I am studying Transistors from the book - "Arts of Electronics" by Paul Horowitz and Winfield Hill (second edition).
In the text, there is the following circuit (under title - Transistor Switch) on page 63:
I understand that the transistor would work in the saturation state, because otherwise the collector voltage would be -84V, i.e. less than the emitter voltage (which in this case in zero).
However the text mentions,
Overdriving the base (we used 9.4mA when 1.0mA would have barely sufficed) makes the circuit conservative.
Incidentally, in a real circuit you would probably put a resistor from base to ground (perhaps 10k in this case) to make sure the base is at ground with the switch open.
My questions: 1) What does the first line (from the text) mean? What is the meaning of a circuit becoming conservative?
2) Why would we have to place a resistor from base to ground? How do we decide the base voltage in absence of current i.e. switch is open? Because in that case collector voltage would be 10V and emitter voltage would be 0V, how do we determine base voltage?