Non-Ideal Op Amps: Input Bias Current

Input bias current is an unavoidable feature of op amps that arises because the input impedance of op amps is actually not infinite and there is some current that must flow into the inputs when the op amps are operating.
This video describes why input bias arises, the effects that it has on the circuit, and ways to mitigate the effects of input bias currents.
Input Offset Voltage Video: • Non-Ideal Op Amps: Inp...
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LM741 Schematic: Daniel Braun, CC BY-SA 3.0 via Wikimedia Commons

[@Charlie-mf3ly]

Tks you share, I suffer IB current measurements issue now

[@jacobs7281]

Thanks for this video. @4.30 , Will you share a link or add an explanation about purpose of 2nd stage opamp(open loop)

[@jacobs7281]

Never mind ..you did say watch your other video..thank you

[@corruptofficial7794]

Hi there, i guess it's a buffer , that is a current amplifier

[@MotoG-ub3lw]

Thank You for Your explanation of this topic. 1) I was wondering what is the purpose of the second amplifier (that one in the feedback loop of D.U.T) and I think it is used as an inverting amplifier to change the feedback type from the positive feedback into the negative feedback. The feedback loop of the second amplifier is probably responsible for ensuring a stability. 2) I am curious if the measurement precision of this circuit is dependent on (or independent from) the second amplifier's input offset voltage. Presumably because it is placed inside the feedback loop of D.U.T. amplifier, its DC performance has no impact and even an amplifier with poor DC performance could be used (maybe the only requirement is that the open loop gain of the D.U.T is enough - I am not sure). If that's the case, this circuit is beautifully designed, because it is not dependent on the 2nd amplifier input offset voltage. What do You think about it? 3) I derived the exact formula for the V_out of this input bias current measurement circuit for the 3 different cases presented. After neglecting some of the terms which are close to zero, I got almost identical expressions. Almost. The difference is in the sign of the terms -(1+R2/R1)*V_io and +(1+R2/R1)*I_bp*Rs or -(1+R2/R1)*I_bn*Rs. The most important difference is that the V_out has always a minus sign before it. I put V_io before the noninverting input of the ideal opamp (in series with the noninverting input). Lets begin with the first case, when S1 and S2 switches are closed. From obtained equation, it is obvious that R1 should be small to neglect the term R1*I_bn (the negative terminal bias current), so that is why R1 is selected to be only 100 ohms. We can also neglect the term I_offset*R2 - although R2 is rather a large value resistance, the I_offset is very small, smaller than the I_bn. So, in the first case, when switches S1 and S2 are both closed, we end up with formula -(1 + R2/R1)*V_io (V_io is the input offset voltage). And this does not agree with formula presented in the video in terms of sign (negative output voltage). In the second case, when S1 is closed and S2 is opened, the sign of the term connected with V_io is opposite to that connected with I_bp (positive terminal input bias current). In the third case, when S1 is opened and S2 is closed, the signs of both terms: the term connected with V_io and the term connected with I_bn (the negative terminal input bias current) are the same. And that is different from equations presented in the video. Why is it so?

[@thepresident2781]

Do you think a dual supply is necessity in case of rail-to-rail op-amps, if so what about the non rail to rail ones ?

[@abheetl9743]

2:15 If there is current going into the non-inverting port, it must come out from the output as well, right? And if it does wouldn't it cause a potential drop in Rf and Ri?

[@electronevice]

the current into the non-inverting port (the "+" port) flows into the op-amp and out through one or both of its power rails, which flow through their voltage sources back to ground. It doesn't flow out the output

[@fauzannurhadi4704]

I made a difference amplifier to measure 4 lion battery 18650 using quad op amp Lm324 as we know the resistor of circuit must be same value to get difference output from 2 input signal. I use 1 Meg resistor in my circuit. i using single supply for the op amp from batt series terminal about max 16.8 v when full capacity. when i measured using voltmeter the ouput is work fine. But When i want to measure using adc of arduino their is small current flow, indicate with internal led of arduino little glow up. I confused how to block a current so i can measure using adc. Now i just solve the problem with put 1500 ohm resistor between gnd from the op amp and gnd arduino pin. Can you help me to solve my problems sir. Thanks in advanced

[@thepresident2781]

16.8V of output voltage is well above the tolerable limit of ADC (which is only 5v), i am assuming you did't connect directly to the adc and use a resistor divider at the output of opamp. Since the adc type in arduino is SAR it draws a little bit more current than sigma deltas. I think you can divide the output to the compatible range of adcs input specifications and add an extra unity gain amp in between adc and differential measurements output just to buffer voltge. Adding a resistor to ground connection shifts the reference baseline and your output voltage will vary with varying current demands of op-amps which is not a good idea if you want precision measurements.

[@lucascheater9536]

First comment!

[@ElectronXLab]

Yes you are!

[@aminnikpay5172]

2nd is better