Electronic – arduino – opt101 (photodiode) adjust voltage output

You can use a differential amplifier with the negative input V₁ connected to a fixed voltage (12V or 5V) and V₂ connected to the output of the circuit.

For this to work you need an opamp that can go rail-to-rail (not a 741).

If I were you I'd stick with the voltage divider, though, unless there's a definite need for the extra bit of precision.

As this is only your second time working with electronics, I'll try to keep my terminology simple.

It's hard to see exactly how you have wired up the sensor and LED, but I can take a guess. (If I'm wrong, then everything below probably makes no sense).

The sensor is connected between + power and the Arduino input, while the LED is connected between - power and the Arduino input.

When you press the sensor, electrical current can flow from the + side of the power the Arduino sensor pin, charging it up and giving it a high voltage. Current also flows through the LED, causing it to light up.

Now, what happens When you release the sensor? The electrical charge inside the Arduino sensor pin which was giving it a high voltage, will now flow as current through the LED to - power, bringing the voltage down, so that the Arduino sees you've let go.

But what happens if you don't have an LED in there? The electrical charge in the Arduino sensor pin has nowhere to go, and so it just stays there, and the voltage doesn't change.

The reason the Arduino's sensor pin behaves like this is because it behaves like a tiny capacitor. It can store a small amount of electrical charge, and thus 'remember' the voltage that was placed on them by the sensor.

So, how can you fix it? You'll need to have somewhere for this charge to flow. If not an LED, then a resistor should do. Any value between 1k and 1000k will probably work fine.