMS at the input, and you measure 0.7V RMS across the known value of Rb (let's say 1K), then you have a current of 700uA RMS (VRb / Rb). You also know that VRc = 1V RMS - 0.7V RMS, which is 0.3V RMS, therefore 0.3V RMS / 700uA RMS = 428 ohms:

    Rc = 428 ohms in this example, and at whatever is the test frequency.

    If the voltage across Rb changes with frequency, with the same value RMS input voltage, then you will know that Rc is reactive. And, a little experimentation and measurement should tell you in what way.

    You could also try placing an LCZ meter across Rc (out of circuit), and see what readings you get.
     
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