NPN and PNP are two complimentary types of BJTs. NMOS and PMOS are the two complimentary types of MOSFETs. THat is a BJT.
So, yes, current is your problem. For it to be used as a switch, you must be above the base threshold voltage (0.7V in this case), and be able to supply enough current so that the BJT can do it's job as a switch- amplifying the base current across the collect-emitter until it can do no more, at which point the current levels off at a maximum.
You don't need a resistive divider though. Think of the base-emitter pins as a diode connecting them. What happens when you just connect a voltage straight across a diode? Short-circuit (in reality, the resistance in the divider will limit it, but still not needed). Instead, just take a voltage higher than the BJT's saturation threshold voltage and connect it to the base through a resistor Choose this resistor so that a current large so that
(worst case current gain)*(base current) > (maximum collector-emitter current).
so the BJT can act as much like a switch as possible.
The current across this base resistor can be found with V=IR:
Ibase = (VDC - Vb_sat)/Rbase
This assumes emitter and VDC have the same reference (ground) or else you must take that difference into account as well. In this case, Vb_sat is 1.2V (base emitter voltage required for saturation). The 0.7V is for the collector-emitter voltage when the BJT is fully on (saturated).
Do you see how this works? It's exactly the same as a current limiting resistor used to drive an LED.