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enter image description here

As seen in the figure above, a current regulator is providing a constant current looping through the series resistor Rshunt. Input impedance of data acquisition device is 10 GigaOhm. Can we say that almost no current passes through the probe resistance Rcable(since input impedance is huge) and therefore there is no voltage drop across the probe cables? How can I relate this resistance to input impedance of DAQ device and can I just neglect the probe resistance?

Can I quantify the error caused by Rcable here in relation with DAQ inout impedance?

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An equivalent circuit could be this one

enter image description here

Can we say that almost no current passes through the probe resistance Rcable(since input impedance is huge) and therefore there is no voltage drop across the probe cables?

Absolutely yes

How can I relate this resistance to input impedance of DAQ device and can I just neglect the probe resistance?

In your drawing, simply replace the DAC amplifier with resistor of 10 GOhms. Since the resistance of the series of R_cable and 10 GOhms is much bigger than R_shunt you can say that the error introduced by R_cable and the DAQ is very small.

Can I quantify the error caused by Rcable here in relation with DAQ inpout impedance?

Solve the circuit: the current in the R_DAQ branch is

I_DAQ = I * (R_SHUNT) / (R_CABLE + R_DAQ + R_SHUNT)

Thus the voltage read by the DAQ is

V_DAQ = I_DAQ * R_DAQ = I * [(R_SHUNT) / (R_CABLE + R_DAQ + R_SHUNT)] * R_DAQ

which can be rewritten as

V_DAQ = I * (R_SHUNT) / [ 1 + (R_CABLE + R_SHUNT) / R_DAQ]

Ideally you would expect to read V_DAQ = I * R_SHUNT but actually you get I * (R_SHUNT) / [ 1 + (R_CABLE + R_SHUNT) / R_DAQ]. The relative error then is given by (R_CABLE + R_SHUNT) / R_DAQ.

As long as R_DAQ is much greater than R_CABLE + R_SHUNT you can safely say that the DAQ does not affect the shunt.

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@Lelesquiz has given you an answer that is applicable to low frequency changes in the signal, however to answer the question about current you really need to know the bias current of the amplifier, which (when modeled) appears as a constant current source at the ideal amplifier input.

The bias current can be much larger than the input voltage divided by the input resistance. For example a precision op-amp such as an LT1013 might have an input bias current of 30nA but an input resistance of several G ohms. That will cause an error 30nA in the reading, even at 0mA from the transducer, which might be significant (or not).

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  • \$\begingroup\$ Do you mean that the error introduced by the "bias current" might be more significant than the error introduced by the probe and shunt-resistor resistances? Can I find this bias current info in data-sheets? \$\endgroup\$
    – user16307
    Commented Jun 28, 2015 at 12:21
  • \$\begingroup\$ @user16307 Yes, usually under that name. It could be more significant, yes. At zero current in it has to be more significant since the other terms disappear. \$\endgroup\$ Commented Jun 28, 2015 at 14:35
  • \$\begingroup\$ there is a fixed value for the bias current in the data-sheet. should I subtract this from the current I measure? \$\endgroup\$
    – user16307
    Commented Jun 29, 2015 at 11:19
  • \$\begingroup\$ The bias current will change with temperature, but if you have a 'typical' number (not a maximum!) correcting for it by adding or subtracting (depending on whether it flows in or out) will give you a bit more accurate reading. \$\endgroup\$ Commented Jun 29, 2015 at 12:16
  • \$\begingroup\$ one more thing from your nice explanations. i just wondered as a thought experiment, if the shunt resistor value is extremely high(as high as almost the daq input impedance) than do you think the "bias current" still would be as in the data-sheet? there must be a maximum shunt resistor value to make sense of data-sheet? \$\endgroup\$
    – user16307
    Commented Jun 29, 2015 at 12:21

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