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In the textbook I use for teaching Instrumental Methods to Analytical Chemistry, there is a discussion about current follower circuits based upon the following figure:

enter image description here

After describing the ideal relationship between input current (ii) and output voltage (vo) they provide a "more exact relationship":

$$v_o=-R_f(i_i-i_b)\frac{A}{1+A} \tag {1}$$ $$v_0=-i_iR_f+i_bR_f\frac{v_o}{A} \tag {2}$$

Equation 1 can be found using a KVL/KCL analysis; however, it is unclear to me how one arrives at equation 2 and I am wondering if there is a typographical error in the textbook.

Additionally, one of the end of chapter problems mentions the percent relative error for a current follower circuit with a given open-loop gain, input bias current and feedback resistor if the input current is doubled. I assume that the authors are looking for a comparison between the exact output defined by equation 1 to that of the ideal output \$v_o=-i_fR_f\$ but I have not found this terminology in another source so it is difficult to confirm my assumption.

Question 1 Is there a typo in equation 2 and if so, what it is? If not, how does one get from equation 1 to equation 2?

Question 2 Is my assumption about the definition of percent relative error in the output correct? Is there another source (text/website) that supports this definition?

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  • \$\begingroup\$ I clicked on the question just to see what a strange beast a "current follower" was. What you're describing would normally be called something like a transimpedance amplfier or simply a current-to-voltage converter. Never heard it referred to as a current follower since you're not getting a current out of it. \$\endgroup\$ – pipe Nov 23 at 1:18
  • \$\begingroup\$ @pipe See Electrochemical Methods Chapter 15 or Principles of Instrumental Analysis Chapter 3. I cut my electronics teeth on these texts and wouldn't presume to know the etymology of electronics terminology from the viewpoint of chemists. \$\endgroup\$ – bobthechemist Nov 23 at 23:55
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I suspect that the second equation was supposed to be:

$$v_0=-i_iR_f+i_bR_f-\frac{v_o}{A} \tag {2}$$

In other words:

$$v_0=-R_f(i_i-i_b)+v_s$$

This emphasizes the fact that in the non-ideal case, \$v_s\$ is no longer exactly 0.

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  • \$\begingroup\$ That makes sense. \$\endgroup\$ – bobthechemist Nov 22 at 16:09
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1) Yes there is an error in (2) . A quick check of the units tells me the 2nd term is inconsistent. There is no [V²]. Just use (1)

2) percent tolerance error is done by derivatives for Sensitivity Analysis of the output w.r.t. selected input error. i.e. \$dv_o/di_b=R_f\dfrac{A}{1+A}\$ I think...

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