I'm having trouble understanding how much current flows through the base of a BJT. With a MOSFET the answer is easy: 0. How do I calculate the amount of base current in a BJT and am I doing it wrong if my circuit cares?
In this circuit, the base current would be approximately (Vs - 0.7) / R2. The BE junction (with the arrow) behaves like a diode, so as long as Vs is greater than 0.7 V, then current will flow through the diode and there will be a drop of about 0.7 V (depending on the part) and then the current is just limited by the resistor according to Ohm's law.
A functional explanation
The base-emitter junction behaves, like a diode, as a "dynamic resistor" that decreases its resistance when the voltage across it increases and v.v., increases its resistance when the voltage decreases.
So, the input circuit consists of two resistors (constant RB and dynamic RBE that is seen from the input source as differential resistance rbe) in series driven by the input voltage source... and the "amount of base current in a BJT" is determined by the dominant resistor having a higher resistance. There are two typical cases:
Voltage-driven BJT. If RB << rbe, the "amount of base current in a BJT" is determined only by rbe according to the input IV characteristic... and it can be graphically solved by the so-called "load line" technique.
Current-driven BJT. If RB >> rbe, the "amount of base current in a BJT" is determined only by RB... and it can be easily calculated by Ohm's law (look at Glorfindel's answer).