Minimal voltage drop over a 2m distance - Page 1
Forum section: electronics / beginners

author: robdejonge
author_rank: Regular Contributor
timestamp: October 20, 2021, 12:39:07 pm
content: Recently I boughthttps://www.eevblog.com/forum/testgear/entry-level-power-supply-korad-ka3005d-or-riden-rd6006/]a new power supply. For its depth, I have it up on a shelf so I keep my workspace clear.To get the power down to the workspace, I’d like to install a permanent wire from the output to a pair of binding posts. This wire would be maybe 2m/6ft in length.My question: how do I ensure there is minimal voltage drop over this distance, that the binding posts output as close to the PSU output as possible?My PSU outputs 5A max, so we’re not talking crazy thick wires here. But generally speaking best I could do is use much thicker cable, maybe 2 in parallel for each? Is that right?

author: Manul
author_rank: Super Contributor
timestamp: October 20, 2021, 01:08:01 pm
content: For minimal drop, you use as thick wire as possible. That's all. Nothing else you can do in this case. High end power supplies usually have remote voltage sensing, so they can effectively compensate any voltage drop. Not in this case.Depending on what kind of circuits you power, I would slightly worry about 2m of wire, because of it's inductance. I would probably add small electrolytic cap, like 22 - 47uF. Electrolytic cap helps suppress possible ringing.To figure voltage drop you can search resistance values of various gauge wires and calculate. You can do it experimentally too, just take a wire and run some current (in CC mode). Use multimeter to see actual voltage drop.

author: radiolistener
author_rank: Super Contributor
timestamp: October 20, 2021, 04:39:44 pm
content: Quote from: robdejonge on October 20, 2021, 12:39:07 pmit depends on what voltage drop you're want to see.For example 5 Amps, 2 meters (total length 4 meters) solid round pure copper wire:- diameter 0.1 mm, R = 8.546 Ω, voltage drop = 5 * 8.546 = 42.73 V- diameter 0.5 mm, R = 0.3418 Ω, voltage drop = 5 * 0.3418 = 1.71 V- diameter 1 mm, R = 0.08546 Ω, voltage drop = 5 * 0.08546 = 0.43 V- diameter 2 mm, R = 0.02137 Ω, voltage drop = 5 * 0.02137 = 0.11 V- diameter 3 mm, R = 0.009496 Ω, voltage drop = 5 * 0.009496 = 0.05 V- diameter 4 mm, R = 0.005341 Ω, voltage drop = 5 * 0.005341 = 0.03 V

author: Siwastaja
author_rank: Super Contributor
timestamp: October 20, 2021, 04:50:39 pm
content: Good lab supplies have separate SENSE pins which you can wire to the end of the power wires, eliminating the voltage drop (so that the supply automagically outputs higher voltage to compensate, exactly tracking the voltage drop in power wires).Failing that, just use thick enough cable for your needs.Wikipedia has a good table on American Wire Gauge which lists typically available wire gauges and their resistance per meter. Use Ohm's law U = R*I.When calculating, remember you have 4 meters of wire because the current is flowing back and forth and voltage is dropping both ways.Then the next question would be, how much voltage drop is tolerable? For a typical hobbyist working with a cheap supply, which isn't very well regulated anyway, I'd say 0.1V drop would be acceptable. Usually. For higher-end stuff, 10mV would be nice but then you'd be using a supply with SENSE wires anyway, not only to bypass the resistance of the cables, but resistance of theconnectionsas well, i.e., connecting the SENSE directly into your circuit.

author: TimFox
author_rank: Super Contributor
timestamp: October 20, 2021, 04:51:40 pm
content: If the voltage drop is critical (as in a precision power supply where you want to know the exact voltage at the binding posts), you should use external voltage sensing (if available on your power supply). The current-carrying leads should be reasonably heavy (see calculations in post above), but the sense leads only carry maybe 1 mA and are not critical. The restriction on many power supplies capable of external sense is that there is often a pair of protection diodes between the output and sense terminals inside the power supply for safety reasons if the sense leads are not connected properly, so the voltage drop across the current-carrying wires must be much less than the diode forward voltage.

author: Vovk_Z
author_rank: Super Contributor
timestamp: October 20, 2021, 05:12:52 pm
content: 2 meters long wires are quite long to possibly make large voltage drops. As for me, the maximum convenient wire cross-section is 2.5 square millimetres. Two 2 meter long 2.5 mm2 wires will have 0.027 R resistance. If it is too much or not for TS I don't know.

author: rstofer
author_rank: Super Contributor
timestamp: October 20, 2021, 06:41:57 pm
content: All wire has resistance and Ohm's Law holds.A #10 AWG copper wire at 25 deg C has a DC resistance of 1.018 Ohms/1000 feet. Your 2 meter extension is about 13 feet (round trip) and has a resistance of:1.018 / 1000 * 13 or 0.013 Ohms.10 Amps through 0.013 Ohms will drop 0.130 Volts.https://philatron.com/cable/copper/dc-copper-resistance.phpYou may find similar tables with more convenient units.BTW, a #10 AWG wire is usually rated for 30A in the electrical world.

author: Picuino
author_rank: Super Contributor
timestamp: October 20, 2021, 09:00:38 pm
content: I use audio cables, with many small wires. That improves resistance at high frequencies.Something like this:https://www.amazon.com/Bullz-Audio-BPES10-25-Gauge-Speaker/dp/B0149VC6KC/ref=sr_1_5?

