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8243
GUARANTEED
WORST CASE
CURRENT
SINKING
CAPABiliTIES
OF
ANY
I/O
PORT PIN vs.
TOTAL
SINK
CURRENT
OF
All
PINS
o
L--L
__
~~~-L
__
~~
__
-L
__
~~
__
-L
__
~~
__
~
o
2
3 4
5
6
7
8
9 10
11
12
13
MAXIMUM SINK CURRENT ON ANY PIN@.45V
MAXIMUM
10L
WORST CASE PIN (mA)
Sink Capability
The 8243 can
sink
5
mA@.45Von
each
of
its
16
I/O
lines
simultaneously. If, however, all lines are not sinking
simultaneously or all lines are not
fully
loaded, the drive
capability
of
any individual line increases as
is
shown by
the accompanying curve.
For example,
if
only
5
of
the
16
lines are to sink current
at one time, the curve shows that each
of
those 5 lines
is
capable
of
sinking
9 mA@.45V (if any lines are to
sink
9 mA the total
10l
must not exceed
45
mA
or
five 9 mA
loads).
Example: How many
pins
can drive 5 TTL loads
(1.6
mAl
assuming remaining pins are unloaded?
10l
= 5 x 1.6 mA = 8 mA
dOL
= 60 mA from curve
#
pins=60
mA
...
8
mA/pin=7.5=7
In
this
case, 7 lines can sink 8 mA for a total
of
56
mAo
This leaves 4 mA sink current capa-
bility
which can
be
divided in any way among
the remaining 8 I/O lines
of
the 8243.
Example: This example shows
how
the use
of
the
20
mA sink capability
of
Port 7
affects
the sink-
ing capability
of
the
other
I/O lines.
An 8243 will drive the
following
loads simul-
taneously.
2 loads -
20
mA@
1V
(port 7 only)
8 loads - 4 mA@.45V
6 loads - 3.2 mA@ .45V
Is
this
within the specified
limits?
d
Ol
=(
2 x
20)+(8
x
4)+(6
x 3.2)=91.2
mAo
From the curve: for
10l
= 4 mA,
dOL::
93
mA
since 91.2 mA
<
93
mA the loads are
within
specified limits.
Although the
20
mA@1V loads are used in
calculating
dOL,
it is the largest current
re-
quired@.45Vwhich determines the maximum
allowable
dOL'
Note: A 10 to
50Kn
pullup resistor to +5V should
be
added
to
8243 outputs
when
driving to 5V
CMOS
directly.
6-61
AFN-00214A-Q5