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The differential amplifier inside the OpAmp have offset current and offset voltage which is said to cause an output error when there is no input.

This makes sense to me if the output of the differential amplifier is a differential one (double-ended output) in which each output is referenced to one another. So that if one side is higher than the other, it will cause an offset voltage at output.(so output is not zero anymore)

But I don't get why that also applies to diff amp inside the OpAmp that has a single-ended output. The output of singled-ended diff amp is referenced to ground so that if one side of diff amp is in offset compared to the other side it won't matter since I'm getting the output in only one side (I think)

This is a simplified schematic diagram of the 741 OpAmp in which the diff amp is single ended:enter image description here

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The output of singled-ended diff amp is referenced to ground so that if one side of diff amp is in offset compared to the other side

That's incorrect. The output of an op-amp is: -

$$A_0\cdot(V_{IN+} - V_{IN-})$$

  • \$A_0\$ is open-loop gain
  • \$V_{IN-}\$ is inverting input voltage
  • \$V_{IN+}\$ is non-inverting input voltage

And, it is how you implement the feedback and connect the input pins (\$V_{IN-}\$ and \$V_{IN+}\$) that determines where the output voltage is referenced. In the below example the output is referenced to ground by virtue of \$R_G\$ but it needn't be - the bottom connection of \$R_G\$ can go to any reasonable voltage within the positive and negative supply rails of the op-amp: -

enter image description here

So now you can see that \$V_{OS}\$ can add or subtract from the input signal and the output error will be \$V_{OS}\$ multiplied by the closed-loop gain.

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