🚀 go-pugleaf

In article <6220013@hpcupt1.HP.COM> dclaar@hpcupt1.HP.COM (Doug Claar) writes:
>I'd like some ideas on a circuit to drive a high power infra-red LED from
>TTL. I know that standard LEDs can be driven straight from TTL, but this
>little beastie wants a maximum of 1.5V forward at 50ma. (3V reverse if it
>matters). For proof of concept for my mostly digital hack, I used an AS640
>buffer driver (sinks 64ma) and a resistor, but it sure is ugly when that's
>all I use the chip for, and all I think that I really need is a transitor
>and some resistors. Any hints would be appreciated. Also, any hints on
>practical transistor fundamentals book(s) would be appreciated. (At my
>school, they taught us how IC wafers were made, which is MUCH more useful
>than how to use transistors...of course, I WAS CS :-()

Well, I was EE :-D.

Use an emitter-follower configuration, so that "logic HI" gives "LED on",
just for convenience.

                Vcc = 5v
                 o
                 |
              |  /
           Ic |  \ 
              V  /  Rc
                 \ 
        Ib       /
       ---->     |  Q (Si, npn)
                 | 
         Rb   b |/ c +
  Vi o--\/\/\---|      Vce   = 0.2v
              + |\ e -  (sat)
      Vbe  = 0.7v|
       (on)     -|
               -----  +
                \ /     Vd   = 1.5
            |  -----  -  (on)
      Id=Ie |    |
            V    |
               -----
                ---
                 -

Typical silicon npn transistor in saturation ("full-on")
has Vce=0.2v, Vbe=0.7v.

The current in the diode is 50ma, the voltage across it (Ve)
is 1.5v.

Therefore Vb = 2.2v, and Vc = 1.7v.

It's a simple matter to find Rb and Rc now from Vi-Vb/Ib and Vcc-Vc/Ic.
(Vi is the TTL "HI" voltage, supposed to be 5.0v, but more like 4.7v)

First find Ib and Ic from Ie.  (Ie is current flowing out of the
emitter and into the diode, and is 50ma; Ib is current flowing
out of the TTL chip and into the base; Ic is current flowing out
of the power supply and into the collector.)

Find the transistor's "beta" (or H  ) on the data sheet.  (Beta is
                                  FE
the ratio of Ic to Ib.)

Since Ie == Ic + Ib, you get the two eq's

    Ib = Ie/(1+Beta)

    Ic = Ie * Beta/(1+Beta) = Ib * Beta

For a good transistor, beta is about 100, and I'll finish the example
with that.

Ib = 50mA/(1+100) = 495uA

Ic = Ib * 100 = 49.5mA

Now,

    Rc = (Vcc - Vc)/Ic = (5 - 1.7)/49.5e-3 = about 67 ohms (use 68)

    Rb = (Vi - Vb)/Ib = (4.7 - 2.2)/495e-6 = about 5.05k ohms (use 5.1k)

The standard-sized resistors you use (68 ohms and 5.1kohms)
are only a little larger than the exact resistances.  This
prevents overdriving the diode, and if you use 10%
tolerance resistors it's plenty close.

Just to check to see if we'll fry anything:

    P  = Pce + Pbe = Vce*Ic + Vbe*Ib = 0.2*49.5e-3 + 0.7*495e-6 = 10.25mW
     Q

	         2               2
    P   = Rc * Ic  = 68 * 49.5e-3  = 167mW (use 1/4W or larger)
     Rc

	         2                2
    P   = Rc * Ic  = 5100 * 495e-6  = 1.25mW (use 1/16W or larger)
     Rb

    P  = Id * Vd = 50e-3 * 1.5 = 75mW
     D

The diode is obviously big enough to handle its rated current.
Be sure not to use anything smaller than 1/4-watt for the
collector resistor, or it'll turn burn.  Even a 1/4-watter
might get hot.  The base resistor can be almost any rating.

You can probably find a suitable transistor easily, but remember to
plug-in its beta and repeat the calculations of the resistances
and power.  Rc won't change too much with beta, but Rb has a
controlling influence on the input, and you want it close to the
correct value (10% tolerance is okay, but an order-of-magnitude
error is bad).  You might also use the real Vbe(on) and Vce(sat)
values, which might vary 50-100% from what I used here.

				--Blair
				  "How'd I do?"