I am starting my EE degree. I must purchase a myDAQ and a kit. The kit lists many logic ICs and I have most of them, but some of what I have are 74S instead of 74LS series.
Will the extra power draw on the 74S ICs cause issues with the myDAQ?
or do I need to bite the bullet and buy the parts?
(on digikey, I am NOT paying their markup on their stupid kit).
The parts kit consists of things like or, and, xor etc…. gates and can be found here: https://www.studica.com/us/en/NI-Hardware-Only/und-student-ni-mydaq-bundle-ee202-electric-lab/796087.html
Edit In case anyone else runs into this in the future, it turns out not to matter for the UND EE201L course. We had minimal fan-out and most of the labs were just to use specified inputs to create specified outputs and make a led blink/turn on/ turn off. There were no high speed requirements and I wasn't driving high current loads, it was just an intro to logic gates and how to build circuits with them so I doubt it would have been an issue to use the 74S parts instead of the 74LS parts. It was a good learning experience though and the myDAQ is a decent tool to have in my tool box since I don't own a quality function generator (yet). Just in case, I ran power from my bench top power supply instead of the myDAQ
Best Answer
Good question. Here is a high level comparison of current consumption.
Note the difference in the values of \$I_{OL}\$ of two devices (for example). 74S has \$I_{OL}\$ of \$20 mA\$ where as 74S has \$I_{OL}\$ of \$8mA\$. The drive ability is different. If you are only doing this for a low speed simple circuits this will be okay but if there are experiments planned considering high drive strength of 74S, you might have to use a buffer.
I would suggest to really go ahead with the S version alone. the LS version are low power and better for a product design, but the S version will do just fine for basic circuits, which is planned. This is a hands on kit and hence i strongly recommend to use what you have and buy only those which you really need in future. IF there are circuits, which are pushing the LS version to their limits, you can handle it with buffers or buy.
I will compare the differences in the performance of the two devices SN74S04 and SN74LS04 and how and where it matters:
Current Consumption SN74LS04 consumes a maximum of \$6.6 mA\$ when outputting a logic zero compared to \$54 mA\$ max of SN74S04. It also means that, the internal temperature of the component will be higher. It also adds to overall current consumption of the product. It reduces battery life (wastes power) or demands higher capacity battery.
Ability of FANOUT the LS part has about \$8 mA\$ of \$I_{OL}\$ and \$0.4 mA\$ of \$I_{IL}\$. It means, when connected in a system, an LS IC can support upto \$ 8/0.4 = 20\$ devices. The S part on the other hand has \$I_{OL}\$ of \$20 mA\$ but \$I_{IL}\$of \$2 mA\$. Hence, the part can drive upto \$10\$ devices theoretically. Hence, LS part is better when there are multiple inputs driven by the single output pin. The fanout may demand perhaps a new buffer to support the fanout needed in S family when compared to LS family, adding to cost and size on the PCB.
Below is a simple comparison from http://www.ti.com/lit/sg/sdyu001ab/sdyu001ab.pdf which compares the applications of LS and S parts from TI (in general).