I am using an analog MUX/DMUX (74LV4051D,118) for enabling/disabling a transistor that is driving a reed relay.

The MUX_DMUX selection lines are controlled by Arduino Nano. The common input/output is connected to Vcc using a pull-up resistor of value 3.32K.The relay P/N is 9011-05-11.

My circuit diagram is given below. May I know the below circuit is fine or it will cause any trouble?

enter image description here

  • \$\begingroup\$ The Arduino should be able to drive R47 directly - shouldn't need the mux (or R48 for that matter) at all. \$\endgroup\$
    – td127
    Dec 27, 2021 at 18:41

1 Answer 1


This will work just fine although you should keep in mind that R57 will be in series with R47. You can leave out R57 and just connect the Z pin directly to Vcc.

An even simpler solution would be the 74HC238, which is a digital demultiplexer (instead of an analog one). You could even eliminate the transistor with a 74HC238; the chip is powerful enough to drive your relay directly. Just connect it between the chip's output and ground.

Of course, don't forget to connect GND and VEE to ground (they're unconnected in your schematic).

  • \$\begingroup\$ Thank you.Driving a relay using 74HC238 means connecting one end of the coil to 5V supply and other end to output of 74HC238.please correct me if I am wrong \$\endgroup\$
    – HARI T O
    Dec 27, 2021 at 16:55
  • 1
    \$\begingroup\$ With the 74HC238, you have to connect the coil between the chip's output and ground. A free-wheeling diode is not needed; the chip's built-in protection diodes can handle the ~10mA free-wheeling current from your relay. \$\endgroup\$ Dec 27, 2021 at 17:21

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