DC vs. AC voltage drop

System

I have a question about the voltage drop on a DC circuit vs. an AC circuit.

If the runs are the exact same length, and the wire gauge is the exact same size, and the voltage in both circuits are the same, will the voltage drops over the DC circuit and AC circuit be the same?

Thanks.

BB.

Re: DC vs. AC voltage drop

You did leave out one parameter... Current.

For DC, Vdrop=Current*Resistance of wire/circut.

Assuming you are talking just about the wiring (no capacitors, inductors, transformers, etc.) for low frequency AC in "short" runs (like home RE systems--not major utility sized power systems), the voltage drop between AC and DC, at the same current, will be the same.

However, there are many other issues that affect home AC/DC questions... The biggest is that DC current is much more difficult to turn off/interrupt. DC sustains arcs much better than AC--so DC switches are typically larger/heavier duty vs a similarly rated AC switch/breaker. Or, if you have a switch rated for both AC and DC--the DC voltage and current rating will be much lower.

There are other issues--such and electrolysis... If you have DC leakage current, it can corrode underground piping and such.

-Bill

BB.

Re: DC vs. AC voltage drop

I guess I should add there is also another issue with AC that you don't normally see with DC...

The fact that AC power has a "phase angle" between voltage and current... For example, you can put a 1 hp motor on a DC circuit and the P=I*V works out just fine.

For an AC circuit, the Current and Voltage have both amplitude and angles. That 1 HP motor may take (excluding efficiency losses):

P=I*V
I=P/V=746 watts / 120 volts = 6.21 amps for a "DC" system

However, because motors are "inductive" (look like coils or inductors), the Voltage and Current are not exactly "in phase" with each other--the end result is that your circuit will need to carry more current than you would have assumed.

A typical motor can have a "Power Factor" (PF=Cos of phase angle) as low as 0.60 or, InvCos (0.6) = 53 degrees (Voltage sine wave vs Current sine wave is 53 degrees "out of phase" (instead of 0 degrees or in phase like a pure resistive device would be--electric heater, filament light bulb, etc.).

Or:

P=I*V*PF
I=P/(V*PF)= 746 watts / (120v * 0.6) = 10.36 Amps for an AC system

So--the AC wiring will have more voltage drop because there is more current being passed (by a factor of 1/0.6=1.67 times in this example).

Also, non-linear devices such as CFL light bulbs, computer power supplies, AC to DC converters/battery chargers take electric current only at the "peak" of the Voltage Sine Wave--This can also be expressed as a Power Factor too (CFL's may have a PF=0.5 or worse).

In the end, a PF=1 is good--no "extra current" required to run the device. PF<0.9 is "bad" as the lines (and inverter) have to supply "extra current" which does not do any "work"--but requires larger wiring and you have more losses (because of the higher current when PF is not 1.0).

And this is a real issue that needs to be taken into account. You will see ratings for devices in Volts, Amps, Watts, VA (volt*amp), etc... If you are looking at smaller 120 VAC 15 amp devices--a Kill-A-Watt meter is real handy to measure VA and PF. To measure AC Watts, PF, and VA--requires a special meter, a normal DVM cannot do it. The K-A-W meter is the cheapest/best device I have ever seen for home use to do this.

This is not a simple subject--Do you have a specific question? It is probably easier to answer that then to try and do a year or two of electrical courses here for the "complete" answer.

-Bill

Solar Guppy

Re: DC vs. AC voltage drop

From a wiring loss prospective, AC is never better than DC ... it might be close, but never better

The Main reason AC is the choice of transmission is due to the ease of converting the voltage up/down via transformers, DC conversion is much more complex. For Long Distance transmission, DC is better and there are many high voltage DC long distance transmission networks in the US

Windsun

Re: DC vs. AC voltage drop

While DC transmission is better just from a loss standpoint, the difference is pretty marginal, around 2% max, which is more than offset by other difficulties - aside from the step up/step down, the large circuit breakers and line switches for DC are several times more expensive.

System2

Re: DC vs. AC voltage drop

Thanks for all the info. My question relates to a 2400w pole mounted PV array.

I've seen pole mounted systems place the inverters in two different locations.

1. The inverter is an exterior rated inverter, and it is mounted to the pole, and sends AC current underground to the main breaker panel located in the house.

2. The inverter is mounted in the house, and DC current is run underground from the pole mounted array to the main breaker panel.

My inverter output is 240v AC single phase. It is exterior rated. My 2400w pole mounted array's Vmp = 334 VDC, and my Imp = 7.1amps. The distance run is more than 100'.

My question is where should I place the inverter? Hence the question: are the voltage drops in a DC circuit and AC circuit the same if the voltage is the same?

BB.

Re: DC vs. AC voltage drop

A Grid Tied Inverter is designed to output with PF=1.0 (good power factor)... So, the AC/DC question is not really an issue here...

The next thing to look at is the operating voltage of your system... The AC (assuming typical North American Home) is 240 VAC.

