> This WILL cause the transistor to be operating in it's linear region,

Yes.

> since
> otherwise the voltages at the emitter would be Vunreg-Vled-Vcesat, which
> could easilly be greater than Vout from the PIC, right?

The voltage at the emitter is driven by the PIC voltage.  It will be the PIC
voltage minus the B-E junction drop.  The transistor is acting as an emitter
follower driving the resistor voltage from the PIC output voltage.  Since
the emitter load is a resistor, the emitter current will be the emitter
voltage divided by the emitter resistor value.  Assuming reasonable gain for
a small signal transistor, this also becomes the collector current for
practical purposes.

> This shouldn't be
> a problem for typical LED currents, but it does mean that for higher
> current loads (light bulbs?) you'll probably end up needing to go back to
> base and collector resistors instead of the emitter resistor (or risk
> blowing up your transistor due to excessive power dissipation.)

Let's say the unregulated supply was 20V (pretty high for linearly
regulating to 5V), the LED requires 2V and 20mA, and the B-E drop is 600mV.
The C-E voltage accross the transistor will be:

  Vce = 20 - 2 - 4.4 = 13.6V

At 20mA the power dissipation in the transistor is 270mW.  Yup, that's
pushing a TO-92 package but still doable.  However that represents another
300mW that the linear regulator doesn't have to dissipate.  Especially with
multiple LEDs, this can make a big difference in the regulator requirements.
The wasted power to run the LEDs is spread over multiple small transistors
and resistors instead of being concentrated in one package.

Also note that 20mA LED current is rarely needed anymore.  Nowadays for
simple indicators I use decent LEDs and run them at 2 - 10 mA even though
they are rated at 20mA.


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