How Far Can You Run 24Vdc Before You Lose Voltage

<...100ft long ... Can I run the dc voltage that far and upscale my wire size by a few gauges?>

Yes. 100ft is no distance for 24VDC control signals.

<Is there a good way to figure out voltage drop over a distance other then looking into wire resistance.>

No. That's the only way.

<The line is going to Canada and I am in the US>

Telemetry might be a better bet than cables over that sort of distance.

Does distance make a difference in the control voltage with a short or long distance?

That depends on the impedance of the control equipment. The higher the impedance the less the voltage drop. Most control circuits require a 4 to 20 milliamp supply and suffer an inconceivable voltage drop over any normal distance.

The only way to answer this question is to know the distance, the size of the wire,, and the inductive/capacitance resistive forces in effect.?

Quite.

Invitation declined.

It's not that hard:

http://www.powerstream.com/Wire_Size.htm

Another possibility is to use a DC/DC converter at your point of load. We use them to get a stable 28V output from an input that varies from 18 to 32V. Or send 120VAC across the room and put a small 24V power supply at the point of load.

A greater concern to me than voltage drop would be the inductance of that amount of cabling. I would definitely test this idea before implementing it.

Hello Guest,

You're not the first person to encounter this kind of problem. It is common enough that power supply manufacturers make power supplies which directly address this problem (for a small additional cost).

If you're willing to run four wires from the 24V DC supply to the load, there are power supplies commonly available that provide so-called 'sense' inputs. Two of the wires are, of course, the usual current-carrying wires connecting the supply to the load. The additional wires connect the 'sense' inputs to the load at the load. These wires carry negligible current and only that demanded by the 'sense' inputs, which have a fairly high input impedance. And as the sense lines carry little current, the voltage drop across them is insubstantial. 22-26 AWG wire for the sense lines is usually sufficient, although for long runs in an electrically-noisy environment they probably should be shielded (the shield being grounded at the supply end only).

A power supply which does not provide sense inputs regulates the output voltage as seen at its output pins. It knows nothing of the voltage which actually appears across the load. If the supply is connected to a distant load over a long length of wire, the load will see a possibly-significant voltage drop due to the wire resistance. Worse, the voltage drop will vary in step the load's current demands and in some cases will vary right out of spec. The supply, of course, knows nothing of this and can't do a thing about it.

However, a supply having sense inputs connected at the load regulates the voltage as seen by the load regardless of the voltage drop. Thus, if the load requires (say) 24V and the voltage drop is 2V, the load will still see 24V. To ensure that the load sees 24V and not 22V, the supply increases its output voltage - as seen at the supply's output pins - by two volts, to 26V. If the current demanded by the load varies, the voltage drop will vary as well, but the load will never see the variation (or very little of it) because the supply is regulating the voltage as seen directly at the load itself. It is this voltage that the supply is regulating, not the voltage at its own output pins. The supply does whatever it can to ensure that the load voltage remains within spec.

What this all means is that you needn't know (nor care ) about the actual wire resistance so long as the wire can safely carry the maximum expected load current (plus some, say 50% more) and so long as the voltage drop does not exceed the supply's ability to compensate for it (another term for this is 'compliance').

Check it out.

It is the current that matters not the voltage....

Upping the gauge of wire (also the number of strands) will reduce the voltage drop.

Is there a good way to figure out voltage drop over a distance other then looking into wire resistance.

NO! It's that nice Mr Ohm's law, volt drop is just Resistance x Current and there isn't anything you can do that will change that...

This is an easy problem but before I advise I am curious.

What is it you are doing that you are running wires and signals between two countries?

Have you considered legalities?

i am sure the question meant that the conveyer is being built in USA and will be shiiped to Canada...that makes your doubts clear..PLM !

At 100ft from US to Canada he could toss his goods over.

It is the current that matters not the voltage....

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True, Mr. Cat, and the wire resistance together with the current determines the voltage drop. But, Mr. Cat, the voltage does matter - at the load! (Lest we confuse our Guest) both current and voltage matter for the system as a whole. (I do believe you have a furball, Mr. Cat! )

Mr. Guest, a larger-dia wire will reduce the voltage drop, but it won't eliminate it entirely. Consider also that larger dia wire is more expensive. For longer runs the wire costs may exceed alternatives.

Let's say your load draws 3 amps, max, and is located 100 feet from the power supply. Also that your load's input voltage is spec'd at 24 V ± 5% and your power supply's output voltage is spec'd to be within 24 V ± 3%.

Designing for worst case means you don't have much headroom here. Tolerances themselves have implicit ranges, i.e., 3% could mean anything from 2.5% to 3.5% as far as you know. I might also point out here that a 3.0% tolerance is tighter than a 3% tolerance even though the numerical values are the same. A 3.0% tolerance implies a range of 2.95% to 3.05%; a lot better, IMHO.

