Re: base resistance
Well you are wrong with your calculation... Since you need to have BJ in saturation mode, well you cannot use hfe, because in saturation hfe<10. And it varies, so you cannot use it at all...
But there is another way.
In saturation you have
Vbes=Vbe=0.6V, usualy
Vces=0.2... You should check it in datasheet for transistor.
In that case you will get:
Ve=Vc-Vces,
Vb=Ve+Vbes
ib=(Vbb-Vb)/R
Where Vbes-saturation voltage on base-emitor
Vces-sat voltage on collector-emitor
Vc-collector voltage
Ve-emitor voltage
Vb-base voltage
Vbb-voltage on the other side of the base resistor.
ib-base current.
If you said that pins are on 3V, whitch i dont think is correct since that circuit is on 6V to 9V, but check it in data sheet, you get Vbb=3V-Vd, where Vd is a voltage on diode Vd=0.6V...
=>
ib=(3V-Vd-Vb)/R
=>
ib=(3V-Vd-Vc+Vces+Vbes)/R
That is from where you get your R, but if you can figure out ib...
Oh, and it just came to my mind, why do you think BJTs should be in saturation, and with 3V on pins it is not possible, since for saturation it must be that Vb>Vc.
Your Vc is on 6V to 9V, and pins on 3V... Therefore, it is not possible for BJT to be in saturation...