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enter image description here There is an integrated circuit(xtr111 datasheet) with a 0 to 5 volt input range (The 0-5 volt signal is applied as DC. That is, the signal does not change constantly and quickly.)and the pin has high impedance input. In the sample diagrams (Reference design ), a 10k resistor is connected serial in front of the pin, . Is there a reason for this?

My thoughts are that it does not related about current limiting because of the input has high input impedance. ı think its related about Filtering and Noise Reduction:

Together with any capacitance at the input, the resistor forms a low-pass filter that can help reduce high-frequency noise and interference from reaching the sensitive input of the IC. Could this be due to the parasitic capacitance on the PCB providing a low pass filter and input stability? am ı wrong?

Signal is VOUT pin come from dac5311

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That's impossible to tell without reading the actual datasheet of both ICs.

However, I've seen such resistors being used out of "I always do that!", but also (and now sensibly) when there's the chance that one chip might already be pulling that line high, while the other has still a low supply voltage. In that case, the voltage on that pin would be higher than the supply voltage, leading to the protection diodes inside being conducting. And these can and will burn if the current flowing into the pin, through the protection diode, out of the supply pin, is higher than what's listed under "absolute maximum ratings" in the data sheet. A resistor limits the current that can flow in that situation, and its effects might be negligible during regular operation.

TL;DR probably protection against high current flowing when right IC is powered, but left IC is not yet.

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    \$\begingroup\$ The XTR111 datasheet contains the following for the VIN pin Place a small resistor in series with the input to limit the current into the protection if voltage can be present without the XTR111 being powered., which agrees with this answer. \$\endgroup\$ Commented Dec 28, 2023 at 8:57
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    \$\begingroup\$ To cite the famous electrical engineer H. Smith: "I love it when an answer comes together" \$\endgroup\$ Commented Dec 28, 2023 at 9:00
  • \$\begingroup\$ If a voltage exceeds the voltage limited by the clamp diodes, current will flow through the clamp diodes and we add a resistor to limit the current to prevent damage to the clamp diodes. I understand. "Place a small resistor" "Consider a resistor value equal to RSET for bias current cancellation." Did it mention here that we should put a resistor equal to or approximately equal to the Rset resistor on the Vin pin? (Maybe I don't understand because of my bad English) In the reference design, the rset resistor is 1.47k ohm and the resistor connected to the vine is 10k ohm, am I wrong? \$\endgroup\$
    – Electronx
    Commented Dec 28, 2023 at 9:16
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It helps stabilize the output of the DAC - it doesn't like capacitive loads. The resistor decouples the capacitive loading from the "VIN" pin on the chip on the right. Same trick is done with op-amps driving capacitive loads. Without it, you will see ringing when the DAC output changes voltage state.

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    \$\begingroup\$ DAC data sheet says 470 pF is OK; unlikely that any IC input would have higher capacitance. \$\endgroup\$ Commented Dec 28, 2023 at 9:02
  • \$\begingroup\$ @ConstantineA.B. No an input would not have 470pF, for example an ADC input might be in the 10 to 20 pF range, but the cap is switched to the pin for taking a sample, so if you are measuring at 10 kHz, the output needs to handle charging that capacitor being switched in to charged it and then disconnected during conversion. \$\endgroup\$
    – Justme
    Commented Dec 28, 2023 at 9:07

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