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Help to determine actual DC voltage drop

M

Mark

Jan 1, 1970
0
I have a device that draws 1.5A at 12VDC. What will I need to start with if
I'm pushing this through between 750' and 1000' of 24AWG solid copper ?

I tried a few online voltage drop calculators but they varry so much. Is
there a formula I can use to figure it out myself?

Thanks.
 
A

Andrew Gabriel

Jan 1, 1970
0
I have a device that draws 1.5A at 12VDC. What will I need to start with if
I'm pushing this through between 750' and 1000' of 24AWG solid copper ?

I tried a few online voltage drop calculators but they varry so much. Is
there a formula I can use to figure it out myself?

Thanks.

According to http://www.mogami-wire.co.jp/e/puzzle/pzl-07.html,
direct 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...
 
M

Mark

Jan 1, 1970
0
According to http://www.mogami-wire.co.jp/e/puzzle/pzl-07.html,
direct 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.

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

Ugh. That's not a good thing.
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...

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?
 
A

Andrew Gabriel

Jan 1, 1970
0
I actually measured it to be 23.2 ohms with my meter.

That's near enough.
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?

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.
 
M

Mark

Jan 1, 1970
0
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.
 
A

Andrew Gabriel

Jan 1, 1970
0
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).

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

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.)
 
M

Mark

Jan 1, 1970
0
1.5A sounds quite high for just the camera, but maybe not if it
also drives IR LEDs, wiper and washer pump, etc.

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.
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.)

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......
 
A

Andrew Gabriel

Jan 1, 1970
0
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.
 
A

Anthony

Jan 1, 1970
0
[email protected] (Andrew Gabriel) wrote in
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.



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.

--
Anthony

You can't 'idiot proof' anything....every time you try, they just make
better idiots.

Remove sp to reply via email
 
Mark said:
What about using 14g romex to carry the DC .... ? What do my numbers look
like then?

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.
 
D

Dave M.

Jan 1, 1970
0
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.
 
M

Mark

Jan 1, 1970
0
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.


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.
 
Mark said:
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 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.
 
M

Mark

Jan 1, 1970
0
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.

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.
Secondly, you're almost guaranteed to have interference.

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.
 
M

Mark

Jan 1, 1970
0
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.

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.

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.

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