"Christopher Ott" spamtrap at ottelectronics dot com wrote in message
...

"tempus fugit" wrote in message
...

"Christopher Ott" spamtrap at ottelectronics dot com wrote in message
...
"tempus fugit" wrote in message
...
Hey all;

I want to drive LEDs directly from the outputs of a 74LS373N latch,
but
I
can't figure out which parameter is the maximum current I can get
from
it.
I've enclosed the data sheet. On p. 6, there's a then a different 1
on
p.
9
IOH High-level output current

and

IOL Low-level output current


then a different 1 on p. 9. Can anyone tell me which parameter I
should
be
looking at? Or am I looking at teh wron g thing altogether?

Thanks

Looks like you can source 2.5ma, sink 24ma with a maximum of 40ma
combined
total per device. The 74HC373 can do much better, sourcing or sinking
35ma
with a total of 70ma per device. I typically use a darlington driver
array
to buffer LEDs from logic IC's. Lookup Digikey part number
TD62083AF-ND.
It's cheap, TTL compatible, and can easily drive 8 high current LED's
without straining your logic IC.

Chris

Thankd guys.

So just to make sure I've got this straight - if I connect one lead of
the
LED to the output of the latch and the other to ground, I've got 2.5mA,
but
if I connect one lead to +5v and the other to the output of the latch,
I've
got 24 mA? (Incidentally, I didn't even know you could connect the LED
this
way...)

Thanks

That's correct, you have to respect the polarity of the LED however.
You'll
also need a series resistor. And when sinking the LED, you will turn it on
by making the output low (switching ground instead of +5). If you're using
all 8 outputs, you'll need to use high efficiency LED's or a buffer IC as
I
mentioned previously.

Chris

OK thanks again Chris. I'm currently using a Darlington IC, but am changing
a couple things in the design and thought I might be able to get away
without it.