It's VERY, VERY non-critical - it has to provide enough current to switch the transistor on fully, but not enough to overload the PIC (or damage the BE junction of the transistor). So it's basically realted to the gain of the transistor - which itself varies over a huge range.

Yes it's related to the transistor.
You want (Vdd-Vbe)*(transistor gain)/R > LED current.

BE junction is a diode, so 0.7v for Vbe is generally reasonable. At higher currents it will be larger and will be somewhat lower for very small currents.
Transistor gain varies significantly with temp, part variations, and age. Vce(sat), the voltage loss on the transistor when acting as an on/off switch, is lower when the left hand side of that eq is much larger than (like twice) the right hand side.

Ideally you want to chose a resistor value that will just barely cause the transistor to saturate, that will allow you the least amount of delay during turn off, using too much drive current is just going to slow turn off times and stress the controller's I/O driver. Too little current and the transistor will stay in an ohmic state and waste power.

Originally Posted by Sceadwian

Ideally you want to chose a resistor value that will just barely cause the transistor to saturate.

I would suggest otherwise, certainly calculate that value - THEN increase the current substantially, to guarantee the transistor is hard on. A lot though depends on what you're trying to switch? - if it's a high current load, use more than one transistor - or a darlington.

If you use an emitter follower, you only need one resistor.

And this arrangement reduces the load on the PIC i/o compared to a common emitter circuit.

See attachment.

Originally Posted by ljcox

If you use an emitter follower, you only need one resistor.

But it's less efficient, and less useful, than the other way, which is why it's rarely seen - and usually it's a sign of poor design.

And this arrangement reduces the load on the PIC i/o compared to a common emitter circuit.

In what way?.

Why is it less efficient? I know that Ve is about 0.7V less that Vcc, but there is still enough voltage for the LED unless it is one that needs 3V or more. But the normal LEDs only need about 1.8V.

It reduces the load since the common emitter config needs Ib to be large enough to cope with the minimum beta in order to ensure saturation. In fact, the rule of thumb is that Ib should be one tenth Ic.

Whereas, in the emitter follower case, Ib = Ic/beta.

eg. if Ic = 10 mA and Beta = 200, Ib = 50 uA.

Otherwise, using the rule of thumb. Ib = 1 mA.

Originally Posted by Pommie

Surely, all the current flows through the LED and therefore this must be the most efficient circuit?

Because of the greatly increased voltage drop across the transistor, admittedly the base current will also flow in the load - but that's likely to only be tiny percentage of the load current, particularly in the case of driving a 40mA LED.

Originally Posted by PeterDove

My LED needs 3.6V.

Peter

In that case, you need either an NPN transistor or an N channel MOSFET.

What current do you need through the LED?

For example, if it requires 10 mA then the base current needs to be about 1 mA. So the base resistor should be (5 - 0.7)/10 = 430 Ohm.

If you use a MOSFET such as the 2N7000, the gate resistor is not critical.

Use 100 Ohm.

Originally Posted by ljcox

For example, if it requires 10 mA then the base current needs to be about 1 mA. So the base resistor should be (5 - 0.7)/10 = 430 Ohm.

I'm all for making sure the transistor is 'hard on', but for driving an LED at 40mA it's hardly of any importance - anything between 1K and 10K would be fine.

BTW, why did you calculate the base current as 10mA above? - a little slip perhaps?