# NPN transistor energy drain from base to emitter?

I have the following example circuit:

I do not understand why this lights the LED. For my main project I am building a circuit that has a button connected to an output as well as the collector of the transistor as shown in the diagram. The circuit works as expected when I put the LED on the collector instead of the emitter, but this solution will not work for my overall setup. Even a 1M Ohm resistor allows current through. Am I just using transistors wrong?

• Draw a proper circuit and not a cartoon. – Andy aka Jan 23 '17 at 8:46
• @Andyaka fixed. – Elijah Seed Arita Jan 23 '17 at 9:02
• So, roughly speaking you are going to see $I_{led}\approx \frac{5\:\textrm{V}-V_{led}-V_{BE}}{R_1}$. You have a resistor, in series with a BJT diode, in series with an LED. And you have enough voltage present to overcome the diode drops. So current flows. No surprise. Why can't you put the LED into the collector leg? – jonk Jan 23 '17 at 9:10
• If you replace that BJT transistor with a MOSFET transistor then yea it shouldn't light up. – Bradman175 Jan 23 '17 at 9:49
• I think I made the LED backwards – Elijah Seed Arita Jan 24 '17 at 2:07