Input Bias Current in Op-Amps (Amplifiers #9)

What is the input bias current and how do we fix it? Ideally an op-amp won't have any current flowing into its input pins because the input impedance is ideally very large, but in reality there is current. What does this affect?
Aaron Danner is a professor in the Department of Electrical and Computer Engineering at the National University of Singapore.
danner.group
Video filmed and edited by Cheryl Lim.
@randomcheryl

[@Enigma758]

Thank you for this video Aaron, I was recently wanting to learn more about this phenomenon and then your video suddenly appeared in my feed.

[@61bhushandeshmukh78]

Sir you are best !!!!!

[@comeflywithme1694]

Very useful video! Please keep up. thank you.

[@kaleoride]

Perfect, thank you!

[@amarmail8033]

Superb.

[@drtidrow]

Well, the input bias current depends on the type of op-amp - BJT-based op-amps will have some bias current, JFET-input op-amps will have a lot less (reverse-biased diode leakage current), and MOSFET-input op-amps have essentially none (pico to femtoamp range) unless they have protection diodes on the inputs. That said, even FET-input op-amps will have some transient currents in the inputs as the gate capacitances charge or discharge.

[@instytutfotonowy2637]

In the video on Input Offset Voltage you assumed no Input Bias Current, in this video you assume no Input Offset Voltage. How convenient ... 😄 In real life they are both present and influence the output voltage. A video that would analyze them both would be cool! Great job anyway!

[@stefano.a]

In general, is it true that the resistance seen from the inverting input of the op. amp. has to be equal to the resistance seen from the non inverting input, to reduce the effect of the bias currents?

[@adanner]

Unfortunately they will rarely be exactly equal. The input bias is mainly a problem affecting op-amps using BJTs where each input leads to the base of a transistor. But due to nonlinearity in the transistor curves, the effective input impedance has a bit of dependence on the current, making the input bias variable. But since the transistors are usually fabricated near one another on the same chip, it's still a reasonable approximation to assume they're just equal and constant.

[@stefano.a]

@@adanner thank you for the answer; sorry, I meant the resistance from the inverting (or non inverting) input *and* ground, not the resistance of the op. amp. input but the one created by the external resistor network around it. In the example of the video the resistance seen from inverting input is the parallel of R1 and R2 and selecting for the value of the “balancing” resistor, connected to the non inverting input, a resistance of value R1//R2, the base current (IB) effect can be minimized. I wonder if this is valid in general