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In this article here:

https://www.analog.com/en/technical-articles/optimizing-precision-photodiode-sensor-circuit-design.html#

It states:

enter image description here

However, I don't understand the rather critical line "instead of looking at the voltage on the output of the amplifier, switch, Switch S2 connectsthe output of the circuit directly to the gain resistor. This eliminates any gain errors due to current flowing through Switch S1."

What exactly is it trying to say here? Because I don't understand what it's trying to do with that second switch. The same resistance is still in the feedback loop, and the same currents flowing in the feedback loop still flow through the switch. All the second switch seems to do is isolate the unused feedback path even more, but that doesn't seem to be the intention.

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  • \$\begingroup\$ Also, if you do use CMOS switches look into charge injection. (Unequal amounts of positive and negative charge injection from mismatching in the switch.) The switches used on the ACF-2101 are pretty good. Look at the specs. Then see if you can find other switches as good or better. \$\endgroup\$
    – jonk
    Apr 3, 2021 at 5:00
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    \$\begingroup\$ @jonk egads. $60 a chip. They better be good. \$\endgroup\$
    – DKNguyen
    Apr 3, 2021 at 5:10
  • \$\begingroup\$ They are very good. I used them for femtoamp currents. Damned things are so good in fact that my first experiments (running over a few weeks of collecting data) showed sudden events that were, at first blush, inexplicable. But I quickly considered the idea of cosmic rays and subsequent particles. I was able to work out what was expected at my location (elevation was most important) and what to expect given the PD area and thickness. I rotated the die and observed the right changes. I then brought in different radioactive sources and put the problem to bed. \$\endgroup\$
    – jonk
    Apr 3, 2021 at 5:20
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    \$\begingroup\$ Isn't charge injection only really a problem when you are switching frequently like during chopping? \$\endgroup\$
    – DKNguyen
    Apr 3, 2021 at 5:24
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    \$\begingroup\$ Ah! Found an Analog Devices page on charge injection. Maybe there are more of them. But that's one, anyway. \$\endgroup\$
    – jonk
    Apr 3, 2021 at 5:31

1 Answer 1

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Imagine a series resistance Rs in S1 normally closed contact. The current If flowing through the feedback resistor Rf1 will result in the voltage at the op-amp output being higher than the ideal voltage by Rs*If.

But the voltage at the right hand side of Rf1 will be unaffected (because it's inside the feedback loop).

If you add the second switch you can pick off that voltage and provided there is negligible current through the resistance of S2 (to the load) you will have eliminated all the error caused by the switch resistance.

For illustration, consider the below example where you have two switches with different resistances in each position.

schematic

simulate this circuit – Schematic created using CircuitLab

This dual-switch configuration is frequently useful when you have high resistance switches such as CMOS analog switches.

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  • \$\begingroup\$ What do you mean by "pick off" that voltage? It the current flowing through R5 and R6 are negligible I don't see how that affects the voltage output. \$\endgroup\$
    – DKNguyen
    Apr 3, 2021 at 4:30
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    \$\begingroup\$ 'Pick off' as in select it without significantly influencing it. Not very technical, I guess. \$\endgroup\$ Apr 3, 2021 at 4:32
  • \$\begingroup\$ Oh, I think I get it....it's pretty subtle. Are you saying that the node you are taking as the voltage output is not directly on the output of the opamp where the switch resistance is affecting the gain, but "somewhere within the control loop" where the switch resistance is unseen? \$\endgroup\$
    – DKNguyen
    Apr 3, 2021 at 4:34
  • \$\begingroup\$ Yes, exactly... \$\endgroup\$ Apr 3, 2021 at 4:35
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    \$\begingroup\$ Yes. There's no way to select that voltage at R3/R1 junction vs. R2/R4 junction without a second switch. \$\endgroup\$ Apr 3, 2021 at 4:39

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