dclaar@hpcupt1.HP.COM (Doug Claar) (03/21/90)
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 :-() Doug "closet hardware hacker (and do I mean hack)" Claar HP Computer Systems Division UUCP: mcvax!decvax!hplabs!hpda!dclaar -or- ucbvax!hpda!dclaar ARPA: dclaar%hpda@hplabs.HP.COM
mark@mips.COM (Mark G. Johnson) (03/22/90)
Doug Claar asks how to drive >50mA into an LED without dedicating
an entire 74AS640 IC package to the task. He asks for a ckt using
resistors and a transistor.
Get the Digi-Key catalog (1-800-DIGI-KEY). Go shopping in the
"transistors" section.
I recommend you use an NPN Darlington device like the MPSA14 ($0.23)
or MPSA13 ($0.22). The reason for preferring a Darlington is
that a Darlington needs twice as much base voltage to turn ON, so
your noise immunity will double.
anode +--------------+ cathode +---------------+
+5V ---------| Hi Power LED |-------------| R2 39.0 ohms |------+
(+) +--------------+ (-) +---------------+ |
|
|
|
emitter collector |
GND -------------\ Q1 /----------------+
\ /
----- NPN Darlington
TTL INPUT base |
(LOGIC-1 +--------------+ |
TURNS ON -------| R1 4.7Kohms |-------------+
THE LED) +--------------+
The current through the LED is set by R2. Resistor R1 is for biasing.
The circuit isn't particularly fast; presumably that isn't necessary
if all you want is a winky-blinkey light for humans to look at.
Another approach would use a normal (non Darlington) transistor but
have two resistors in the base circuit. This lets you choose from
the much bigger universe of normal transistors, but means you have
to solder in another component. It also saves a nickel on the
transistor cost.
--
-- Mark Johnson
MIPS Computer Systems, 930 E. Arques, Sunnyvale, CA 94086
(408) 991-0208 mark@mips.com {or ...!decwrl!mips!mark}bph@buengc.BU.EDU (Blair P. Houghton) (03/22/90)
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?"