how to find minimum acceptable current through voltage divider driving transistor

Question

I'm using a BC557 transistor to control a LED and a uC to control the transistor. The transistor itself is controlled using a voltage divider.

It was suggested to me to use resistors around 100K + 10K for the divider but I'm curious as to how do I figure out what the best resistor values would be? The system is supposed to run on battery so I guess that the lower the current draw, and if that is the case then why not use essentially infinite Ohm resistors for the divider?

I have looked through the datasheet but it does not seem to mention a minimum base current.

Answer 1

No, the datasheet won't give you a minimum base current, because that's determined by the collector current you want. \$H_{FE}\$ is the parameter you need. For the BC557 (no suffix) that's minimum 125, which means that for a LED current of, say, 20mA you'll have

\$ I_B = \dfrac{I_C}{H_{FE}} = \dfrac{20mA}{125} = 160\mu A\$

If \$H_{FE}\$ would be higher (can be up to 800) the resistor in series with the LED will limit the current.

You can use a divider to set the base current, but this will cause unnecessary current parallel to the base-emitter junction. Just use a series resistor:

\$ R = \dfrac{\Delta V}{I} = \dfrac{5V - 0.7V}{160 \mu A} = 27k\Omega \$

That's the maximum resistance to get a guaranteed 20mA collector current. (I'm presuming your BC557 is connected to 5V, if it's 3.3V just redo the calculation.)

---

Note that the BC557 is a PNP transistor. If you want an NPN choose the BC547.