Basically that's the question: Does this IC let me connect LEDs directly to the outputs or should I use transistors?
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1\$\begingroup\$ You accept too quickly. Questions with an accepted answer get less new answers, which otherwise might be interesting too. Give it some time to go around the world, or until you got some other answers. (Oli was awake, but the rest of Europe wasn't, for instance.) \$\endgroup\$– stevenvhAug 6, 2012 at 3:44
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\$\begingroup\$ Well thank you for the tip. Although for this kind of question what Oli answered was more than enough. I'll keep that in mind for next questions! \$\endgroup\$– rMaeroAug 6, 2012 at 12:29
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\$\begingroup\$ Don't be too sure! There may exist 4017-compatible products that don't require the ten transistors + base resistors... \$\endgroup\$– stevenvhAug 6, 2012 at 12:52
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1\$\begingroup\$ The 4017's outputs are active-high. Don't you mean source instead of sink? \$\endgroup\$– stevenvhAug 7, 2012 at 17:55
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1\$\begingroup\$ That's right, when sourcing the current flows from Vcc to the output (from the IC to the outside). When sinking it flows from the outside into the IC's output to ground. \$\endgroup\$– stevenvhAug 9, 2012 at 17:49
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Looking at a couple of the datasheets (there are a few variants with slightly different specs), you would be much better to use a transistor as the source/sink capability is not very high.
For example the recommended operating conditions for the ST version give only 1mA source/sink typical at 5V and 25degC (whilst remaining within 0.4V of rails)
If you power it from 15V it rises to 6.8mA (3mA min), but that's still not great, considering a typical 5mm LED can be run up to 20mA (30mA in some cases)
Here's the relevant bit of the datasheet:
answered Aug 5, 2012 at 23:45
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\$\begingroup\$ I find it unusual that the datasheet lists both "Output Drive" and "Output Sink" currents, where the drive values are just inverted values of the sink currents. Why would they bother to do this instead of just listing Output Sink current only? \$\endgroup\$– JYeltonAug 20, 2013 at 21:46
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1\$\begingroup\$ @JYelton Because they can sink AND source current, and not all versions of the cd4017 have equal values for both. The TI/NatSemi version ti.com/lit/ds/snos357a/snos357a.pdf can sink ~3x more current than it can source, at equal voltages. Also notice that there is two 0/5 Drive rows, showing that when Vo is lower, drive current is higher. Voltage Sag/Droop. To keep Vo within 0.4v of a 5v VDD, current should be limited to 1mA at 25°c, but if you need more, like 3.2mA, Vo will drop to 2.5v. \$\endgroup\$– PasserbyAug 20, 2013 at 22:31
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\$\begingroup\$ @JYelton and the TI cd75hc4017 version can do 6mA at 6v VCC with the same 0.4v difference. ti.com/lit/ds/symlink/cd74hc4017.pdf \$\endgroup\$– PasserbyAug 20, 2013 at 22:42