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I don't understand why this resistor and capacitor are parallel in a noninverting pin in Opamp. Why do we need this circuit? What is the purpose of this circuit?

enter image description here

This circuit was taken ADS8555 (pg.33)

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  • \$\begingroup\$ >>> The value of component C1 is chosen as 0.1 µF to provide a low-impedance path for noise signals that can be picked up by R1 ; the 0.1-µF capacitance value improves the EMI robustness and noise performance of the system. <<< page 34. \$\endgroup\$
    – Antonio51
    Aug 3, 2023 at 8:48
  • \$\begingroup\$ bb0667 are we done here? Can you choose an answer and accept it or, are you still not understanding something that might be helped with an additional bit of information. If so, please add a comment. \$\endgroup\$
    – Andy aka
    Aug 4, 2023 at 9:34

2 Answers 2

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Page 33 of the data sheet (implied in the question) shows the image in the OP's question but, if you read the words in the DS below that image, they refer to another diagram (figure 36) where the op-amp is explicitly named: -

enter image description here

And, if you read the datasheet for the OPA2277 you'll see two things: -

  • Ultra-Low Offset Voltage: 10 µV
  • Low Bias Current: 1-nA Maximum

Well "low" is a matter of opinion because with 1 nA flowing through (say) a 100 kΩ resistor, the input offset voltage created is 100 μV. This then becomes the bigger error when compared to the natural input offset voltage of 10 µV.

So clearly, TI are recommending the 50 kΩ resistor in the non-inverting input to balance out the effects of input bias currents for both inverting and non-inverting input pins. The 50 kΩ resistor is the same Thevenin impedance as the two 100 kΩ resistors in the inverting input side of the op-amp.

As for the capacitor, it's there to reduce the thermal noise produced by the 50 kΩ resistor. Personally I don't think it's needed for this type of application but, I see no other reason for it (the OPA2277 has only 3 pF input capacitance).

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Although an ideal op-amp will have zero input currents, a real op-amp will always have some current, called the input bias current on data sheets. For CMOS and other FET amplifiers, it can be very small, sub pA, and often possible to ignore. For bipolar op-amps it can be many nA, which can be significant in some circuits.

This bias current flowing through the gain setting resistor Rin will cause an offset voltage.

Op-amps are usually designed to make the bias current at both inputs equal. R1 is chosen to have the same value as Rin Rin in parallel with Rf (thanks to Electrocol's comment), so the voltage offsets are equal, and cancel each other out.

A virtual ground amplifier is intended to have no signal on the +ve input. Now that the +ve input doesn't go to ground, but goes through a resistor, stray capacitive coupling to that node can create a signal on it, as can resistive noise from the resistor. C1 is to short-circuit AC signals on that node to ground. It has to be large compared to the stray capacities.

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  • \$\begingroup\$ R1 should be equal to the sum of Rin AND Rf \$\endgroup\$
    – Electrocol
    Aug 3, 2023 at 8:12
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    \$\begingroup\$ @Electrocol thx, corrected, I have credited you \$\endgroup\$
    – Neil_UK
    Aug 3, 2023 at 8:17

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