I am studying a common emitter amplifier. I am using a 2n4401 npn if I'm reading the datasheet correctly, it has a hFE of 40 at .001 amp. I have chosen that as my quiescent amperage. I'm using a Vcc of 10v my emitter resistor is 1k ohm which should give me a 1 volt drop at the emitter. the remaining 9 volts will drop through a 4.5k ohm collector resistor drawing 2mA when transistor is saturated. with an Ic of .001 and beta of 40, that makes my desired base current of .000025 amps at the quiescent collector current.

I am biasing the base with a voltage divider with a desired voltage of 1.7 volts. My studies have told me that it is best to make the biasing divider network 10 times the base current. This is wherein my question lies. One source has stated that i should try to draw 11 times the base amperage between the base and Vcc while drawing 10 times the current between base and ground because one times is going through the base, but for the sake of simplicity (and to keep my head from exploding) I choose to keep it at 10 times. I chose 33.2k ohm between base and Vcc and 6.8k between base and ground (10 volts/ 40000 ohms = .00025 amps)

Thevenin resistance is 5644 ohms at the base. Thevenin voltage is 1.7 volts. 1.7v/5644ohm=.0003 amps.

Now, if I'm trying to keep the base current at .000025 amps, why has what I've tried to learn tell me to supply the base with .0003 amps? what am i missing here? Thank you for your time, Michael.