Non-Ideal Op Amps: Input Bias Current 

Tks you share, I suffer IB current measurements issue now

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

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?

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?

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

First comment!

2nd is better