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I am using an IC which is an RS422 driver chip. There are 4 drivers in silicon and each can be enabled using a mux kind of enabling system. The screen shot is below - Chip enable My doubt is - the input comes in Ain,Bin,Cin and Din. The differential outputs are given outside at D0,/D0,...etc. Now, each driver is enabled via the enable and /Enable pins. This means at any given instant, I can enable only 1 driver,right ?

For eg - 00 - driver A,01 - driver B,..and so on and so forth .

Hence-forth, If I want to drive 2 different signals I will need 2 different ICs.

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From the schematic you posted I would read it as:

If either enable pin is high OR (not enable) pin is low, all of the outputs are enabled.

The enable inputs are just wired in parallel to the output of the OR-gate. If the function you are imagine is implemented, the schematic is very misleading and there must be a clearer description around in the datasheet.


Update:

The datasheet is indeed very limited, but it gives a link to a more detailed description where you will find a truth-table:

truth table of the device in question

So you can see, there is no distinction between the different inputs or outputs made. So it will behave just as I read it from the schematic above and all outputs are enabled or disabled.

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  • \$\begingroup\$ The chip is HS1-26CLV31RH. The datasheet also states -"The HS-26CLV31RH, HS-26CLV31EH accept CMOS level inputs and converts them to differential outputs. Enable pins allow several devices to be connected to the same data source and addressed independently. These devices have unique outputs that become high impedance when the driver is disabled or powered-down, maintaining signal integrity in multi-driver applications". From the looks of it we cannot independantly control drivers. Itrs like either all on or all off as the output of the OR gate is connected to all. \$\endgroup\$ – Board-Man Oct 29 '15 at 10:37
  • \$\begingroup\$ @VUK that (and a link to the datasheet to make things easier for the helping people) will help greatly to provide better answers if you ask about specific devices, so put it in the question. I have updated my answer which proves my point. \$\endgroup\$ – Arsenal Oct 29 '15 at 10:49

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