ULN2803 current limit

[David Harmon]

ULN2803A: The datasheet says the 500 mA limit applies to "one device
in the package". Does that really mean that the whole package can
pass 4 Amps at once using 8 pins? Or do you only get the full 500mA
while using only one device at a time.

 

[Chris]

ULN2803A: The datasheet says the 500 mA limit applies to "one device
in the package". Does that really mean that the whole package can
pass 4 Amps at once using 8 pins? Or do you only get the full 500mA
while using only one device at a time.

Hi, Dave. Great question. Here are two data sheets for the 2803:

http://www.onsemi.com/pub/Collateral/ULN2803-D.PDF
http://www.allegromicro.com/datafile/2801.pdf

The first data sheet is remarkably opaque on this really basic
question. I would guess that's because it's kind of a "gumball" part
-- it's been around for decades.

Both data sheets tell you that each output is rated for 500mA.
However, the Vce of the outut darlington is typically between 1.0V and
1.5V for high output currents. Assuming worst case, eight outputs of
500mA with 1.5V across each output, you've got 6 watts being dissipated
by the IC package. Not happening in this world. The IC will glow red
for sure.

The second data sheet is more descriptive. It has a graph that says if
you've got a 100% duty cycle (the output is always on), you can't even
run two outputs at full current. This is because of power dissipation
limitations. The maximum output per pin goes from 500 mA (one output
on at one time) to about 470mA with two on, down to about 160mA with
all eight outputs on at 100% duty cycle. Look at the graph on page 6.

Actually, since these are maximums, you should derate the outputs on
the data sheet accordingly. I would think good engineering practice
would suggest no more than about 1 watt power dissipation maximum for
the IC package. That would mean 500mA for 1 output, 330mA for two
outputs, 220mA for 3 outputs, 167mA for 4, and so on. But how
conservative you want to be is up to you. Don't exceed the maximums
shown on the graph, though.

If you need 500mA from more than one output, use TO-220 NPN
darlingtons, and discrete resistors to give you 1mA of drive current at
the base of the transistor. Almost any CMOS or TTL output can drive
1mA. And the 500mA collector current is small enough so you won't need
heat sinks for the transistors.

Good luck
Chris

 

[Pooh Bear]

ULN2803A: The datasheet says the 500 mA limit applies to "one device
in the package". Does that really mean that the whole package can
pass 4 Amps at once using 8 pins? Or do you only get the full 500mA
while using only one device at a time.

There will be a thermal restriction.

Essentially it's dissipation limited. It's up to you to figure out how
that affects your useage.

Don't forget to derate the package dissipation vs ambient temp btw.


Graham

 

[David Harmon]

On 6 Oct 2005 19:21:46 -0700 in sci.electronics.basics, "Chris"

http://www.onsemi.com/pub/Collateral/ULN2803-D.PDF
http://www.allegromicro.com/datafile/2801.pdf

The first data sheet is remarkably opaque on this really basic
question.

You noticed that too! But I've got the Allegro datasheet as well
now, and that with your help has straightened me out. Thanks.

 

[Jasen Betts]

ULN2803A: The datasheet says the 500 mA limit applies to "one device
in the package". Does that really mean that the whole package can
pass 4 Amps at once using 8 pins? Or do you only get the full 500mA
while using only one device at a time.

That depends the amount of power disipated, as
long as it's below the limit for the package you should be fine.

you have the data sheet (figures for voltage drop etc...), you work it out.

Bye.
Jasen