Eric,
Just to test my understanding –
If I have a simple circuit – a 2N2222 switching the current for a RED LED (SAY)
A red LED requires a voltage of at least 1.6V, and I want to minimise the current through the LED to 15 mA then –
12-1.6 = 10.4 V
Ideally you should also subtract the Vce sat voltage , say 0.2V for you example.
12-[1.6+0.2]
10.4V / 0.015 A = 693 ohms ( this is the resistor leading into the collector)
A 680R would be the nearest preferred value.
So for the transistor base current to saturate for the load it needs to be at least a tenth of the load current?
Its usually assumed that the transistor gain is 10 when the transistor is in Vce saturation.
One tenth of the load current is 0.015/10 = 0.0015A
The transistor needs a voltage of at least 0.7 volt to overcome the bias of the transistor.
This means my base needs 0.7v & 0.0015A to enable saturation for the desired load.
You would not normally tackle the problem this way.
The supply is again 12v, so the resistor between the 12V supply and the base will be
12V – 0.7V = 11.3V
This is the method in common use.
11.3V/0.0015A = 7553 ohms.
Would this work??
Yes, I would use a 6800R , preferred value
Regards
Mark