> 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.

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

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