What happens if the base current goes out of the base of an NPN BJT?
Electronic – Negative base current in NPN BJT
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1In all the cases you cannot say that the base voltage is irrelevant for the transistor's operation or as you said transistor;s saturation operation.
If I consider your case the the transistor is mainly dependent on the base voltage and also if you have not taken care of the transistor for its ability to sustain the huge base current it might even blow off.
Now consider when the switch is suddenly opened the voltage induced across the inductor is huge and it is seen by the 10k resistor. This gives rise to a huge base current which might even blow the transitor. So you need to check the maximum Vbe which can be given to the base of the transistor before switching on the circuit.
Solution: If you can place a zener of some voltage lesser than the supply voltage then I think your circuit will work.
One input: Placing inductor at the base of the transistor is always not a good thing to do. Better try to place it in the collector with a free wheeling diode across it.
If you think your transistor can sustain a huge base voltage of 100s of volts and base current of 100s of mA then you can go with the circuit you have else No.
Understanding transistors is a bit like peeling an onion- there are many layers. At the simplest large-signal level you can consider the transistor as a current sink that's controlled by the current through the base-emitter junction. The latter behaves like a forward-biased diode. Not much current until you get to some hundreds of mV, and way too much current if you put volts across the junction. As you say, the transistor will conduct excessive current and will be destroyed if you simply connect (say) 5V to the base with emitter grounded. This is in stark contrast to the behavior of a MOSFET.
At a more sophisticated level of understanding (which is required if you want to predict how most amplifiers work) and for small signals the base-emitter junction behaves like a resistor of Vt/Ib where Vt is the thermal voltage, about 26mV at room temperature. So if your base current is 2.5uA (say the beta is 300 and the transistor is biased with 0.75mA collector current), the base-emitter junction looks like about a 10K resistor for small signals. You can consider the transistor as a (somewhat imperfect because of r0) voltage controlled current source with an input resistance of Vt/Ib. This is the hybrid-\$\pi\$ model. Note that the transconductance gm (and thus the voltage gain in a common emitter configuration) is a function of the collector bias current and temperature and beta does not enter into it at all.
I must emphasize that this model is a linearized model about a bias point and is quite invalid if the (change in) input voltage is large (more than some millivolts). In other words we're talking about relatively small changes on top of a fixed base-emitter voltage of perhaps 600 or 700mV.
Best Answer
you just reverse biased both diode(B-E & B-C) . Now question about the magnitude of the current.
If you have very little amount of current like micro ampere , then you just reverse biased both diode, on the other hand emitter and collector voltages are higher then base voltage.
If current is high like mA or more, you not only biased them reverse but also reverse bias voltage is more then break down voltage. And most likely your transistor is going to burn.