author: TimFox
author_rank: Super Contributor
timestamp: October 20, 2021, 09:44:40 pm
content: Multiple uninsulated wires forming a "stranded wire" do not have less resistance at high frequencies than solid wire with the same total cross-section of copper. There are other advantages, such as mechanical flexibility."Litz wire" (q.v.) uses multipleinsulatedwires to form the cable, where the individual strands are only connected together at the ends of the wire.At frequencies where skin effect is important, Litz wire does have a larger useful area than solid wire.None of this is important at DC.At mains frequencies (50/60 Hz), skin effect can rear its ugly head at diameters exceeding, say, 20 mm. (The calculation is left as an exercise for the reader.)

author: radiolistener
author_rank: Super Contributor
timestamp: October 21, 2021, 12:48:31 am
content: Quote from: TimFox on October 20, 2021, 09:44:40 pmfor pure copper with conductivity σ = 5.96e7 [S/m] and relative permeability μ = 0.999991 at frequency f = 60 :ρ = 1/σ = 1 / 5.96e7 = 1.67785e-8 [Ω/m] (resistivity)δ = sqrt( ρ / (pi * f * μ0 * μ) ) = sqrt( 1.67785e-8 / (3.14159 * 60 * 1.25664e-06 * 0.999991) ) = 0.00842  = 8.42

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 21, 2021, 01:51:26 am
content: Put a capacitor in series with the resistor and then charge the capacitor (use a good 1 milifarad cap, just to get rid of all leak), then check the volt reading on the capacitor.That will tell you the voltage going through the resistor, the peak volts of the cap.

author: TimFox
author_rank: Super Contributor
timestamp: October 21, 2021, 02:49:40 am
content: Or use a normal procedure and measure the voltage across the wire with a voltmeter.Be sure to practice proper probing locations.No capacitors needed.What 1000 uF capacitor would you recommend that has negligible leakage?

author: TimFox
author_rank: Super Contributor
timestamp: October 21, 2021, 02:54:08 am
content: Quote from: radiolistener on October 21, 2021, 12:48:31 amTo get the current distribution within the wire with that skin depth, one goes to Bessel functions. There was an old rule of thumb that conductors (such as bus bars) used in 50/60 Hz distribution should be less than 3/4 inch thick, but rectangular cross sections are preferred over circular for that purpose.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 21, 2021, 02:55:17 am
content: I mean the leak through the megohm resistor of the voltmetre, when its connected to the cap, it will drain the cap, so you need a fair whack of farads to make sure u have enough time to read the value properly.What happens if you end up with one result using the cap (measuring the cap directly with the voltmetre), and another result without the cap, is that going to be wierd?

author: TimFox
author_rank: Super Contributor
timestamp: October 21, 2021, 02:59:23 am
content: No, the voltage measurement (done properly) will be correct and the result of your procedure is meaningless.

author: WattsThat
author_rank: Frequent Contributor
timestamp: October 21, 2021, 03:02:26 am
content: Quote from: Capernicus on October 21, 2021, 02:55:17 amThat’s called ESR. Perhaps weird to you. But then you’re the one suggesting the use of a capacitor where it looks suspicious like you’re trolling again.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 21, 2021, 03:19:59 am
content: Why dont you try it, you might be pleasantly suprised.I dont see how me suggesting things from my personal technique set is being a troll, I'm just trying to help!!!

author: WattsThat
author_rank: Frequent Contributor
timestamp: October 21, 2021, 04:30:06 am
content: Fair enough but as already stated, the method is impractical and that’s being kind. There is a time and place to use pulses to determine physical properties of a quantity under test but this isn’t one of those cases.Lets stick to simple, practical, real world examples that result in an solution for the OP, not a trip down a gopher hole that needs a heavy dose of advanced component behavior to explain, which voltage drop due to resistance rather than capacitor leakage, ESR, inductance and other things not yet considered. No need for $1 solution to a $0.01 problem.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 21, 2021, 04:35:01 am
content: I dont understand.Theres virtually no change in the setup, you just have to add a cap to the 2m or so wire, and check the volts on the cap. (Just make sure its 240v) seems pretty simple to me.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 21, 2021, 04:52:49 am
content: woops sorry... thats only if its dc. not ac. maybe test it with 9v instead. still pretty easy.

author: timenutgoblin
author_rank: Regular Contributor
timestamp: October 21, 2021, 05:06:05 am
content: Would it be worth considering using a twisted pair arrangement for the extended wire leads? What about adding a TVS diode at the output of the wiring to clamp any transients should they present themselves at any time?

author: james_s
author_rank: Super Contributor
timestamp: October 21, 2021, 06:27:04 am
content: If you can find a schematic for the power supply it might not be too hard to modify it to have remote sensing. Otherwise there is no magic that will help you, just Ohms law. You can easily calculate the drop you will get at 5A over various lengths of wire, don't forget to include both wires.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 21, 2021, 07:11:07 am
content: excuse me, I just have to apologize, I've got it messed up here. This is voltage drop due to current overblasting the resistor, and then the capacitor would go down in voltage, its too hard to test with ac, u need dc to test it out.sorry... I apologize.

author: TimFox
author_rank: Super Contributor
timestamp: October 21, 2021, 03:05:27 pm
content: Remember, the OP had a legitimate question about the DC voltage drop in 2 meter wires from the rear terminals of a laboratory power supply to a pair of binding posts located at a convenient location.The range of wires discussed in answers to his post had resistances around 0.01 to 0.1 ohms, which at 10 A would give a noticeable voltage drop of 0.1 to 1 V.The input resistance of a reasonably-priced hand-held DVM is 10 megohms, which has a negligible effect on the resistance for the 0.1 ohm case (10-7of total resistance or 0.1 ppm).On low-voltage scales (appropriate to < 1 V), a better DVM (such as a bench unit) has a very high input resistance of about 10 gigaohms.