How to reduce the voltage dropped in a mosfet used as a switch?
Let's analyze this. Assuming the MOSFET is conducting and the LED is getting, say, 10mA with voltage drop of around 1.5V, the voltage on the Source pin will be 1.5V+220*10mA = 3.7V. While the gate voltage is 3.3V. So your Vgs is -0.4V. Which, of course, will never make the MOSFET conducting. Contradiction. To overcome this you need to make sure Vgs is sufficient by making it known and constant by connecting source to the ground and moving the load (resistor and LED) in series with the drain.