Help to determine actual DC voltage drop

According to

current resistance of 24 AWG copper conductor is 0.0842 Ohm/m, which is 0.0257 Ohm/ft, which is 25.7 ohms for 1000ft.

At 1.5A, this will drop 38.55V, so you'll need to push 50.55V into the cable to leave 12V at 1.5A at the output, or 89.1V if you have a double run (flow and return) in that cable length.

You might want to step the voltage up for the cable transmission, and reduce it back at the other end. The step up and down would have to be very inefficient before they got anywhere near the cable losses of doing it at low voltage.

Let me know what mark I get for the homework question...

I actually measured it to be 23.2 ohms with my meter.

Ugh. That's not a good thing.

No homework and thank you for answering. I was waiting for a barrage of people to accuse me of this or telling me they won't do my job for me.

I'm trying to actually hook this thing up. It appears the more efficient way is to run some 120VAC over romex and use the DC power supply at the other end.

Or...

What about using 14g romex to carry the DC .... ? What do my numbers look like then?

That's near enough.

Not being American, I don't know what the characteristics of 14g romex are. I had to do a web search for 24 AWG characteristics.

If you explain at a higher level what you are trying to do, there may be several alternative suggestions. For example, if your load is short term and intermittent in nature, rechargeable batteries charged at a much lower current over the cable might be an option.

The application is for a remote security camera. This will be a constant

12VDC 1.5A draw every day all day (probably not a true 1.5A draw, but let's assume it anyway).

I'd prefer not to run AC out at that distance since it will be outside and far from the house. If 14g provides much less DC drop, that may be my answer. I can pick up a cheap 20V (or thereabouts) power supply real cheap. This would be nice if doable.

1.5A sounds quite high for just the camera, but maybe not if it also drives IR LEDs, wiper and washer pump, etc.

One thought would be to go with 48VDC, and use a DC to DC converter at the camera. 48V input converters are common as they're used in the telco industry a lot -- you can probably find a tiny PCB mount one easily in an electronics surpless store which would give you 12V 1.5A. At 48V, your 18W load will only draw 0.375A, which would be an 8.7 or

17V drop in the cable. The DC-DC converters usually have a wide input voltage range. Find one with at least a 17V range between the lowest and highest input voltage, and feed the cable with the max allowed At max load where it drops the full 17V on the cable, the input voltage will still be in spec at the converter and hence so will the output voltage. (I'm ignoring losses in the convert itself, but they're probably small anyway.)

It runs over 50 led's ! So, come to think of it, because of the photocell those are only on at night therefore the draw during the day is minimal.

Can you clarify this a little for me? If I can take 48VDC and still have something left to convert to 12VDC at the end of the run, why were the earlier numbers indicating something like 89VDC was needed to end up with 12VDC ?

I'm missing something......

If you run 1.5A in that cable, you are dropping 77V in the cable, which is over 6 times the voltage you actually want at the far end. If you run 0.375A in that cable, you are dropping 19V in the cable which is only half the voltage (48V) you actually want at the far end. In the first case, the transmission losses are 7 times the load. In the second case, the transmission losses are around half the load.

Transmission at higher voltage and lower current is vastly more efficient in terms of transmission losses (although insulator costs go up). That's why power is supplied to your neighborhood at several kV and transformed down, rather than directly at 120V.

Your example problem is a good demonstration of this effect.

snipped-for-privacy@cucumber.demon.co.uk (Andrew Gabriel) wrote in news:ccmm0d$t81$ snipped-for-privacy@new-usenet.uk.sun.com:

I feel, since you are going to run romex out there anyway, the best approach would be supply it with 120AC and use a 12V power supply at the camera. For that distance, I would run at least 12, if not 10 instead of

14.

A 500 foot run of #14 romex would have a bit over 3 ohms resistance - which would drop about 4.5 volts at 1.5 amps.

Voltage drop really isn't your problem - installation of the 500 foot run (or whatever it actually is) is the backbreaker. You want to install the cable so that it is well protected - which means burying it. You want it down at least 18 inches - more could be needed in some situations. If you have to invest the effort/money to dig an 18" deep trench 500 feet long, you want to do that once and only once. And you want to put wire in that trench that will unquestionably serve the load - and any planned future load. If you use #14 uf at 120 VAC, you'll have ~180 watts available at the far end. You could run 8 security cameras from that, figuring

18 watts per camera and some loss in the power supply. That may seem like overkill, but #14 is the smallest size you are permitted on a 120 VAC circuit. If you feed the #14 with 16.5 volts DC, you can run 1 camera.

Mark,

This is a classic problem with low voltage high current power. I have a thermoelectric cooler in my car. In series with the power cable near the cooler is a low voltage detector. It disconnects power from the cooler when it measures 11 volts. However, the voltage at the battery is 12 volts. There is a 1 volt drop in the power cable, making it necessary for the compensation. I could have used remote sensing, by sending two conductors back to the battery and measuring the actual battery voltage.

Dave M.

Well, I've decided to go with 120VAC out to the camera and will tackle that this weekend. I already purchased the romex that was on sale - $59 for 500' of 12/2 which was pretty good IMO. I'm not going to dig a trench. Rather, I'll be using 1/2" electrical pvc conduit running through heavy woods and will shallow burry it before it enters the house. I will do some testing first to make sure the AC doesn't interfere with the cat5 camera signal since I plan on stuffing both in the same conduit.

Thanks for all the help.

On Tue, 13 Jul 2004 13:59:24 -0400, Mark put forth the notion that...

I wouldn't put low voltage cat 5 wiring in the same conduit with Romex carrying 120 volts. First of all, it's illegal. Secondly, you're almost guaranteed to have interference.

I hate to burst your bubble, but before you do it, at least understand that that installation will not comply with the electrical code. Table

300.5 in the 2002 NEC requires a minimum of 18" burial for your installation. You could reduce that to 12" and be code compliant if you feed it from a GFCI.

Personally, I would not do it. I'd look into a solar/battery source to power the camera, or a time slicing arrangement, where the camera draws power for rapid but brief intervals. Say you set it up to be on

1/10 of the time - 9/10 of the time the (low voltage) power from the house charges the battery. So as an example, the camera turns on once each tenth of a second for 1/100 of a second. I have no idea what the time slices should be for your camera, but it seems worth investigating. You would not need a lot of current to keep a battery charged that way. In ten hours, with the camera drawing 1.5 amps each time it turned on, you would use 1.5 AmpHours. In the same ten hours, you would have 9 hours of charge time. If you charged at .2 amps, that's 1.8 AmpHours - about the right amount. Your 500 foot run of romex would drop about .6 volts. An 18 volt supply at the house feeding a charge circuit at the camera would work. In addition, there is a built in "extra charge" period in daylight when the camera doesn't use the LEDS and consumes a lot less. The issue in doubt is whether your camera will work well at the time slice mentioned.

I wasn't planning on pulling a permit. ;)

Seriously, I know some of these laws are in place for valid reasons, but this application isn't one of them. The draw on both sides is negligible.

I am concerned about this, however since there will be the lowest possible draw on the VAC line, I'm guessing there will be no interference. I will do some testing before committing to this however. If I have to run 2 pipes, I will.

I figured it needed to be buried, but I had no idea that deep. Since I'm going to be using pvc and it will be waterproof end to end, I'm not going to get too worried about it.

As intriguing as this sounds, it's a little more than I can handle or, honestly, care to investigate.