So, let's say the lowest input voltage your load will accept is really 24V-4.5%, not 24V-5% as given (see foregoing). The minimum voltage is then 22.92V. Any lower than that and your load kicks the bucket.

Let's also say your (marginal) power supply output voltage is 24V-3.5%; that is, 23.16V. Both of these conditions together mean that you have a measley 0.24 volts to work with if you're gonna design for worst case. And just to be safe, your voltage drop at maximum load current should not exceed half that value, i.e., 0.12 volts! That ain't much, Bud. I'm guessing that you're gonna need some big wire just in case fate decides to conspire against you.

That being said, let's find the smallest gauge wire that, at 3 amps, incurs a voltage drop of 0.12 volts or less. Since the load is 100 feet from the supply, we're talking 200 feet of wire.

Wire resistance is often spec'd in units of ohms/1000 ft. (I'm deliberately ignoring metric specs here). 200 feet of our wire must have a resistance of 0.12V/3A = 0.04 ohms or less. Consequently, we need a wire gauge having 0.04 ohms x 1000 ft/200 ft ohms per 1000 feet, or 0.2 ohms/1000 ft. or less. We find from our wire table that the smallest-dia wire meeting our requirement is a #2 AWG at 0.15 ohm/1000 feet. At 3 amps this wire should incur a voltage drop of 0.15 x 3A x 200/1000 = 0.09 volts; less than our max of 0.12 volts.

200 feet of #2 wire is a lot of copper, Mr. Guest. Worse, (and even more importantly) you have to be sure to terminate the wire using very-low-resistance connections. What a mess! For my part, I'd be in the market for alternatives right about now. Look for power supplies having 'sense' inputs. I'd start there.

Hope this helps.

Cheers!

Nicely stated, clear & concise post, europium. The voltage drop problem is one reason why most nominal 24VDC instrument power supplies actually produce 28VDC when lightly loaded. You end up with a lot more flexibility on the sensor end of the line, and high voltage is not usually a problem.

I'm running 24 VDC on a Panalarm annunciator which monitors breaker positions at remote substations. Some of the contact points are 2-3 miles (4-6 round trip) from the alarm panel, across 24-26 AWG leased telephone lines.

If the sensor is a discrete signal (a limit switch or other binary device), and the I/O card is simply looking for bias voltage, there's very little current to cause significant voltage drop. On the other hand, if you intend to use a loop-powered potentiometer transducer to drive a 4-20 mADC loop, the voltage at the input terminals on the I/O card will be woefully inadequate.

Europium,

I have worked in a nuclear power plant where the extremely over-sized cable you described was required on a motorized valve. But, that application needed the rigorous engineering you also described. On a conveyor system, using typical components for sensors and discrete I/O to a PLC, your approach is way "out of specification", although its approach is valid.

Yes, good design will require looking at the specifications of the components. But typical 24-VDC input cards have a threshold nearer 16-VDC before they become uncertain in their "ON" response, and typically a lower threshold below which their response will be "OFF" (leaving a range of input voltages which will give an unpredictable response). The power drawn by typical photo-eyes or proximity switches is fairly low, and I have had no problems with using 12-AWG or 10-AWG for my 24-VDC supply to 20+ sensors at distances up to 300 feet from the panel. The individual signal wires back from these sensors can be #14 or even #16 AWG because the "ON" current drawn by the input card is very few milliamps. For outputs, to solenoid valves or similar devices, the common return wire chosen must be over-sized also because it is carrying the load for multiple outputs simultaneously. Here, the inrush as well as the sealed current for the valves must be considered.

If I were to draw up this control system, I would have a separate pair of power wires for every group of 16 input devices, and have them numbered uniquely. I would also have a separate common return wire for every group of 8 or 16 output devices, also numbered uniquely.

Another approach worth considering would be distributed I/O, in which the DC power supplies can be remotely mounted with the I/O and the only required signal cabling to the PLC would be a multi-drop serial link running the appropriate protocol.

--Thanks, JMM

jmueller writes: "...where the extremely over-sized cable you described was required on a motorized valve. But, that application needed the rigorous engineering you also described. On a conveyor system, using typical components for sensors and discrete I/O to a PLC, your approach is way "out of specification", although its approach is valid."

-----

Let us first compare apples to apples before we begin throwing around terms like "extremely over-sized" and 'way "out of specification."

