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What is the use of the differential RC filter at the end of this op amp (at least I'm seeing R100+R98+C28 as that)?

P3_3V_REF is a reference voltage for the 4 ADCs in the TMS320F28377D-EP MCU. An article from LT explains to me the reason for using the buffer after the reference voltage IC when providing a reference voltage to multiple ADCs from a single source. I could understand why you might want an LPF on your reference voltage to cut out any high frequency noise, but I dont see the need for it to be differential. So is there any reason for R98?

Thanks

enter image description here

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  • \$\begingroup\$ What do you mean "differential RC filter"? And R99 is 0Ohm, i.e., short? \$\endgroup\$ – V.V.T Jun 25 '20 at 6:01
  • \$\begingroup\$ Yes R99 is essentially a short, but give potential to remove/change value in the future. Differential filters are used to filter, well.. differential signals.. more on that: Stack Exchange LT Article \$\endgroup\$ – Michael Jun 25 '20 at 6:18
  • \$\begingroup\$ Removing R99 kills DC feedback, and single-supply opamp application with grounded inv input is quite.. unusual. So maybe an application with a non-zero R99 value is pertinent to your question. Wonder what will you report back, Michael(s).. if ever \$\endgroup\$ – V.V.T Jun 25 '20 at 7:44
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R100 + C28 form a low-pass filter to reduce noise in the ADC reference, as you say.

R98 is a bit more interesting. It's going to bias the output stage into sourcing a considerable amount of current (33mA) at DC and will almost certainly lower the output impedance of the amplifier over some range of frequencies. I suspect it is there to give better noise performance of the G=1 amplifier, at the expense of power consumption.

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See JRE's comment below, 100Ω resistor (R100) is used for stability and filtering.

This doesn't really answer the question, but I don't know if this design you are looking at is good for an ADC reference. ADC ref inputs are susceptible to large, high bandwidth current spikes. Having a 100Ω resistance on the output of buffer is a strange design decision. Buffers are used because they have a low output impedance and can help source transient current spikes. The 100Ω resistor will drastically affect output voltage if reference current spikes. The µC you are using has SAR ADCs which can require transient currents. The image below shows a typical ADC reference driver.

enter image description here

This design seems like it's more for a reference that is built for a low noise, no current kind of reference (input to an op-amp for example).

Where did you get your schematic from?


Image source: https://www.ti.com/lit/ug/tiduck3a/tiduck3a.pdf?ts=1593065295985&ref_url=https%253A%252F%252Fwww.google.com%252F

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  • \$\begingroup\$ You know, i was thinking the same thing. The article from LT that i referenced talks a lot about evaluating specific op-amps as buffers and the two things they stressed were to make sure they could source enough current and to make sure they had a low enough output impedance, typically below 1 ohm. I dont know why the 100 ohm is there.. 100 ohm is standard differential impedance.. I got this from a schematic one of my coworkers drew up haha. Not sure where he got it from but I'll ask and report back. \$\endgroup\$ – Michael Jun 25 '20 at 6:40
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    \$\begingroup\$ @Michael: The 100 ohm resistor is there to keep the opamp from oscillating. Operational amplifiers and large capacitive loads don't mix. "Large" is usually nanofarads and smaller, but there's a 10 microfarad capacitor there at the output - that'd make the opamp do terrible things. That 10 uF is also the reason why you don't have to worry about the spikes the ADC draws. \$\endgroup\$ – JRE Jun 25 '20 at 7:45
  • \$\begingroup\$ @JRE I'm curious, I ran AC some sims with the op-amp, and found that with an isolation resistance of just 1Ω results in a phase margin of ~60 deg. Transient sims confirmed stability. Do you know why a 100Ω is chosen over a smaller value? I guess the smaller resistor only really factors into initial cap charging which may not be a concern. \$\endgroup\$ – Michael Jun 25 '20 at 9:19
  • \$\begingroup\$ Sorry, I don't know. Taking a guess, it could be to protect the opamp output stage from high current on start up (the capacitor looks like a short circuit before it is charged.) Alternatively, it's to protect the opamp from damage when the capacitor is charged but power to the opamp is off. \$\endgroup\$ – JRE Jun 25 '20 at 9:23
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    \$\begingroup\$ Hmm. Another possibility: the resistor and the large capacitor form a low pass filter - more filtering to get rid of noise. \$\endgroup\$ – JRE Jun 25 '20 at 9:25

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