Here is the data-sheet: http://www.mccdaq.com/pdfs/manuals/PCI-DAS6034-35-36.pdf

A data-acquisition board's differential amplifier measures a constant voltage across a resistor R1 shown in the Figure below:

enter image description here

Ib+ and Ib- are input bias currents and their value is +/-200pA given at the data-sheet page 30.

I'm trying to quantify the error introduced by the input offset current in terms of voltage.

I mean ideally lets say the true value of the potential difference across the R1 is V1, in that case what would be the error introduced by the input offset current in terms of V1?

  • \$\begingroup\$ I don't see the this schematics at page 30. Anyway I don't understand if current sources are needed for this purpose. You would need current source if you measure the resistance, not the loop current. Look at RCV420 current loop receiver or similar. \$\endgroup\$ – Marko Buršič Oct 22 '15 at 11:52
  • \$\begingroup\$ schematics is not from data sheet it is from my setup \$\endgroup\$ – user16307 Oct 22 '15 at 11:54
  • \$\begingroup\$ current sources Ib+ Ib- is just there to represent input bias currents \$\endgroup\$ – user16307 Oct 22 '15 at 11:54
  • \$\begingroup\$ I1 is just a current regulator which generates potential difference across R1 \$\endgroup\$ – user16307 Oct 22 '15 at 11:55
  • \$\begingroup\$ My question is about V1 the voltage across R1. Will the Ib+ and Ib- have effect on voltage reading on dataq end? How to quantify it? How quantify the error introduced by input bias currents in terms of V1? \$\endgroup\$ – user16307 Oct 22 '15 at 11:56

Input offset current is an input referred term that develops a voltage across the input resistors, and is usually calculated for a no-input condition (so a static analysis).

The input offset current is precisely what it says: the difference in input current between the two inputs. In this case, that difference in input currents will develop a voltage across the input resistance.

In your case, assuming Rdiff is very large compared with the sense resistor, you have a worst case offset current of 200pA * R1.

That yields a +/- 50nV voltage difference at the inputs of the amplifier. This will be multiplied by the gain of the amplifier to yield an output offset error.

See this article for a thorough discussion (search for offset).

Update: made notes about the effect in this circuit.

Your circuit is measuring a 4 - 20mA loop. Even at minimum current of 4mA through the sense resistor, the error due to input offset current is -146dB (or 0.05 parts per million, if you prefer). With such a small error relative to the current you intend to sense, it is a non-issue here.

You are doing the right thing, though, as input offset voltage, input offset current and input bias current are all sources of errors and they can be a cause of difficulty in high gain circuits.

| improve this answer | |
  • \$\begingroup\$ yes but how can I quantify this? I couldnt see the gain the data sheet. \$\endgroup\$ – user16307 Oct 22 '15 at 12:07
  • \$\begingroup\$ @user16307 It doesn't matter what the gain is - there is potentially a 200pA error in the 4-20mA signal - relative to full scale (20mA) this is an error of 0.01 parts per million. \$\endgroup\$ – Andy aka Oct 22 '15 at 12:09
  • \$\begingroup\$ btw will the error be input bias current or input offset current? \$\endgroup\$ – user16307 Oct 22 '15 at 12:11
  • \$\begingroup\$ can this be (+/- 50nV voltage difference at the inputs of the amplifier) quantified without knowing the gain? i would like to see how it contributes to the offset voltage \$\endgroup\$ – user16307 Oct 22 '15 at 12:13
  • \$\begingroup\$ But wait, since this is a daq input channel. Isnt gain supposed to be 1? \$\endgroup\$ – user16307 Oct 22 '15 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.