The 2803 outputs provide a path to ground.The UDN2981 is a "high-side driver".Adding link --They're still available from Jameco.http://www.jameco.com/Jameco/Products/ProdDS/1762681.pdfYes, resistors are (still) required.
No, the 2003 is a low-side switch, too, equivalent to a 2803 only the 2003 has 7 "switches" and the 2803 has 8.Can't you sketch out your circuit and take a picture of that with your cell or something and post that? You haven't provided a schematic for what you're on about, but I think your maths are off.
You need to post the schematic of how you are driving this, it sounds like you are trying to use the transistor array incorrectly.Just for anyone reading this thread later note that connecting LEDs in parallel like this is wrong.
Yes that circuit is right. Note that the transistor shown is a PNP transistor. The arrays you were asking about are darlington drivers made from NPN transistors. These are not the same thing.
So I checked how much one LED pulls without a resistor. 2.5V 30ma LED unrestricted on 5 volts pulled 235ma and didn't burn out.
So with a PNP transistor, I'd be looking at (5.0-0.7)/0.03 = 4.3/0.03 = 143 ohm resistor.(5-.07)/330 = 4.3/330 = 13ma using test resistors.
So doing it right, I would still have to account for the 1.2V saturation voltage of the darlington?
That'll be 840mA per segment to display an 8- not sure the Darlington's can push that much