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