Voltage drop on ground - long cable

Design and Layout Schematic and layout
1

Consider a power supply physically located at 4 meter away from the PCB. The power cables consist of the main power cable and the ground cable for return current.

If the PCB needs 2 A at 3.3 V. How do we account the voltage drop across the cables.

Do we only consider the voltage drop through 4 meter power cable ? or also the voltage drop through 4 meter power cable ?

The output voltage at the power supply has to be 3.3 V + the voltage drop through 4 meter cable.

Do we also need to consider the voltage drop through the ground (return current) cable ?

2

The current interacts with the resistance of the cable and creates a voltage drop that is directly proportional to the current. This is just as true of the 3.3V supply as it is of the 0V return, and when you do the maths it is obvious that it always works against you.

So let’s do some maths. Suppose that the 3.3V wire has a resistance of 0.1 Ohms (this is only an illustration, the real figure you will need to look up on-line), and also the 0V wire has a resistance of 0.1 Ohms. If the current is 2A, then the voltage lost across the length of the wire is 0.1 * 2 = 0.2V.

If the PSU is set to 3.3V, then the PCB will see 3.3 - 0.2 = 3.1V on the 3.3V input. But there is the same current flowing back into the 0V on the PSU, so at the PSU, 0V is 0V, but at the PCB it sees 0 - (-0.2) = +0.2V (the -0.2 is negative because the current flow is in the opposite direction).

But what voltage did your PCB see? 3.1V - 0.2V = 2.9V which may not be enough.

What can you do about it? One answer is increase the output voltage on the PSU. In this example you have lost 0.4V in the cabling. If you turn up the PSU to 3.7V, then your circuit will see 3.3V.

But there’s a problem. When you engage “sleep” mode on your board, the current drops to practically nothing, so the voltage drop in the cables becomes virtually nothing, which means your circuit now sees 3.7V. This may not be desireable!

There are solutions to this. If your cable has a resistance of 0.1 Ohms over the 4 metre length, a thicker cable will have a lower resistance, which results in a lower voltage loss across the length of the cable. Changing to a 0.01 Ohms cable means your circuit now sees 3.26V which may be “good enough”.

The best solution is that lab PSUs also have voltage sense inputs for both the positive and negative power outputs. These are usually joined together by a metal strap. You can disconnect the metal strap and now these pins become available to remotely sense the voltage at the load. For this to work you need 4 wires (not 2) to link your board to the PSU. You will have 3.3V to PSU positive Power terminal, 3.3V to PSU positive Sense terminal, 0V to PSU negative Power terminal and 0V to PSU negative sense terminal. This way the PSU compensates for the loss in the cable automatically and thus keeps the voltage at 3.3V on your board.

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3

Thanks for your reply. How do we design the remote sense at the power supply. The power supply I have in the deign is not a standard commercial power supply. Actually the power supply which is located at 4 meters from the PCB is DC-DC converter. How can I add the feature of remote sensing at the DC-DC converter such that the voltage at the PCB remains at 3.3 V and the ground at 0 V for different currents through the long cables.

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4

If you have designed your own DC-DC converter, there will be a feedback signal so that the controller can adjust the voltage to the right level. This feedback is normally internal to the design on the assumption that the losses between the point where the feedback is taken from and the point where it is used can be ignored. However in your case, they can’t be ignored.

The principle is that you take the feedback from the target device, but this is not as simple as it sounds. You need to know how much you are losing in both the supply cable to the load, and also how much you are losing in the return cable back to the power supply. Whatever voltages you see here have then got to be subtracted from the output level at the feedback network so that the controller will raise the voltage to the correct level for that amount of current.

The bit that makes this tricky is that the feedback network is part of a closed-loop negative feedback system, and long cables and the associate time delays will make most designs unstable. Then there is the problem of overcoming the possibilities of what could occur if one of these feedback cables became disconnected (usually the output voltage rises in an uncontrolled way). The solution to this is to have a high impedance bias to the main output as seen at the PSU, but have a far lower impedance route from the remote sensing point, thus the remote point dominates when it is connected. Much the same is needed for the 0V return.

To make things worse, sensing at 4m distance means you cannot use a 4-core cable (i.e. power, return, feedback+, feedback-) because the coupling between the adjacent cores within the cable affect the feedback signals. That means that the feedback needs to come back as a separate shielded 2-core cable, and this must be well made (pigtail shield connections will ruin it).

In short, this is a non-trivial thing to design. If there is no way of moving the DC-DC power supply close enough to the point where the 3.3V supply is used, another solution is needed. The only solution left to you is simply to reduce cable loss by using Ohms law. Reducing the voltage loss by reducing the resistance of the wire becomes the only easy answer. Of course, you will need thicker cable which will cost more, but unless you’re going to redesign your power supply for remote monitoring, or relocate the DC-DC converter, there aren’t many other options.

One other option that might exist if you can change the design for the load is to make it adaptable to whatever input voltage it receives (within limits). If your circuit is a 3.3V circuit and does not have a regulator at the power input, you could modify it so that you put a SEPIC converter in the load at the point where the power is going to be used. A SEPIC converter is a type of switching regulator that takes an input that could be lower than the required level, in which case it raises the voltage; or higher than required, in which case it lowers it, or it could be just right, in which case it stays the same. These converters are readily available from Analog Devices, Texas Instruments and many other companies. Of course there will still be power loss in the cable, and power loss in the SEPIC regulator, so the actual current drawn from the supply will be higher than 2A to make up for this.

The one other obvious remark is that the SEPIC converter does not compensate for the fact that 0V at the DC-DC power supply (4m away) will not be the same as 0V at the point where the power is used. This may affect things if you have any DC-coupled signaling that pass between these points. The IR drop in the original cable is still present, and only you can answer whether that is going to be an issue or not.

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