Originally Posted by MrAl

Hi,


The switch could be looked at as simply a saturated amplifier.
The equation Ic=B*Ib is also used to simplify that, even though it
might be used as a switch. Of course then we would want to take
a close look at the saturation characteristics too as i think you noted.

For example, how to drive a relay that requires 40ma coil current
from a 10v supply?
Say we have a transistor with min gain of 40. How much base
current do we need to drive the NPN transistor properly and what
resistor could be used as a pullup to drive the transistor base from
the 10v supply?

The simple equation to use is Ic=B*Ib again, and we know Ic=0.040
and B=40, so rearranging the equation:
Ic=B*Ib
Ib=Ic/B
so
Ib=0.040/40=0.001 amps
so we double that to make sure the transistor stays in sat, to 0.002 amps.

Now to answer the question about what value resistor to pull up the base
with to the 10v supply in order to keep the relay on.
Since the voltage is 10v, the resistance would be 10/0.002 or 5k.
Note here we didnt even consider the base emitter voltage and still
came up with a reasonable design.
If we wanted to consider the base emitter voltage though, it wouldnt
be hard as we take that as a constant equal to about 0.7v, or to be
sure and make things even simpler, 1v...
Now we have 10v-1v=9v, and 9v/0.002a equals 4.5k base pull up.
Here, we estimated the base emitter voltage and didnt have to consider
it's entire equation.
If we wanted to make sure this was going to work, we could look at
what happens when the base voltage changes by say 0.1v and see
what effect this has, but we can already see that the small changes in
base current wont have too much effect so this is mostly ignored.

Now repeat the above process using the exp(Vbe/VT...) equation and
see the difference.

Happy Fourth to you all too!

Originally Posted by alphacat

Thank you very much
I just love reading your posts here