Voltage drop on return line

[joniengr081]

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

If the PCB needs 2 A at 3.3 V. Do we consider only the voltage drop through 4 meter cable ?

The voltage to be set at the power supply would be 3.3 V + voltage drop through 4 meter cable. Do we also need to consider the voltage drop through the return current cable ?

[Siwastaja]

Of course. Other things to consider:
* Is there ground-referenced signalling to other devices? If yes, is it acceptable that ground potentials are so much different?
* If you set the power supply to higher voltage to compensate for the drop, can you guarantee your load current is always on, and constant? If not, the device is seeing overvoltage during lower consumption (e.g., before starting operations). Can it handle this overvoltage?

For very low voltages such as 3.3V, local regulation is usually the solution. Make the board accept 5V (e.g., accept range of 4 to 6V), which is also easily available from USB power supplies, and use 3.3V regulators on-board.

[joniengr081]

With 4 meter cable between the PCB and the power supply. Would there be a voltage drop on the cables. The answer is yes. Now the question is how much in numbers.

V_drop = Current (2 A) * Cable Resistance (4 meter or 8 meter)

[Siwastaja]

U = R * I, where R is the resistance of the 8 meters of wire.

And, for example, if the total voltage drop is 0.4V, then the device would see 2.9V over it, with the negative input 0.2V higher than the power supply -, and positive input 0.2V lower than the power supply +.

Note that in reality cable resistance is not a finely controlled parameter - even if you buy wire rated to exact same cross-sectional area, resistance could easily vary by tens of %, so you get the exact answer by taking a measurement.

Also remember that contacts (connectors, screw terminals etc) also have resistance, something that might produce a significant voltage drop, too.

[joniengr081]

I am getting it now.

Let's suppose there is 0.4 V total voltage drop for 2 A current on a 2 meter cable of given given resistance. The 0.2 V will drop on +ve supply cable and the remaining 0.2 V will drop on the return current cable.

In other words on the PCB the positive supply connector will be 0.2 V lower then the power supply +ve and the return current connector will be 0.2 V higher then the power supply -ve.

What setting do I need at the power supply to get 3.3 V on the PCB.

3.3 V+ 0.4 V= 3.7 V ?

This way the positive supply connector on the PCB will be 3.5 V and the return current on the PCB will be 0.2 V. The voltage difference on the PCB will be 3.3 V. But the absolute zero voltage will be on the power supply connector, not on the PCB.

[Ian.M]

Does the power supply support remote sensing, and is its output floating with respect to ground? 

If so you simply run a pair of thin sense wires next to your + and - power cables, connect them to the power cables at the far end (your PCB), and to the PSU's sense terminals and the PSU will take care of adjusting its actual output voltage to maintain a regulated 3.3V (or whatever you set) at the far PCB.  See the PSU's manual for details, including any jumpers or links that may need to be removed to use remote sensing.  Don't forget to put them back afterwards so you don't get a nasty surprise next time you use the PSU *without* remote sensing.

If the PSU is *NOT* floating with respect to ground, and your PCB connects to anything else grounded or you try to test it with an ordinary scope probe, you can run into the ground potential difference issue Siwastaja mentioned.  This can vary from mildly compromising noise immunity, through stopping stuff working, up to causing actual damage to your PCB or grounded stuff connected to it.

[joniengr081]

I am not sure if our power supply have remote sensing option.

[Ian.M]

So examine it looking for sense terminals (usually clearly marked) or RTFM and see if it mentions remote sensing anywhere.

[selcuk]

Is 2A load a resistive heater or a decent circuit? 4 meter cable is not a pure resistor. It is resistor + inductor + capacitor + RF antenna. You may get interesting effects on your circuit.

[tunk]

What's your device?
If you know the square area of the cable, then it's simple
to calculate it's resistance using the resistivity of copper.

[Ian.M]

Is 2A load a resistive heater or a decent circuit? 4 meter cable is not a pure resistor. It is resistor + inductor + capacitor + RF antenna. You may get interesting effects on your circuit.

Yes.  Even with remote sensing, if your PCB is anything with logic ICs or a MCU, you may need additional decoupling capacitance at the PCB end if it hasn't been designed with adequate on-board bulk decoupling.    Keep the power leads loosely twisted together to minimise induced EMI pickup.

What's your device?
If you know the square area of the cable, then it's simple
to calculate it's resistance using the resistivity of copper.

If you don't know the Cross Section Area (CSA) of stranded wire, you can measure the diameter of one strand with a micrometer, calculate the CSA of the strand (=D²𝝅/4), and multiply by the strand count.  Alternatively, due to how the strands pack together, its approximately 90% of the area calculated from the overall core diameter.

Unfortunately nowadays there is so much dodgy wire that contains low grade copper or even that isn't copper but has a thin copper coating to fool the unwary, so unless you bought your wire from a reputable distributer that can trace each individual lot back to the manufacturer, you may well find that using the standard resistivity for electrical grade copper seriously underestimates the voltage drop!    If the calculated and measured voltage drops don't match up, your DMM is not out of calibration and you've accounted for thermal EMFs, you've got fake or crappy copper wire, not fit for purpose!

[joniengr081]

The PCB will have analog and digital ICs and MCU. The remote sensing at the power supply will set the power supply output higher then 3.3 V, right ?  But the ground on the PCB can not be at 0 V. How can we make it at 0 V potential.

[PGPG]

The PCB will have analog and digital ICs and MCU.

So it will probably not consume continuously constant current. With current fluctuations the voltage drop at cables will also fluctuate.

The remote sensing at the power supply will set the power supply output higher then 3.3 V, right ?

Yes. It will set supply output to ensure 3V3 at the place where sensing wires are connected. With changing current the output voltage will be adequately changed to ensure 3V3 where it is needed.

But the ground on the PCB can not be at 0 V. How can we make it at 0 V potential.

Where we have 0V it depends on our decision. There are no something like absolute 0V. Even local Ground potential changes (sometimes high and rapidly - consider what happens during thunder strike).
So you need only assume that you have 0V at your PCB and you have some negative voltage at power supply output.

[TimFox]

Remote sensing can be very useful for high currents.
However, watch out for the details of the four-terminal connection:  many power supplies have a diode between each sense terminal and its corresponding output terminal, to prevent the supply generating too much voltage when the sense connection is open.
With such a circuit, you must avoid excessive voltage drop along each high-current wire that results in its diode going forward-bias.
Normally, the current in each sense lead is small (maybe 1 mA), so its wire can be much smaller than the high-current output wire.

[Siwastaja]

The PCB will have analog and digital ICs and MCU.

Mixed signal (digital + analog) PCB drawing 2A at 3.3V, supplied with 4 meters of cable - forget about directly supplying it with 3.3V. Redesign the board with a voltage regulator, and supply with 5V (or 12V or whatever). Higher supply voltage to the board reduces current, reducing voltage drop, but also allows you to have more variation in the voltage.

Most 3.3V rated ICs are 3.6V recommended maximum. If you try to compensate for voltage drops by outputting higher voltage, you easily exceed that. Plus parasitic inductance of the supply cable can be a problem.

Local regulation makes life easier.