Georg Hager wrote:

> On Thu, 18 Jun 1998, Thomas McGahee wrote:
>
> > A better solution is to use an NPN transistor with a base resistor of about
> > 330 ohms. Connect the emitter to ground (Vss), connect the cathode
> > of the IR LED to the collector. Connect a current limiting resistor
> > from the anode of the IR LED to +5. A high out from the PIC will
> > cause the IR LED to emit infra-red light.
>
> I see. But I think one could spare the base resistor by using an emitter
> follower, right? I mean connecting base to PIC output, collector to +5V
> and placing diode/resistor between emitter and GND. This cuts the drive
> voltage for the diode/resistor combination by 0.7V, but has better
> switching characteristics as the transistor is not put into saturation.
> Right?

Time for a lesson ?

Let's say you need a pulse of 100mA. Let's assume a gain of at least 200.
You only need a base current of 100mA/200 = 500uA! If you have
5V to run into the base, you need a series resistor of (5-0.7)/0.0005
= 8600ohm. The next value down is 8k2. With this, you get a full 5V,
(@ at least 100mA) over the load.

But now you want to switch an IR LED, and you don't want to pop it,
so you need some current limiting. You can put a resistor in series with
the LED, or you can do it the fancy way by putting it in as an emittor
resistor. This is called feedback. Let's say you want to hit the IR LED
with no more than 100mA pulses. Let's assume again that you have
5V to go into the base. The emittor resistor will see 4.3V, right?
Now 4.3V/0.1mA = 43ohm minimum.

How can this limit the current ? - if more than 100mA flows through
the emittor, the emittor voltage goes higher than 4.3, effectively
switching the transistor off.

The same priciples apply when you want to get the base resistor as in
situation A.

Right! Let's take a 47ohm emittor resistor, 8k2 base resistor, and a
NPN transistor with a gain of more than 200. Hook the collector to your
LED's cathode, and the anode to any voltage you like (remember it is
current limited, so any voltage above the LED drop will do), say 5V.

You will get 4.3V/47ohm = 91mA (I'd say that's close enough!)
To control if the base resistor is not too large : (5-0.7)/8200 = 524uA
base current, that can give you at least 104mA.

                        _ Positive
                        |
                        |
                       _|_
                       \ /  IR LED
                        V
                       ---
                        |
                        |
                        |
                        |
          8k2 ohm     |/c
PIC------/\/\/\/\-----|     NPN
                      |\e
                        |
                        |
                        /
                        \ 47 ohm
                        /
                        \
                        /
                        |
                        |
                      -----
                       ---
                        -

BTW make sure you don't leave the LED on - it
will pop @ 100mA.

--
Friendly Regards

Tjaart van der Walt
mailto:tjaart@wasp.co.za

Add your voice !!
Vote at the Great G Com Public Vote. Go to :
http://www.wasp.co.za/~tjaart/gcomvote.html

|--------------------------------------------------|
|                WASP International                |
|R&D Engineer : GSM peripheral services development|
|--------------------------------------------------|
|SMS mailto: 0832123443@wasp.co.za  (160 chars max)|
|     http://www.wasp.co.za/~tjaart/index.html     |
|Voice: +27-(0)11-622-8686  Fax: +27-(0)11-622-8973|
|          WGS-84 : 26¡10.52'S 28¡06.19'E          |
|--------------------------------------------------|