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I'm analyzing this circuit to condition a PT100.

enter image description here

I clearly understand what the circuit does, but the current flowing through R7 is very small and negligible.

So in theory it could be any value (generally from 1k to 100k), but I don't understand why the value should be 49.9k.

There is no information on why the value of R7 where said circuit is described.

Any suggestion or comment is welcome.

The circuit is described here: Precision Temperature Sensing with RTD Circuits

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    \$\begingroup\$ It is there because of something called the amplifier bias current. Note that it is about equal to the parallel of those two \$100\:\text{k}\Omega\$ resistors? There's a reason for that. If curious, take a look at a typical bipolar opamp input stage (diff-amp, current mirror, currrent source/sink, and a few bits and pieces around that area.) \$\endgroup\$
    – jonk
    Commented May 9, 2021 at 23:39
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    \$\begingroup\$ So 100kΩ || 100kΩ and there is no 50kΩ, so not strange at all. And I though I saw a cosmic black hole pass through here. \$\endgroup\$ Commented May 9, 2021 at 23:56
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    \$\begingroup\$ Yes, you're right, I remember I learned that when I studied op-amps, but had forgotten. Thanks @jonk \$\endgroup\$ Commented May 10, 2021 at 1:04
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    \$\begingroup\$ @FabiánRomo I'm so glad to hear that all this settled back in, after some years! It's wonderful when just a reminder or two is all that's needed to get things back into place! And it's a good reason that justifies this site and the time that people put into it, too! \$\endgroup\$
    – jonk
    Commented May 10, 2021 at 2:12

1 Answer 1

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In theory it’s to balance out offset due to bias current so the resistance is equal to the parallel combination of R5 and R6. 49.9K is the closest value to 50K in the E96 series that many 1% resistors are manufactured in.

In practice, the voltage drop due to the entire bias current is less than 4uV at up to 85°C for the particular op-amp chosen so it won’t make much difference, given the offset voltage and offset voltage drift of that part.

A secondary reason to have a relatively high value resistance there is to protect the op-amp input against transients. With 50K, the 2mA maximum input current would allow a +/-100V transient safely.

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  • \$\begingroup\$ Thank you very much for the information, I learned it when I studied op amps, but had forgotten. \$\endgroup\$ Commented May 10, 2021 at 1:06
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    \$\begingroup\$ @FabiánRomo IMO you can mount a resistor with a value close to 50k that doesn't need to be accurate as the R5,R6, for example 47k or 56k should be good enough. \$\endgroup\$ Commented May 10, 2021 at 8:34
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    \$\begingroup\$ I agree with @MarkoBuršič and was thinking of adding a similar comment, however using a cheap carbon film resistor (for example) might add more thermal EMFs than using similar construction metal film precision resistors for all positions. The exact value is not important in this case, in fact at room temperature the typical bias current is the same as the typical offset current so there is no improvement. Typically at higher temperatures the leakage in the protection networks (I assume) starts to show up so more matched resistors are beneficial. In any case, 1% resistors are cheap these days. \$\endgroup\$ Commented May 10, 2021 at 13:31

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