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Suppose there is a really long track (not infinite) running over a 2-layer PCB with the bottom plane being the ground plane (just to complete the reverse path of the current). Now, on one end, 5V is supplied. But since the track is really long and the only one existing track on the PCB, there will be voltage loss with the length.

1. So, what can be done to maintain the voltage on the other end?

Can capacitors be used?

There won't be current loss since it is a single track.

But due to resistance of the tracks, there will be voltage loss to maintain the current and hence power loss too.

2. What can be done to maintain power?

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    \$\begingroup\$ Make the track bigger. Reduce current. Make the track shorter. Double the track on other layers. \$\endgroup\$
    – Hearth
    Commented Jan 18, 2023 at 22:40
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    \$\begingroup\$ The voltage drop also depends on the current. Don't make long thin traces that are supposed to carry high currents. \$\endgroup\$
    – Eugene Sh.
    Commented Jan 18, 2023 at 22:41
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    \$\begingroup\$ Is this a theoretical or more practical question? \$\endgroup\$
    – datenheim
    Commented Jan 18, 2023 at 23:03
  • \$\begingroup\$ You have defined no load on the track therefore the current is zero, the voltage is constant, and the power is zero. \$\endgroup\$
    – Tim Williams
    Commented Jan 18, 2023 at 23:33
  • \$\begingroup\$ If a single trace is inadequate to support the current load w/o voltage droop, and there's no room on the PCB for a fat, wide trace, you can simply use low-tech WIRE to jump the gap. You can drive quite alot of current through, say, an 18-gauge wire before the voltage loss becomes noticable. \$\endgroup\$
    – Kyle B
    Commented Jan 19, 2023 at 3:39

2 Answers 2

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  1. Capacitors are useless. You have the PCB trace resistance, and you have a load which can be modeled as resistance. That's a resistive divider and thus under load the capacitor would not even charge up to the voltage fed on the PCB trace input. A capacitor would help only with pulsed loads, assuming the capacitor is large enough to keep the voltage high enough for long enough to run the load.

To maintain voltage at the far end, you must use active electronics.

a) You can compensate by applying higher voltage to PCB trace so even after voltage drop it will be correct. This is a problem if the load is not fixed but variable.

a1) you can use active circuitry to measure current, and feeding in just the right amount of voltage which depends on used current, and for all currents the voltage at load is constant.

a2) you can use active circuitry to sense/measure the voltage at load and feed in higher voltage to end up with constant measured voltage at load.

b) you can feed in a voltage and regulate voltage at load, it can be stepped down or up.

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98% of the time, you just make the track bigger until the voltage drop is as low as you need it to be.

If you don't have space to make the track bigger, you may solder a wire in parallel, or even a busbar, or just deal with the fact your PCB is too small and make it bigger anyway.

This is almost always how excessive voltage drop is dealt with, even though there are alternatives that could work in theory, like putting a boost converter at the far end.

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