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I'm confused about the source of Input Bias Current, is it generated by op amp, where the current "leak" from the op amp. or it's a must to operate the op amp? for example if I have in a data sheet a 1pA input bias current, should I provide a 1pA to the inputs of op amp? I know that the bipolar op amp needs a current to operate, but what about FET and CMOS op amp?

In short question: Should I build circuit to "provide" this current, or it's just given by the op amp itself?

Also, what is the difference between input bias current and the leakage current?

Note: if the type of op amp make a difference, I'm interested in LTC6078/LTC6079

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  • \$\begingroup\$ It is just because no real amplifier has an infinite input impedance. In this case, 1pA may flow into or out of the inputs as a non-ideal or parasitic property, which can be a problem if you have a very high impedance source. Leakage and bias currents have different physical origins, but they can be considered to have the same effect as far as the opamp user is concerned. I think this question has been addressed well enough on this site already so I won't say any more than this--please see the related questions in the sidebar. \$\endgroup\$ Commented Jan 30, 2016 at 16:23
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    \$\begingroup\$ Short answer: the direction of the bias current is not defined (usually). The opamp may sink or source it, and this may change over the operating range (e.g., with respect to temperature). You cannot know which will be the case and must be prepared for both possibilities. \$\endgroup\$ Commented Jan 30, 2016 at 16:36
  • \$\begingroup\$ @OleksandrR., so I must take care to not making a voltage error(voltage drops due to this current), and the op amp will take care about this current? \$\endgroup\$
    – hbak
    Commented Jan 30, 2016 at 16:38
  • \$\begingroup\$ @hbak Exactly, yes. \$\endgroup\$ Commented Jan 30, 2016 at 16:38

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Input bias can flow both into or out of an op-amp's inputs, and in certain op-amp topologies may do either. The consequence is that you need to design your circuit accounting for the current, i.e. if your source impedance is 100kΩ and you've got 1pA of input bias current, a non-inverting buffer will have a voltage error of (100kΩ * 1pA) = 100nV.

As you can see, input bias current has a negligible effect for low impedance sources, and can have a very significant effect for large impedance sources. This is also why you cannot capacitively couple op-amp inputs and expect them to work; the inputs must sink or source this tiny current for the input topology to work.

Edit: To more directly answer your question, input bias current is not something you need to "provide" explicitly, rather, consider your circuit as sourcing (or sinking) this current. In the process, an error term will be generated. Another way to think of this is that the op-amp has a finite input impedance, and therefore your circuit needs to account for that. If your source has a very high impedance and the op-amp input impedance is too low (which is the same as the op-amp having a relatively large input bias current), then the voltage divider formed by the source impedance and the op-amp input impedance will be further away from 1, which results in a voltage error.

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  • \$\begingroup\$ Also, Olin has explained this better than I can here. \$\endgroup\$
    – uint128_t
    Commented Jan 30, 2016 at 16:25
  • \$\begingroup\$ I saw this question yesterday, But Please check again the edit. My question is simply: Should I build circuit to "provide" this current, or it's just given by the op amp itself? \$\endgroup\$
    – hbak
    Commented Jan 30, 2016 at 16:33
  • \$\begingroup\$ I edited my answer to more directly answer your question. Hope that's more clear. \$\endgroup\$
    – uint128_t
    Commented Jan 30, 2016 at 16:38
  • \$\begingroup\$ Why I should provide a path for this current? \$\endgroup\$
    – hbak
    Commented Jan 30, 2016 at 16:56
  • \$\begingroup\$ Because if you don't, your op-amp won't work the way you expect. \$\endgroup\$
    – uint128_t
    Commented Jan 30, 2016 at 17:10

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