Consider the voltage across a diode and the current that flows. Below are the curves for an old germanium diode (1N34A) and a silicon diode (1N914): -
Concentrate on the silicon diode (1N914). With 0.6 volts across it, the current is about 0.6mA. Now drop that voltage to 0.4 volts. The current falls to 10 uA and, with 0.2 volts across it the current is about 100 nA.
Now, the base-emitter junction in a BJT is a forward biased diode. The forward biasing comes from the voltage you place across it and this is usually via a biasing resistor. In your circuit, R2 and the power supply voltage define the current that can jointly flow into the base and into R3.
When R2 supplies a decent amount of current, most of it flows thru the base emitter junction because you are on that part of the diode curve and that part of the diode curve has a dynamic resistance that is much smaller than R3. As base-emitter voltage lowers, its dynamic resistance gets higher and R3 starts to become the "path" to which most of the current from R2 flows.
Dynamic resistance is the small change in applied voltage divided by the change in current. You could look at the diode graph above and pick some points: -
- At 0.60 volts the current is possibly 600 uA
- At 0.62 volts the current is about 1000 uA
Dynamic resistance would be 20mV/200uA = 100 ohms
- At 0.40 volts the current is about 10 uA
- At 0.42 volts the current is about 11 uA
Dynamic resistance would be 20mV/1uA = 20 kohms.
So, when R3 lowers it becomes more dominant that the base emitter junction and rapidly the junction current falls away. Given that we can approximate transistor action to a device with current gain, lowering R3 beyond a certain point means a rapidly falling collector current and, in effect, the transistor is regarded as turned-off.