I have a voltage divider configuration for biasing my transistor. The gain of transistor is lets say 150 which is typical value for BC547.
What I want is limit the current flowing through the base to be 10 micro amps.
Doing the Thevenin's analysis gives me the Thevenin Voltage(Vth) of 6V and the Thevenin resistance (Rth) of 500 ohms. So I redraw the circuit with Rth in series with the transistor and the input voltage being Vth=6V. The B-E on voltage is 0.7V. So doing the calculations gave me the current of 10.6 milli amps.
Instead, I choose to go with the design in figure 2. Since I know that the voltage divider produces the voltage of 6V, keeping a resistor of 530 kohms does the job of limiting the current through the base to 10 micro amps.
So my questions are:
Why do we calculate the Thevenin's resistance and keep it in series with the transistor? We can create a voltage divider to create a voltage of 0.7V and feed it directly into the base. Why don't we do that?
Can't we just calculate the voltage from voltage divider(since the current going into the base is very less) and place the resistor of our choice in series? This would enable me to change the current going into the base just by applying Ohm's law and changing the resistor accordingly. Why bother with Thevenin when Ohm's law is simple enough?
Looking for conceptual explanation rather than mathematical one.