This question is related to this question: why-are-the-pins-of-an-udn2981-related?

In the UDN2981 datasheet, the figure below shows the maximum current (page 14, bottom left picture).

enter image description here

Does this mean if I connect 8 pins (thus using the lowest line in the figure), I can use around 120 mA per pin (thus 960 mA for the total IC), or is it 120 mA for 8 pins together (thus 15 mA)?


A little ambiguous, but this sentence from the first page suggests you can draw 120 mA per pin continuously when all are ON.

Under normal operating conditions, these devices will sustain 120 mA continuously for each of the eight outputs at an ambient temperature of +50°C and a supplyof 15 V [..]


Yes, however I believe that applies only to the DIP package. The SOIC-18 and SOIC-20 packages have about 0.63 or 0.66 the dissipation capability so they are not capable of as much current, all other things being equal.

They may also be making an assumption that the IC is soldered into a board with a certain amount of surrounding copper so a socketed chip or an SMT part with sparse copper might not be as good. All the more reason to be conservative.

Typical voltage drop at 120mA is going to be around 1.65V, so that's almost 1.6W for the chip, meaning that the die (assuming DIP in datasheet reference conditions) will be running at 145°C with a 50°C ambient, which is very hot.

  • \$\begingroup\$ I only use DIP anyway, so this is luckily not a problem for me. \$\endgroup\$ – Michel Keijzers Mar 7 at 14:31

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