# Nonconstant Resistance (newb) Question

#### Ippocrate

Joined Jun 5, 2007
6
Hi everyone,

I feel almost ashamed to ask this - but I am mechanical engineer so hey 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 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 ? #### Ippocrate

Joined Jun 5, 2007
6
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 #### Ron H

Joined Apr 14, 2005
7,014
Why are you using AC+DC? Also, I think your equation is wrong. It should read:

Rc = (Vin/Vout - 1)*Rb

#### beenthere

Joined Apr 20, 2004
15,819
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.

#### Ron H

Joined Apr 14, 2005
7,014
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.
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.

#### Ippocrate

Joined Jun 5, 2007
6
Why are you using AC+DC? Also, I think your equation is wrong. It should read:

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

#### Ippocrate

Joined Jun 5, 2007
6
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.
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

#### nomurphy

Joined Aug 8, 2005
567
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.

#### Ron H

Joined Apr 14, 2005
7,014
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?

#### Ippocrate

Joined Jun 5, 2007
6
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.
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?

#### Ippocrate

Joined Jun 5, 2007
6
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?
I don't have the specifics for the cantilever (but it is really tiny )

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? :|

#### Ron H

Joined Apr 14, 2005
7,014
I don't have the specifics for the cantilever (but it is really tiny )

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? :|
I wouldn't expect resistance of a tiny element to be a function of frequency, unless it is piezoelectric.

#### nomurphy

Joined Aug 8, 2005
567
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.