0
\$\begingroup\$

I have one logic 3v3-input (5v non tolerant) on my microcontroler and I want to interface 4 5v-inputs.

So my idea was to connect a multiplexer supplied in 3v3 but I have seen in datasheet (for exemple 74HC4052) that Vi (input voltage) must be GND and Vcc and Vcc can up to 10V.

What's happen if Vi=5V and Vcc=3v3? It burns or the Vo=3v3?

Should I use a digital multiplexer (5v tolerant)?

What is the best solution with only one component?

\$\endgroup\$
  • \$\begingroup\$ That part (74HC4052) doesn't do you any good at all. It just selects one set or another set of four inputs to four outputs. Plus it is an analog switch, it doesn't do any logic conversion at all. \$\endgroup\$ – tcrosley Mar 21 '15 at 17:52
1
\$\begingroup\$

Since you have only one input available, you will need a 4:1 mux.

The ISL43640 is a 4 to 1 analog multiplexor. It can tolerate inputs up to Vcc + 0.3 volts so you will want to operate it with a +5v supply. You will select which of the four inputs you want using the address lines ADD1 and ADD2, requiring two output lines from you microcontroller. The minimum high voltage for these is 2.4v, with a typical value of 1.4v, so you should not have a problem driving them with a 3.3v output.

enter image description here

The internal resistance of the mux is approximately 100 Ω @ 5v.

To create the level translator on the output, you could use a resistor divider with a ratio of 100 / 150 to get 150/250*5v = 3v. One could use the internal resistance of 100 ohms for the top half of the divider, and a 150 resistor for the bottom, but this would put a load of 250 ohms on the inputs, or 5 / 250 = 20 mA which is probably more than desired (since you didn't give any spec on the available current going into the inputs, which obviously depends on the output stages driving the four mux inputs).

So instead, using a divider of 1000 ohms and 1500 ohms drops the current required to 2 mA. Subtracting off the 100 ohms internal resistance, leaves 900 ohms for the top resistor (closest 1% is 909) and 1500 for the bottom, with the output of 1500/2509*5v = 2.98v taken from the junction (you should measure this first using a 5v in before tying it to your micro).

enter image description here

The miniscule amount of current required by you microcontroller's input pin will not affect the balance of the resistor divider.

The part is available from Digi-Key for $2.64.

\$\endgroup\$

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.