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I have a unity gain differential op amp, using a grounded dual supply, with both inputs AC coupled.

Does the inverting input need a dedicated path to ground or can the DC bias current find its way to ground via the power supply?

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I've searched around and haven't been able to find much regarding this particular configuration. Most of the information I've found talks about single supplies and/or op amps with only one input AC coupled.

If it matters, the op-amp I'll be using is a TL07X.

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  • \$\begingroup\$ Quick answer, No, the extra resistor isn't strictly needed. I do wonder if you've considered an IC that has all the precision resistors built in (THAT124x, INA134, SSM214x). Your discrete resistors will require extra-low tolerance (pricey), if you want good CMRR. \$\endgroup\$
    – Rich S
    Dec 17, 2021 at 23:29
  • \$\begingroup\$ The +input has a 47k resistor to ground and the - input has a 47k resistor to the opamp output that is 0VDC. Then both inputs have a 47k resistor to 0V. \$\endgroup\$
    – Audioguru
    Dec 26, 2021 at 17:05

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Does the inverting input need a dedicated path to ground or can the DC bias current find its way to ground via the power supply?

If the bias current is (say) 10 nA then that flows through R3 and creates a DC error at the input to the device of 470 μV. That will also appear at the output because the DC gain of your circuit is unity due to using input capacitors.

However, because you have a net resistance of 47 kΩ in the non-inverting input, the effect of this bias-induced offset voltage is much reduced (see offset current below).

You should also ideally consider the offset current (usually about ten times lower on this type of op-amp than bias current). This adds another 10% worst case to the numbers above.

But, those offsets all pale to insignificance given the typical input offset voltage (3 mV) that a TL07x produces.

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    \$\begingroup\$ IMO the resistance in the non-inverting input is R4 = 47 k. And since R3 = R4, there is a full cancellation of the voltage drops caused by the two input bias currents (if they are equal). \$\endgroup\$ Dec 17, 2021 at 16:55
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    \$\begingroup\$ Well, I must thank you for that correction @Circuitfantasist \$\endgroup\$
    – Andy aka
    Dec 17, 2021 at 17:19
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You're fine, the inverting input has a path for bias current through the feedback resistor, and of course the non-inverting input has the resistor to ground.

Both inputs have a dc path to ground, which is the important consideration.

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The best way to understand and explain circuits in an intuitive way is to show where currents flow (see my Codidact paper 1) and what voltages are in circuits (my Codidact paper 2).

Here the OP is interested in where the input bias currents flow. That is why I have shown in my Fig. 1 only their paths on a simplified conceptual circuit diagram of a part of an internal op-amp structure.

Input bias currents - paths

Fig. 1. A conceptual op-amp circuit diagram with input bias currents visualized.

It can be supplemented with (thicker) loops representing the collector currents flowing in the input and output stage (see my Fig. 2).

Input bias currents - all paths

Fig. 2. A conceptual op-amp circuit diagram with all currents visualized.

It is interesting to see that the input bias current I- entering the inverting op-amp input is provided by both the positive power supply V+ and negative V- (connected in series) while the input bias current I+ entering the non-inverting input is provided only by the negative power supply V-.

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