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I have two ICs, one on board A and one on Board B. Each of the ICs are powered by 5V on their respective boards that are generated by 12V that each board gets.

The concern is that each of the boards has its own separate 12V to 5V regulator.

How do I make sure the high levels are correct when IC on board A drives a signal to IC on board B?

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  • \$\begingroup\$ Under every circumstance, what is the AC and DC difference in DC, 0V and how is 0V (Ground) bonded together? What is the function, distance, spectrum or BW, current on "signal" \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 25 '18 at 0:44
  • \$\begingroup\$ different ICs might have different logic levels as some IC have internal LDOs, so it would be easier to answer you if you provide the IC models here or you could look into their respective datasheet to see what is the threshold voltage and maximum allowable voltage for the logic inputs. \$\endgroup\$ – Zy Gan Sep 25 '18 at 2:39
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As long as the ground is the same for boards A and B it'll work fine. If not, use optocouplers.

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You need to make sure that the low levels are correct too.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A connection (shown in green) is required between the GND of each circuit to give a common reference for signals passed between them.

Note that while the power supply negatives are directly connected to each other through the ground circuit that there is no direct connection between the supply positives. Therefore there is no risk of high voltages or currents between the two devices.

If your logic levels are the same (3.3 V or 5 V) on the out and in ports there should be no problem with this arrangement. The common ground gives a common voltage reference for the signal being transferred. Assuming a 5 V system, when OUT switches high there will be 5 V between OUT and GND and there will be 5 V between in and ground. As mentioned at the top, when OUT switches low there will be 0 V (or close to) on OUT and 0 V on IN. All will be well.

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    \$\begingroup\$ One risk: If one board is powered while the other is not powered, sending a signal from one to the other could cause damage. \$\endgroup\$ – The Photon Sep 24 '18 at 23:12

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