3
\$\begingroup\$

I have a VBus of 12VDC. I would like to take 14 discrete inputs and Light a single LED for each input that is active.

I was planning on just doing a simple current limiting resistor and having a common anode and allowing the inputs to switch ground.

A new requirement was added: If any of the inputs are active they want to switch on a 500mA load on. (ie a buzzer or light). I would most likely do this with a NFET.

I would like to try and do this as cheaply as possible.

If i use a micro then I need to generate 5v. I could probably do a micro / powersupply / and either optos or resistor divider for the inputs this would cost around $8.00 per board.

Can anyone suggest any easier / cheaper way to mux the LED inputs together?

enter image description here

\$\endgroup\$
3
\$\begingroup\$

If I understand correctly you're still using each of the signals to drive the LEDs and need to combine them to turn on a single buzzer or light. You can do this with buffers (or inverters for active low signals, not shown). Simply pass each signal through a buffer and combine the outputs to drive the single load device. A discrete component view might look something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

You can use three CD4050Bs (HEX buffers). This will provide easy expansion for up to 18 inputs (I would put them all on the PCB and, for now, only populate the 14 that will be needed). If your signals are active low, use three CD4049UBs (HEX inverters) instead and hook up the LEDs as you've drawn them in your schematic.

\$\endgroup\$
8
  • \$\begingroup\$ Thanks! I am going to try this! I will also make sure that I state my questions more clearly. \$\endgroup\$
    – EE_PCB
    Sep 10 '13 at 16:03
  • \$\begingroup\$ The 7404 or 7407 still requires a max vcc of 7vdc is there any parts out there that would be able to take in the 12v bus w/ out modification? I could use a voltage divider for that or a 7805 \$\endgroup\$
    – EE_PCB
    Sep 10 '13 at 16:10
  • 1
    \$\begingroup\$ Oh, my mistake. There are certainly such parts. No need to invoke a regulator for this solution. Just use something like a CD4050B instead. Edited answer. \$\endgroup\$
    – Samuel
    Sep 10 '13 at 16:26
  • 1
    \$\begingroup\$ @Samuel 14 inputs - only 2 hex buffers, what happened to the last two inputs? 7407s only work up to 7V and are open collectors - you need a pull up resistor for each. This circuit will not work. Substitute CMOS 4000 series logic e.g. 4049 hex inverter. \$\endgroup\$ Sep 10 '13 at 16:28
  • 1
    \$\begingroup\$ @Samuel Read them just after I posted. It was simply a head's up to get you to edit your answer and get it sorted before folk on here start mark it down. Once the answer has the green tick you can't delete. \$\endgroup\$ Sep 10 '13 at 16:36
0
\$\begingroup\$

I am assuming (a bit difficulty understanding) that you want 14 LEDs, one for each input? You could use something like this:

http://www.intersil.com/content/dam/Intersil/documents/fn72/fn7288.pdf

Just use 4 of them, they're cheap and reliable. Mouser has some that run 6 drivers off of 12V, but I would go with the 7457 as I have had a lot of success with it.

edit: Because you want one for one, you really don't need a MUX.

\$\endgroup\$
2
  • \$\begingroup\$ I am sorry if i was not clear enough. For each input there is a corresponding LED that should be turned on. Then there is one single output that should be asserted if ANY of the LEDs are asserted. That is what I was looking for the mux for. \$\endgroup\$
    – EE_PCB
    Sep 10 '13 at 15:45
  • \$\begingroup\$ Are they TTL level signals (Or just that 12V bus?) If just TTL, you could just break them off before the quad drivers (or whatever driver you choose), put them into a 14 input OR gate array, then buffer the output through one of the two extra driver gates you would have (if you used my drivers) edit: The others suggesting that a pullup would work are correct. I'm was actually just assuming that your signal bus didn't want to supply all the current and that you maybe had a power bus in addition. You may want to edit your question with more system specs. \$\endgroup\$
    – scld
    Sep 10 '13 at 15:51
0
\$\begingroup\$

If I understand you correctly you have 14 switches, one side to ground, the other side to a resistor, then a LED, then 12V. So when you push a switch, the corresponding LED lights up.

Now you want to add a buzzer that buzzes when ANY switch is pressed?

Just connect 14 diodes, each diode connects to a switch and (all together) to the buzzer, other buzzer wire goes to 12V.

\$\endgroup\$

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