An explanation may help. Ignore the connection of the transistor base
for a moment.
R1 and R2 form a V divider, Vjunction = (5V * R2) / ( R 1 + R2). So for the values
and locations you had Vjunc = ( 5 * .47 ) / ( 10 + .47 ) r values in K ohms
So Vjunc = .22 V
Now we connect base of transistor to that junction, but it wants to turn on around
~ .7V for a silicon bipolar transistor. So it never got turned on. Here is a graph of
Vbe and Ic, current in the collector, same current in your LED, for a "typical" NPN
transistor -
So your voltage divider was prohibiting the transistor from turning on.
But if you swap their values then Vjunction = (5V * R2) / ( R 1 + R2)
= ( 5 * 10 ) / ( 10 + .47) = ~ 4.8 V. But at .7 V that Base Emitter junction
turns on and as you can see from the graph it does not want to change
its Vbe much so acts as a sort of clamp in the base circuit. But its turning
on the transistor hard.
Here is a graph of base current versus its Vbe, as you can see once it
turns on it does not want its Vbe to change much. Thats the nonlinear
behavior of a basic diode junction.
i
Keep in mind in a NPN Ic =~ Beta x Ibase, so in this graph if 80 uA was
shoved into the base and Beta was 100 then the collector current would
be 8 mA.
So we want the transistor to operat as a switch, basically to make Vce as
low as possible. We can do that by forcing much more current into the base,
rule of thump is Ib = Ic / 10. As you can see in this graph the sat region of oper-
ation, cross hatched area (vertical cross hatched area) , shows Vce quite low,
base current quite high relative to I collector.
Note Ib/10 is called "forced beta" in the industry.
The region out of the cross hatched areas is the linear operation of the transistor when
we use it to amplify signals. Where its beta become important. A whole lot of other con-
siderations come into play here.
Regards, Dana.