Keep in mind here that mine is an example of the factors one should consider when selecting a wire gauge - especially under the tight constraints I cite in my example. There I specify a maximum load current of several amperes - not the few milliamps typical of the signalling applications you describe. My example also specifies tight tolerances and how they might combine under worst-case conditions to reduce headroom, and how these factors contribute to the choice of an appropriate gauge. The constraints I cite in my example are far more severe than those you cite in your post. A few milliamps vs several amperes? Eight volts of headroom vs 0.24 volts? 12-14 AWG vs 2 AWG? But "extremely oversized" and "way out of specification?" Hardly. Not for the conditions I describe. Do the math yourself and see what conclusions you draw. Chose a smaller-dia wire and I guarantee you're playing Russian Roulette with all chambers loaded.

Tell you what: instead of applying the conditions I describe in my example, apply the conditions you describe in your post; ie, a few milliamps, seven or eight volts of headroom, 300-foot wire runs, and so forth. Do you think your (far more relaxed) constraints might influence your choice of wire size somewhat? Had I used your examples rather than mine, do you think I would still have selected 2 AWG wire? What do you think? Of course you can get away with a smaller gauge in your application. The reasons for your choice of a smaller gauge are plainly obvious and only a fool would choose "extremely over-sized wire" for such purposes. On the other hand, an engineer worth his salt would still do the math.

Sir,

In no way do I dispute the conclusions contained in your earlier post and more recent reply. I also agree with your statement, "an engineer worth his salt would still do the math."

Your post uses a worst-case scenario to illustrate the valid educational point you make, that voltage and current both are important. As such, your "theoretical" answer neatly side-stepped the questions raised in the original post. My response was intended to be the "practical" reply to the original post.

Thanks--JMM

Let's look at the original post:

Guest writes: "I am designing a converoy (conveyer) line that is about 100ft long by 50ft wide. I am looking at how to run 24Vdc for dc controls ( inputs cards, valves, and so on)."

Is Guest speaking here of powering his dc controls with a 'remote' 24V supply, or is he speaking of low-current (milliamp-range) 24 V signals?

He doesn't actually say.

Guest writes: "Can I run the dc voltage that far..."

How far?

Guest continues: "...and upscale my wire size by a few gauges?"

How many gauges? Where he wants to go depends on where he is. What is his current wire size. And how might he determine the minimum gauge necessary to ensure reliable operation under worst-case conditions? Or should he even consider worst-case conditions?

<flame>

Any engineer who fails to accommodate worst-case conditions in his/her designs is an incompetent fool who should be fired on the spot, IMHO. These engineers are the kind whose designs destroy life and property. These are the engineers who design o-rings that work fine in fair weather and destroy Challengers when it turns a bit cold. The ones whose structural designs fail to take into account stresses and resonances in high winds. Bridges and buildings collapse, killing hundreds of people because the engineer designed for typical conditions only. These are the engineers who make the news, and history is replete with them and the disasters they cause.

</flame>

As the math seems a bit too "theoretical" to be of any practical use here, perhaps Guest should just throw a few more wire gauges at the problem, cross his fingers and pray that it works? Is this how Real controls engineers do it?

Guest asks: "Is there a good way to figure out voltage drop over a distance other then looking into wire resistance."

No. But there are ways to mitigate its impact on a design.

You are absolutely right, Mr. Europium. People who don't even have the slightest idea of what they are doing, should change job. And certainly don't use their (absence of) knowledge when their "designs" are gonna be used or are in the neighborhood of other people. Now in this specific case, probably no one is gonna be hurt directly. Of course when he uses his sensors and other equipment in e.g. a safety circuit then there could be injuries, even death.

Another risk is : when this "designer" sees that his 24V system seems to work, he might get brave and designs 240 or even 400V circuits. Then things could get really dangerous...

yes this is correct, the current flowing when combined with the wire resistance will tell you exactly what the voltage drop will be. Dont forget that it is typically twice the distance because there is a return trip for the current.

You need to look up the resistance per foot or calculate as it is typically given per 1000ft. Then Vdrop = (distance in ft over and back)/1000ft x (resistance per 1000 ft) x (current in Amps)

For a small control system the current might be 10mA (0.010A) , but check your own circuit. Often a digital signal is run though an opto-isolator and the transistor side is pulled up with a resistor - you need to understand this to get an accurate current , or put your DVM in current mode and measure it.

100ft with 18 AWG wire is fine if you have 10mA current for example. Lets test that out with 20mA current. 18AWG wire is 4.4906 Ω per 1000 ft . For 100ft one way run we get Vdrop = (100ft x 2) / 1000ft x 4.4906Ω/ft x 0.020A = 0.018 V drop

So 18mV drop is nothing of course. Even at 100mA which is 5x, its less than 0.1V drop so no problem. Often the wire gauge is selected for physical reasons such as pulling, or to provide extra strength in case it corrodes underground. Insulation etc is a whole other chat.

An amp of current is 50x so that would see 5V of drop which is a 20% loss of 24V which is bad but still might work depending on what you are doing and how good the receiving end is setup.

