I'm looking at high channel multiplexers that can operate at voltages over 10V and for this component it says that the supply voltage is 12V which is okay for my application, but when you read under the section "Theory of Operation" it says "The valid input range for ±12 V supplies is ±5 V".

So what does this mean? Does it mean that when I supply the component with 12V it can only handle inputs of 5V?

Also, this component has specified input and output ports, they can't be both like this one from intersil? Am I correct?


1 Answer 1


Yes, it means that with a +/-12V supply, it can only handle inputs in the +/-5V range.

That seemed strange to me also at first (I did not know this device before your question), because it is specified everywhere that the output signal voltage range is +/10V with 12V supplies. Then, I saw that this device has a gain of 2 on all its outputs (it doubles all output signals). So it is consistent.

And, as you noticed, the AD chip doesn't have bidirectional input/output ports (you can't switch input and outputs). Most certainly because of this x2 amplification gain stage, by the way.

  • \$\begingroup\$ The thing about the intersil chip is that there is no specified inputs or outputs on the chip, the chip just connect the two ports and you may use it as you like. I could use the Y14 port as input and X5 as output, or I could do it the other way around, X5 as input and Y14 as output. The chip simply just connects the two signals and you can use them as you'd like. On the chip from Analog Devices however this doesn't seem to be the case. So if I understand your answer correctly I have to input a 5V signal on the input and then I will receive a 10V signal on a output.. \$\endgroup\$
    – Pethead
    Jul 20, 2016 at 11:13
  • \$\begingroup\$ Yes, if you put 5V on input, you'll have 10V on output. \$\endgroup\$
    – dim
    Jul 20, 2016 at 11:20

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