How to connect LED to output device to show the output state without any voltage drop so that it can be sensed by MCU?
From the NE555 datasheet:



we can see that with a 5 V supply we can expect around 3.3 V at the output when loaded with 100 mA. Your LED of course consumes less current but that will not increase the voltage by much. So the behavior you see is to be expected.

I expect that even without the LED the voltage at the output will not reach 5 V! That's a consequence of the design of the output stage in the NE555 chip.

You might not actually need a proper "5 V" one signal, when a uC is running on a 5 V supply the actual decision point is at 5 V /2 = 2.5 V, so 3 V might be enough. That isn't a robust solution though, I would only rely on this for a prototype or hobby project, not some device which will be mass produced.

Solution 1:

Remove the LED (use a transistor to switch the LED if you still need the LED) and add a pull up resistor (try 1 kohm) to the NE555's output. That might raise the voltage enough.

Solution 2:

Use the CMOS version of the NE555, it is the ICM7555 that chip can pull its output close to the 5 V supply rail provided you do not draw much current from it. You will need to run the LED at less than 1 mA for that, with a modern LED, that will still be enough to see it light up.

If you really need more current through the LED, use an NPN transistor (BC547, 2N2222 for example) or N-channel MOSFET (2N7000 for example) to switch on/off the LED.

Solution 3:

Use a transistor to level shift the voltage, that will invert the signal though.