Is this where I say "Until your legs or Voltage get tired and give up??"

Sapper

I am scared when people who apparently don't know anything about electricity are going to design electrical systems. I would suggest to have a fire extinguisher ready at the first use of your "systems". But, in this case we're only talking about 24 Volts, so there is not immediatly a risk of injuries or severe accidents.

The calculation for determing the voltage loss in cables is as follows :

U = I x (l x Rho) / a

where

U = voltage drop in volts

I = current in Ampères

l = lenght of the cable (in meters)

Rho = the resistance of 1 meter of cable with a surface of 1mm² depends on material used for the cable. Copper has a Rho of 0,00175Ω. (can be found in tables)

a = surface of the used cable in mm²

Good Luck !

a = surface of the used cable in mm²

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More specifically the cross-sectional area.

The question is how far can you run 24vdc before you loose voltage, right? The answer is: you can not! Running it any distance at all will loose some voltage even if it is not measurable unless you are using a super conductor without any resistance at all. However, to operate 24vdc controls you usually don't need a lot of current as this is not the load that does the work, just the controlling. You don't really have a problem.

The answer is: you can not!

Hmmm.

On a philosophical level I would argue the converse...you can run it to the moon and back and lose no volts at all...you only lose volts when you try to measure or load it!

Mind..cable clips would present a bit of a problem .

Del

I stand corrected. In an open circuit you loose no voltage, but then again, would you actually be running it somewher if it were an open circuit?

Actually - even with an open circuit you will have some small current flow due to the charged particles in the solar wind, radiation, and whatnot. Best to shield your wires by encasing them in a water filled pipe of several feet diameter, and add plasma generators to equalize any accumulated space charge.

Of course, that leaves the potential being generated by the orbiting moon dragging the wires through the earth's magnetic field. So, better make those water pipes out of a superconductor.

Hi

Forever, if you don't draw any current.

Think about running 110 or 220 to the device and putting the transformer at the device. For most applications you can then ignore he math

BillZ

That would require a 'forever' open circuit which would make this entire discussion pointless.

I am guessing that the 24V is regulated down from the main power. If that is the case you might be better off running the higher voltage down the 1000 ft of cable at a lower current and stepping it down to a lower voltage at the load. That will reduce the wire guage requirement and the losses in the lines.

You could figure out the voltage drop by going ahead and hooking up the length of wire you think you will need to the amount of load you will be using and measure the voltage. This would not have to be the full length, just a length that would provide you with a measurable difference in voltage. This would give a conversion factor.

You people need something better to do.

This discussion is not even entertaing anymore let alone productive.

I agree it was a simple question and only needed a simple answer!

Engineers should be wary of beauracratic expansion of non existant problems

Engineers should be wary of beauracratic expansion of non existant problems

d'oh! There goes my raison d'être!!!

I guess that just leaves spelling.

I am designing a converoy line that is about 100ft long by 50ft wide. I am looking at how to run 24Vdc for dc controls ( inputs cards, valves, and so on).

REPLY

Only 100 feet movement but 50 feet wide. Sounds like a saw mill kind of conveyor. How much control circuitry on either end? Where is the control booth? Valves? What type as in Hydraulic controls for some things like gates barriers and clamps?

Which of the circuits are safety related as opposed to just operational, as in start, stop, jog, and reject flap diversion etc.

Keep the safety circuits seperated from the other control circuits so a failure in control does not inhibit safety.

BTW I assume you are aware that CSA may sometimes have different standards for your type of equipment than UL. We sometimes find US built equipment does not meet with our Canadian requirements and the inspectors will require the equipment to be modified before it is put into service. Many things are the same, but there are always exceptions where mill, factory, or mining equipment is concerned.

The big question would be how much current has to flow over that distance. You are probably better off placing a mains powered 24V DC supply at the far end for raw control power of solenoid valves etc and keep your data lines separated. More than likely there will be lights at both ends so you are already running power lines to both ends of that 100 foot conveyor line.

Question to your statement "The line is going to Canada and I am in the US"
Do you mean the Conveyor Line you are designing will be built in the USA and used in Canada and that you are sitting in an American Design office ?
As somebody here pointed out that Canadian Regulations are different to US regs. it might be worthwhile hiring in a Canadian electrical / electronics Engineer as a short-term Contractor to design the necessary electrical circuits to ensure they meet the Canadian regulations.
It would be a cheaper solution than you designing everything in the US using US regulated parts and then finding they don't meet Canadian regulations.
Do you even know what Voltage is available on site ? To my limited knowledge 3 Phase Voltage in Canada varies between 208v and 600v.

Where will the Control Panel be situated ? for example if you place it in the middle of the Belt you will only have a 50 foot cable run but if its located in a building somewhere remote you could have 100's of feet cable or more.