0
\$\begingroup\$

I am using Arduino Nano for driving TMUX1208. The address lines, Enable and I/P are coming from Arduino Nano.

The output of the MUX is used for driving a BJT which is used for driving a relay. The Vil of TMUX1208 is 0.87V and the Vol of Arduino nano is 0.9V(IOL=20mA, VCC = 5V).

May I know will cause any issues in my circuit.I want the transistors(MMSS8050-H-TP) to be off when EN is low.

The input leakage current of TMUX1208 is ±0.005uA. So I assume Arduino Nano is Vol will never go to 0.9V. Please correct me if I am wrong.

MY CIRCUIT

enter image description here

DC SPEC of Arduino Nano

enter image description here

DC Spec of TMUX1208 enter image description here

\$\endgroup\$
7
  • \$\begingroup\$ You could’ve used something like a tpic6b595 or 238. \$\endgroup\$
    – Kartman
    Dec 30, 2021 at 8:53
  • \$\begingroup\$ I am not in a situation to change the MUX.If I put a diode in the EN line ,will it be okay \$\endgroup\$
    – HARI T O
    Dec 30, 2021 at 9:00
  • \$\begingroup\$ Using the mux is a strange design choice. Rethink now before you get too far down the road. Why would you want a diode in the EN line? What is the input current to EN? Miniscule. If it were 20mA, then there would be cause for concern. \$\endgroup\$
    – Kartman
    Dec 30, 2021 at 9:06
  • 2
    \$\begingroup\$ The Vol spec is 0.9V MAX with a 20mA load. You are not loading the output anywhere near 20mA, so the output voltage will be less than 0.9V. The mega328 outputs are mosfets, when turned on they appear as a resistance. You can estimate the resistance by solving R= 0.9/0.02 = ~50Ohms. \$\endgroup\$
    – Kartman
    Dec 30, 2021 at 9:21
  • 1
    \$\begingroup\$ That is an estimation. This is not specified or guaranteed by the manufacturer. Use your knowlege of mosfets to draw your own conclusion. You can also perform your own tests to validate the assertion. I think you'll find the value is close enough. \$\endgroup\$
    – Kartman
    Dec 30, 2021 at 10:21

1 Answer 1

1
\$\begingroup\$

Literally, your circuit is right. No diodes needed. I know what you mean, you thought to add a diode before EN of TMUX1208 to step up a little trigger threshold voltage.
Don’t do that, makes no sense. According to the basic circuit theory, if your Arduino is powered by 5V, the the ripple must less than 5%. So here it is, 5V*5%=250mV. But you never get 250mV noise. Make sure TMUX1208 has enough output current to drive a BJT.

\$\endgroup\$
4
  • \$\begingroup\$ ,Thank you. The thing which confused me is The Vil of TMUX1208 is 0.87V and the Vol of Arduino nano is 0.9V(IOL=20mA, VCC = 5V).Hope that won't be a problem.TMUX1208 can provide 30mA output current.That is enough to drive BJT \$\endgroup\$
    – HARI T O
    Dec 30, 2021 at 9:54
  • 1
    \$\begingroup\$ @HARITO, the device is an analog mux, it switches the current not provides it. \$\endgroup\$
    – Kartman
    Dec 30, 2021 at 10:23
  • \$\begingroup\$ @Kartman,Thank you.So this circuit will work properly.Correct me if I am wrong \$\endgroup\$
    – HARI T O
    Dec 30, 2021 at 11:37
  • \$\begingroup\$ Hopefully we provide you with the facts. You make the decision. \$\endgroup\$
    – Kartman
    Dec 30, 2021 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.