I want to build a smart locker using this part https://www.aliexpress.com/item/32845920160/32845920160.html and a CD74HC4067 multiplexer. Essentially I need to turn on only one lock at time using a relay. I have seen that the lock draws 350mA @ 6 V. Am I risking to burn CD74HC4067 chip? Do you know if there is a multiplexer with higher currents than this?

  • \$\begingroup\$ Look for the datasheet of the chip. 350mA is a lot for a chip to drive. If only there was some kind of switch that could be driven by a current/voltage to control a higher current/voltage... \$\endgroup\$ – georgjz Mar 12 '18 at 22:27
  • \$\begingroup\$ You don't want to use anything similar (a multiplexer) for this task. You want a separate power supply for the locks (separate from the controlling electronics), and you either want a mechanical relay or a power transistor/MOSFET to actually switch the lock's current. The relay would automatically give you galvanic separation, and for the other option (transistor/MOSFET), I'd definitely insert an optocoupler into the control path to provide the same galvanic separation. \$\endgroup\$ – Laszlo Valko Mar 13 '18 at 3:26

The 4067 is an analog multiplexer, not a power switch. It is not designed to supply any significant current to its outputs -- the datasheet specifies an "absolute maximum" of:

DC Output Source or Sink Current per Output Pin: ±25 mA

This is because its intended application is for switching analog signals to high-impedance loads, like operational amplifiers or DACs.

A MOSFET or relay would be a more appropriate choice here.

  • \$\begingroup\$ Not to be picky, but what is your source for the quote? \$\endgroup\$ – Sparky256 Mar 12 '18 at 22:58
  • \$\begingroup\$ @Sparky256 From the datasheet. Page 3, near the bottom of the "Absolute Maximum Ratings" section. \$\endgroup\$ – duskwuff Mar 12 '18 at 22:59
  • \$\begingroup\$ Ok, but you should post the link in your answer. I have been 'chewed out' before for not listing the source of my quotes. Just a heads up. \$\endgroup\$ – Sparky256 Mar 12 '18 at 23:02
  • \$\begingroup\$ @Sparky256 Fair point. Edited. :) \$\endgroup\$ – duskwuff Mar 12 '18 at 23:03
  • \$\begingroup\$ The small FET switches in that analog mux, probably 100 micron on a side at most (to avoid wasting silicon area) will be just a few microns deep. The thermal time constant of 2 micron channels is about 50 nanoseconds; hence high currents lasting longer than a few nanosecond will simply charge up the thermal mass ---- the FET channel ---- and cause over temperature and likely silicon melting. \$\endgroup\$ – analogsystemsrf Mar 13 '18 at 3:44

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