Electronic – Detect logic signal in a long distance. What type of interface to use

In general, trying to push final regulated power any distance is not a good idea. In your case it clearly won't work. Yes, the return path adds to the total resistance since it is in series with the load. It is strange that you have connectors in the positive supply but not in the ground. If this is a fixed installation, then why not solder wires from one end to the other?

A better way to deal with the need for distributed regulated power, especially at low voltage and high currents as you have, is to distribute a higher roughly regulated voltage and make the final tightly regulated voltage locally. This does two useful things:

1. The drop in distributing the higher voltage won't matter since is will be regulated anyway to the final voltage. You do have to make sure the voltage at the other is at least the minimum required for that regulator to work correctly, but that headroom is usually easy to build in.

2. In the case of local regulators being switchers, the higher voltage will have less current, which means it will also have less voltage drop accross the distance, with less power wasted and heat that must be dealt with.

So where does your 1.2V supply come from? You probably have some higher voltage with a buck converter somewhere. Send that higher voltage over the distance and put a buck regulator right at the DSP. Note that this relaxes the requirements on the 1.2V supply on the main board. Two smaller buck regulators will still be more expensive than one larger one, but allowing both to be smaller will help somewhat. It also distributes the heat from any losses, which usually makes that easier to deal with.

Added in response to your comment:

If you really really can't put a local regulator by the load, then the next best thing is to have a sense line coming back. This line reports that actual voltage at the far end back to the regulator on the main board. This voltage is used as feedback so that the voltage at the far end is what is regulated. The voltage at the regulator then will automatically be higher as needed to overcome the voltage drop on the way to the load. The sense line doesn't experience these voltage drops since it has very little current flowing thru it. It is just a voltage feedback signal.

If the ground connection also can have significant voltage drop, then it gets more tricky. Sometimes you use two sense lines and treat them differentially at the power supply. Sometimes you assume the forward and backward voltage drops will be about equal and add a little bit of gain in the sense circuit. Sometimes you just set the output of the supply a little higher to compensate for the nominal total voltage drop and not try to actively regulate around it at all.

Your first circuit is perfect... If you use a mosfet instead of a bjt. A mosfet does not pull anything low when it is on, it indeed is a capacitive load but you are going at 20kHz so that should not be an issue at all. Just be careful: your mosfet should support a quite high \$V_{GS}\$, i.e. 12V, that's not something all MOS are happy to do.

Your circuit should look like this:

^{simulate this circuit – Schematic created using CircuitLab}