When in university I used to get confused how a transistor (NPN) can draw current from ground. I'm now studying power amplifiers and again, this same configuration appears. How does Ir draw current from seemingly nothing/ground?
Electronic – BJT base connected to ground and still operates
bjtclass-apower amplifiertransistors
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![Repicating your circuit in LT spice]
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.
The key to it all is the minority carriers in the base.
Your suspicion is correct that if all you had was the CB junction it would just become a diode. Reverse biasing this diode does not give you any current. The p-base of an npn is full of holes and the n-collector is has lots of electrons. In reverse bias the majority carriers move away from the junction on both sides and you do not get any current, just like an normal diode.
The tricky part happens when you forward bias the base-emitter junction. The holes in the p-base move towards the BE junction and the electrons in the emitter also move towards the junction. Some of them annihilate each other but because of the doping inequity a lot of the electrons from the emitter pop through into the base!!! As a result they can keep propagating through the base to the collector and you get the collector-emitter current that you were hoping for.
You should take long look at the diagram labelled Lecture 7 - Slide 12 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-012-microelectronic-devices-and-circuits-fall-2009/lecture-notes/MIT6_012F09_lec07.pdf Holes are green and electrons are blue.
Best Answer
Simply redraw the circuit and you will see from where this base current is coming from.
simulate this circuit – Schematic created using CircuitLab
As you can see the \$Q_2\$ base current is coming from \$\textrm{Vee}\$ voltage source.
And flows this path:
+Vee--->R1---> Q2 base-emitter junction--->-Vee
And remember that in the electronic ground is just a reference point, not a real Earth ground. And we are measure all our voltage with respect to this reference point called ground.
To measure the voltage we need two points in the space. One of this point is treated as a reference point.
We have a very similar situation when we try to measure a height of an object. We need a reference point. The most common reference point is "above mean sea level".
But when you measure the height of the table in your house the floor now becomes your reference point.