Control voltage required to turn on the LED through an NPN
I am trying to figure out the value of V1 DC which will turn on the LED, D1.


about 1.4v: the voltage drop on R2 is 0.7v, so the minimum amount of current going through R1 is 0.7v/R2. As R1 = R2, the minimum amount of voltage drop over R1 is 0.7v / R2 * R1 = 0.7v. thus the answer.

the key here is that the transistor has sufficient current gain so that its base current can be ignored. if not, you just need to add that to the current going through R1 and re-calculate.

But conceptually, it is the same.

edit: here is your circuit in spice.



the current through the led goes from 1ma@V1=1.3v to 14ma@V1=1.5v.