Electronic – Switch Configuration for Variable Gain Photodiode Amplifier

amplifierphotodiode

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.

Best Answer

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.