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I'm dealing with the measurement of impedance sweeping the frequency from 0,1Hz to 100KHz. On the web I found many methods but the most quoted seems to be the one which uses an operation amplifier in inverting low-pass filter configuration:

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Z represent the impedance under test. In this configuration, we can find the module and phase of Z comparing the input and the output of the filter.

Another simplest method I found is the common voltage divider:

enter image description here

Z is the impedance under test again and its module and phase can be found again comparing the input and the output signals.

Now, I know that the first method which uses the opamp should be the better choice since it's used in the most Z-meter, but I don't understand the real advantage using that one.

Can anyone light up my mind?

thoraz
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  • Z is measured under frequency, the filter allows to reduce some noise, so your filter must be on the frequency of usage of device under test (DUT). An LCR can help you – Fernando Baltazar Jun 05 '18 at 20:28

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Because the op-amp when used as an inverting amplifier has a virtual earth at -Vin, the component "Z" is driven by "IN" with respect to ground. This means that the output of the op-amp (barring it being non-ideal and also ignoring the feedback capacitor) is representative of the current passing through "Z".

So you have two voltages; the drving voltage and a second voltage proportional to Z's current.

Therefore the op-amp route is simpler to compute impedance than having a voltage applied across Z in series with Ro and a voltage across Ro that is proportional to current through the series combination of Z and Ro.

Having said all of that and given the non-idealities of op-amps as frequency gets high, the 2nd approach will, inevitably, be more accurate if a little more complex mathematically to solve.

Andy aka
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  • So are you tell me that the voltage divider is not so good of this kind of application because the expression of output voltage show a ratio where the numerator and denominator are not independent due to the presence of Ro in both? I think to got it. But when you say "as frequency gets high, the 2nd approach will be more accurate if a little more complex mathematically to solve", how should I get the frequency point where the non-idealities of opamp start to be important? It's the opamp's bandwidth the limit or the input capacitance? – thoraz Jun 06 '18 at 07:40
  • And last, why are you saying that the divider "will be more accurate if a little more complex mathematically to solve"? I think that it's not really complex to solve (maybe are different point o view). – thoraz Jun 06 '18 at 07:40
  • "a *little* more complex" is true. If you don't regard the slight complexities as a big deal that is fine. For the op-amp, if you look at its open loop gain you'll see that beyond a certain frequency it no longer has gain therefore you have problems. There is also the input capacitance of the op-amp to consider and, when measuring capacitance, that the output voltage may becomes clamped to the power rails at high frequencies. – Andy aka Jun 06 '18 at 08:45
  • Ok I got it. Input capacitances apart, the frequency features of an opamp have significantly changes beyond several MHz, so for my application (0.1 - 100K Hz) the use of an active filter should be quite useful. Thanks. – thoraz Jun 06 '18 at 09:02
  • @thoraz good luck with it. I think I'd go the op-amp route if Fmax is 100 kHz. – Andy aka Jun 06 '18 at 09:05