Electronic – Voltage across transistor

You have a few problems. First, you say the LEDs are 20mA standard? Yet you list 15Ω resistors. That's ((5 - 3.6v) / 15Ω) = 93mA EACH. They won't last long.

Second, transistors are not just on/off. That's only if you use them as a switch, in saturation mode. They can and are designed to have a wide range of use.

Third, when SW1 is opened, you leave the base pin and resistor floating. It's neither 0v, or 5v, ie "Off" or "On". It can float between the two, and cause the transistor to switch on. Basically like a little antenna.

Fourth, you have the floating pin and resistor on a breadboard, which has high capacitance. It acts like a capacitor, storing then releasing energy when it can, which is why it varies when you bring your finger near the resistor. Again like a little antenna.

Try adding a 10k resistor after the 100Ω resistor, to ground. This creates a weak pull-down, which will turn off the transistor unless the SW1 is closed, which will override it.

As for the specific IC you linked to, they essentially do the same. Instead of a NPN transistor, the "typical use/reference design" they provide uses a N-Channel Mosfet, but includes a 180k pull down (lower left). A NPN like you have will work just as well, BUT they don't list a maximum current that can be sourced from the load pin. I'd assume, based on the Backlight pin (tested at 3mA at 5v with a max 0.5v voltage sag, see datasheet page 17), it's only a few mA, probably 5mA at best. You might need a high gain transistor, or a darlington pair.

All MOSFETs contain diodes. In the case of an N-channel MOSFET as you have drawn here, the diode goes from source to drain. So what you really have is this:

^{simulate this circuit – Schematic created using CircuitLab}

So, it doesn't really matter what you do with SW1: there is still a path for current through the body diode of Q1 and your LED (D1) no matter what you do with the gate. This is why N-channel MOSFETs are most usually used with the drain attached to ground and the source connected to the load. That is, what you have here is "upside down" from the most usual application.

Also, the gate of a MOSFET is essentially a capacitor. When you close the switch it will be charged, and when the switch is opened, it will only be discharged by very small leakage currents. (This is the basis of flash memory.) Thus, the transistor may remain on for a very long time after SW1 is opened. The cheap, simple, and common solution to this is to add a pull-down resistor.

With those two changes, your circuit should look something like this:

^{simulate this circuit}