The typical solar panel DC input to the inverter operating point is typically around 200-350 VDC (depending on inverter vendor and panel setup).

So--if your DC voltage is equal or larger than the AC voltage--then run the DC voltage the longer distance. In the typical case, your DC voltage for most inverters will rarely be lower than the AC output voltage. But--many times the voltage differences are not that large--so this may not the deciding factor.

Personally, I would look at where the inverter can be installed that is in deep shade with lots of natural ventilation, sheltered from rain (and direct sun). Easy to read the on-board meter/easy to wire to your computer monitoring system (if you want one).

For me--my garage is in deep shade and pretty cool in the summer. I mounted it inside the garage next to the kitchen door--so it was easy to check on. I did not bother with computer logging (Xantrex GT 3.0 inverter installed about 3.5 years ago). Also--since my panels are on a 2nd story roof--the entire installation is invisible from the street and there is nothing exposed to vandalize/steal at ground level.

Some people do mount in their garage but want/need a bit better cooling... So they mount a cheap fan facing the controller and use a timer to turn the fan on between 9am and 3pm (as a suggestion). Works very well too.

Another issue that may affect you... GT Inverters are relatively high powered devices with internal computers. And, even the well shielded ones can interfere with nearby receivers. In my case, I can get AM radio noise/whine on my intercom's radio (2' away from the inverter--oops) and within about 10' of my car radio. Does not always interfere, and I have not had any other RF issues (FM, TV, WiFi, cell, etc.).

From my little experience, the typical emissions mostly affect AM radio. So if you are out in the middle of nowhere and listen to AM, it can be an issue.

Some inverters (not necessarily GT inverters) have been reported to interfere with Maine Band repeaters and probably various HAM bands too. But those have been rare (from what I have read here). I am not a HAM person--but we have some here that can speak to the issue if you have more questions.

Making sure your GT (or any) Inverter is FCC Class B listed can be a help if RF noise will be a problem.

Others here have more experience on mounting/installing the GT systems--I am sure they will correct any misleading information I may have posted here.

-Bill

RCinFLA

Re: DC vs. AC voltage drop

Looking at it from power loss might be easier then getting too involved with power factor.

Power loss is always I^2 * R with AC being r.m.s. value. There is no power factor to this current drain loss against pure resistive wire loss.

The name of the game is usually not to throw away more then a few percentage of total power. To this, wire resistance to DC is usually required to be lower only because DC is usually dealing with lower voltage then AC circuits.

House A.C. wiring size standards are based on I^2*R heating of wire per unit of length, not voltage drop. It is roughly in one watt per foot range. The more the wire is jacketed with insulation or encased in tubing, preventing heat dissipation, the lower the maximum current rating is. With the heat dissipation limits, the voltage drop is usually not an issue when dealing with 120vac or 240vac supplies over reasonable wire lengths in a house of hundred or so feet.

What is acceptable for A.C. house wiring would not be acceptable for a 24vdc or even a 48vdc inverter battery wiring. To keep percentage loss down, the allowable voltage drop must be much less, and wire size much bigger.

On the other hand, if your talking about a 400-500 vdc solar PV serial stack output, wire size can be a bit smaller then A.C. house wiring.

For A.C., poor power factor means more current to transfer the same power. This means more I^2*R loss and is why power companies are concerned about power factor correction to lower their transmission line losses.

lberhold

System said:

I have a question about the voltage drop on a DC circuit vs. an AC circuit.

If the runs are the exact same length, and the wire gauge is the exact same size, and the voltage in both circuits are the same, will the voltage drops over the DC circuit and AC circuit be the same?

Thanks.

AC and DC for an extremely short run behave similarly. When you get into longer distances, DC will always outperform AC (assuming same Power demand at the load, and voltages). AC suffers from complex impedance (inductance and capacitance). As a result of complex impedance you generate what is known as a Power Factor for AC based loads, likewise AC will have considerably greater parasitic loads relative to the line (all physical things have a certain inductance value and will drain some energy from nearby lines, though many will seem as negligible). Electric fields will transfer power significantly better than a Magnetic field, and electric fields are harder to shield against than are magnetic fields (though a changing magnetic field creates an electric field, and a changing electric field creates a magnetic field). DC (less the ramps) will generate a strong magnetic field, AC will always generate a strong Electric field. The sole reason AC was originally used for power transmission was that AC was able to be stepped up with a transformer, and at the time no such practical technology existed for DC transmission. Since that time, power electronics through the development of semi-conductors have made stepping up and down DC voltages significantly more efficient than US 60Hz, Europe 50Hz AC. A realistic step up or down for a high efficiency semiconductor can easily obtain 98% efficiencies, for a power transformer, that's a pipe dream. DC can also mate well to storage devices (capacitors, batteries, etc...), so in the modern world, DC is the significantly better candidate as charge controllers can be used for inductive loads such as motors to make them even more efficient than US 60Hz and EU 50Hz AC.