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Why doesn't a resistor on the non-inverting wire of the op amp affect dc offset (non-inverting op amp)? And does it have any effect on the dc offset of the inverting amplifier.

The reason I ask is because I tested this by simulating a non-inverting amplifier and the result was: no effect on the dc offset.

But I expected change in the dc offset as we model the dc offset source voltage as a dc voltage source attached to the positive terminal of the op amp. And so if there is a resister attached on the same wire than naturally some voltage should drop on that resistor. And thus the voltage see by the positive terminal of the op amp should change.

Why is this conclusion not correct and also explain the "no-change" in the dc offset. ty.

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  • \$\begingroup\$ non-inverting op-amp? No such thing. Maybe you want to show us a schematic or three... \$\endgroup\$
    – Trevor_G
    May 9, 2017 at 19:43
  • \$\begingroup\$ I meant non-inverting amplifier \$\endgroup\$ May 9, 2017 at 19:46
  • \$\begingroup\$ Briefly, because op-amps have very high input impedance \$\endgroup\$
    – user28910
    May 9, 2017 at 19:54
  • \$\begingroup\$ What is the input bias current spec of your op-amp? \$\endgroup\$
    – The Photon
    May 9, 2017 at 21:11
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    \$\begingroup\$ Please add a schematic to your question. There is a built-in schematic editor. First click edit to edit your question. Then, click the icon on top that looks like schematic symbols. This will bring up a schematic editor which you can use to add a schematic. \$\endgroup\$
    – user57037
    May 10, 2017 at 3:30

2 Answers 2

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If the source impedance(several kilos) of the dc offset voltage source is small compared with the load impedance of the non-inverting input(several megas) of the op amp, there will be no loading effect of the offset voltage. Typically this is true for your circuit, you can try increase the source impedance of your dc offset source, see if dc offset lowers.

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In bipolar opamps, that cancellation resistor is needed for the best accuracy. The offset currents were low microamps or high nanoamps.

Most modern OpAmps are CMOS, with room-temperature input current ----- leakage in ESD diodes and across the high-quality gate oxide of FETS wired as differential amplifers ------ of pico amps.

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    \$\begingroup\$ The unique '------' punctuation in your posts indicate some kind of stream of consciousness but don't help improve the clarity of your answers. I recommend standard English punctuation in full sentences with subjects, verbs and objects. \$\endgroup\$
    – Transistor
    Jul 29, 2017 at 11:32

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