Re: [CAD_CAM_EDM_DRO] explanation needed
Posted by
Harvey White
on 2003-09-14 19:42:26 UTC
On Sun, 14 Sep 2003 20:38:01 -0000, you wrote:
If current flows out of the chip, you are "sourcing" the current.
TTL logic is very good at sinking current.. as an average TTL load
(standard variety) is 1.6 ma sink, and about 40 ua (microamps)
sourcing... as far as the chip driving the load is concerned... (it's
a back biased diode in behavior...)
So asking a TTL gate (specified for 10 loads capacity... called
"fanout" to run an LED (which can take 5 to 10 ma)... is possible only
if the current from the LED goes to the gate output... (read printer
port....)....
That is possible only if there is a separate 5 volt supply, current
goes through a resistor (to limit current)... thence to the LED (or
"input" of an optoisolator)... then through the gate to ground.
THe most that a TTL gate could source would be about 1 ma or so
(milliamps...) and that's not good enough....
Sourcing the current would mean that the LED cathode is grounded, the
gate goes to the Anode (+) lead, through a resistor...
Hope that this gives you an idea....
Harvey
>what is meant by the following ...........If current goes into the chip, then you are "sinking" the current....
>
>" sinking XX ma "
>
>boy do i have a lot to learn !
If current flows out of the chip, you are "sourcing" the current.
TTL logic is very good at sinking current.. as an average TTL load
(standard variety) is 1.6 ma sink, and about 40 ua (microamps)
sourcing... as far as the chip driving the load is concerned... (it's
a back biased diode in behavior...)
So asking a TTL gate (specified for 10 loads capacity... called
"fanout" to run an LED (which can take 5 to 10 ma)... is possible only
if the current from the LED goes to the gate output... (read printer
port....)....
That is possible only if there is a separate 5 volt supply, current
goes through a resistor (to limit current)... thence to the LED (or
"input" of an optoisolator)... then through the gate to ground.
THe most that a TTL gate could source would be about 1 ma or so
(milliamps...) and that's not good enough....
Sourcing the current would mean that the LED cathode is grounded, the
gate goes to the Anode (+) lead, through a resistor...
Hope that this gives you an idea....
Harvey
>
>don
Discussion Thread
Don Thigpen
2003-09-14 19:11:36 UTC
explanation needed
Harvey White
2003-09-14 19:42:26 UTC
Re: [CAD_CAM_EDM_DRO] explanation needed
Bevan Weiss
2003-09-14 21:27:46 UTC
Re: explanation needed
Jon Elson
2003-09-14 22:24:10 UTC
Re: [CAD_CAM_EDM_DRO] explanation needed
Don Thigpen
2003-09-15 07:40:46 UTC
Re: explanation needed