Consider the op-amp integrator circuit below. Calculating its frequency response the result is the following:

$$ \underline{H}(j\omega)=\frac{\underline{V_a}(\omega)}{\underline{V_e}(\omega)}=\frac{-1/j\omega C || R_g}{R_1}=-\frac{R_g}{R_1+j\omega R_1R_gC} $$

If I now assume a non-ideal op-amp and non-negligible input bias current, does that change the transfer function in any way? And why/how?


simulate this circuit – Schematic created using CircuitLab

EDIT: A small signal model approximation


simulate this circuit


1 Answer 1


Yes, it will. The most basic op-amp model you are using presumes the inputs to have infinite impedance, drawing no current, and therefore allows you to ignore it from the transfer function.

To state that the op-amp model draws current means that there is a non-infinite input impedance. For the integrator op-amp circuit you provide this could be represented by a schematic as such:enter image description here

EDIT: for posterity I will leave my previous response. I have updated the image to better represent what I was trying to explain, that the input bias current could be modelled (in this specific case) by an input impedance.

However, if you are presuming that the input bias current is purely a DC effect then it can be ignored in the transfer function. This won't hold under all assumptions, but as noted in the comments, other effects usually dominate the dynamic behaviour of an op-amp, predominantly the internal low-pass. For example, see here for a basic op-amp model.

  • \$\begingroup\$ What does that mean for the transfer function? From my point of view \$ I_{B+} \$ (input bias current on the positive terminal) does not show any effect, since the current will flow directly to ground. This would be consistent with your image, since \$ I_{B+} \$ is connected "below" Rin. However, \$ I_{B-} \$ would cause a voltage drop on R1, correct? But how does the above formula for the transfer function change? \$\endgroup\$
    – Daiz
    Commented Feb 21, 2017 at 14:35
  • \$\begingroup\$ The input bias current is just that a bias current meaning it is (mostly) DC. Therefore it does not influence the input impedance as input impedance is an AC parameter. As long as R1 and Rg have a significantly lower value than the Zin of the opamp, you can still ignore the Zin of the opamp, irrespective of input bias current ! \$\endgroup\$ Commented Feb 21, 2017 at 14:41
  • \$\begingroup\$ does that change the transfer function in any way In my opinion a better answer would be: Yes, in theory it does, in practice it often does not especially if you keep the value of R1 and Rg low enough. And when using an opamp with some input bias current you cannot use very high values for these resistors as that would introduce DC offset. \$\endgroup\$ Commented Feb 21, 2017 at 14:47
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    \$\begingroup\$ Then you need to draw the small signal equivalent circuit, a bit like the circuit above but add a voltage-controlled-voltage-source for the opamp. You could model the input bias (DC) current as DC current sources to ground, since those as DC they will be eliminated in the small signal circuit. You would need to know the small signal input impedance of the opamp. The Zin between the inputs but also between any input and ground (the common mode input resistance). Then calculate the transfer of the circuit. Yes, it's some work ;-) \$\endgroup\$ Commented Feb 21, 2017 at 15:07
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    \$\begingroup\$ I like to mention that thelast expression for the transfer function is NOT correct. For DC (w=0) the result must be (-Rg/R1). More than that, there are more real opamp parameters (parasitics) which have more influence on the real behaviour than a finite input impedance. \$\endgroup\$
    – LvW
    Commented Feb 21, 2017 at 16:01

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