Circuit - http://i49.tinypic.com/mt5lqo.jpg Boylestad example - http://i45.tinypic.com/20jpta9.jpg So the circuit is what i quickly drew as the question states. "The led is to have a forward current of approx 10mA. Vcc=9V, Vf= 1.6V, Vb=7V and Q1 has hfe(min)=100. Take Vbe =0.7V Find suitable resistor values for R1 and R2. What i've done is to say: Ib=Ic/hfe=10mA/100=1mA [current is the same for series right?] R1=(Vb-Vbe)/Ib= 6300 ohm then R2=(Vcc-Vf)/Ic = 740 ohm ------------------------ Okay so my circuit is very similar to the example from boylestad I have found. Is my calculations correct? For resistor R2 I have followed the book. But in the book they just use the voltage over Rc as Vcc. Is this not supposed to be Vcc-Vce? I cant seem to find any other calculations to get Vce... Thanks
If you look at the Vce(sat) specs of almost any BJT, you will find that they are specified for Ib=Ic/10. This is called "forced beta", and it is necessary because the beta of a transistor in saturation is much less than it is when Vce is, say, 10V. Boylestad has probably never had a real job designing circuits. See the attachment.
I saw a graph somewhere from a textbook that when Vce is in saturation mode the voltage varies from between 0V and 0.7 (Vbe).