I have some video equipment i need to power off directly from my
car's
battery. I need a wire that is not too big, because i need to solder
on a special quick-release connector that will be used to easily
disconnect the cable from my video equipment. The cable requires 2
wires inside, one for positive and one for negative on the battery.
The biggest cable i found with 2 conductors, has 2x 22AWG wires
inside. Some tell me that it is too small and some say it will work
no
problem. I don't want the wires to start melting and cause a fire!
The
video gear will draw no more than 1 amp total, directly off the car's
12V battery. My question is, what is the maximum length of cable i
can
use to remain safe? Thnks for your advice.

On Sat, 15 Nov 2008 15:47:57 -0800 (PST), wrote:

I have some video equipment i need to power off directly from my
car's
battery. I need a wire that is not too big, because i need to solder
on a special quick-release connector that will be used to easily
disconnect the cable from my video equipment. The cable requires 2
wires inside, one for positive and one for negative on the battery.
The biggest cable i found with 2 conductors, has 2x 22AWG wires
inside. Some tell me that it is too small and some say it will work
no
problem. I don't want the wires to start melting and cause a fire!
The
video gear will draw no more than 1 amp total, directly off the car's
12V battery. My question is, what is the maximum length of cable i
can
use to remain safe? Thnks for your advice.

---
22AWG has a resistance of about 2 ohms per thousand feet at 20C, so that
means a 2 conductor 22AWG two conductor cable will have a resistance of
about 2 ohms per 500 feet of length, since one 500 foot, one ohm
conductor will go from the battery to the video gear, while the other
one will return to the battery from the video gear.

With the video gear drawing one ampere, a 500' cable will drop:


E = IR = 1A * 2R = 2 volts

And will dissipate:

P = IE = 1A *2V = 2 watts, so there'll be no fire hazard since the
heat from that dissipation will occur along the total length of the
cable and will barely heat it.

More to the point, since the 500' cable will waste 2 volts, then with
12V at the battery the video gear will only see 10 volts if it's drawing
1 ampere, which will shorten the time you can work until your battery
gets exhausted.

View in Courier:

12V |------ 500'-------| 11V
/ | | /
+-----------[1R]----------+
|+ |
[BAT] [VIDEO GEAR]
| |
+-----------[1R]----------+
\ | | \
0V |------ 500'-------| 1V

Since the resistance of the conductors equates to 2 ohms per thousand
feet, that's 0.004 ohms per foot of cable and you can figure the voltage
available for the video gear like this:


El = Vbat - (Il * 0.004 * L),

Where Ev is the voltage into the video equipment with a 1 A draw.
Vbat is the voltage at the battery terminals
0.004 is the resistance of the cable, in ohms per foot, and
L is the length of the cable in feet.


Just trying out a 100' cable gives us:


El = Vbat - (Il * 0.004 * L)

= 12V - (1A * 0.004R * 100')

= 12V - 0.4V = 11.6 volts


Since you'll be recording off a car battery it'll probably have enough
capacity to let you record without worrying about how much time 'til it
goes flat, but what you will need to know is what's the longest cable
you can use before the input voltage to the video gear is too low for it
to work properly.

Let's just arbitrarily pick a number, 9V say, and let's use that as the
absolute lowest input to the video gear which will allow it to work, OK?

Then we can draw:

12V
It-- /
+---------+----Vb
| |
| [Rc]
|+ |
[BAT] 9V--+----Vl
| |
| [Rl]
| |
+---------+----GND


Assuming that the video equipment is drawing 1 ampere with 9V across it,
then the current in the circuit will be the same In Rc (the resistance
of the cable) and Rl (the resistance of the load) since they're in
series.

The drop across Rc will be:

Vrc = Vb - Vl = 12V - 9V = 3V

and, since it has 1 ampere through it, it's resistance must be:

E(rc) 3V
Rc = ------ = ---- = 3 ohms
It 1A

since that represents the resistance of the cable, the length of the
cable will be:


Rc ohms 3 ohms * foot
L = ------------ = --------------- = 750 feet.
0.004 ohms 0.004 ohms
---------
foot

The low-voltage threshold for your video gear will most likely be
different than 9V, but you can use the foregoing to figure out just what
you need.

JF

On Sun, 16 Nov 2008 10:51:02 -0600, John Fields
wrote:

On Sat, 15 Nov 2008 15:47:57 -0800 (PST), wrote:

I have some video equipment i need to power off directly from my
car's
battery. I need a wire that is not too big, because i need to solder
on a special quick-release connector that will be used to easily
disconnect the cable from my video equipment. The cable requires 2
wires inside, one for positive and one for negative on the battery.
The biggest cable i found with 2 conductors, has 2x 22AWG wires
inside. Some tell me that it is too small and some say it will work
no
problem. I don't want the wires to start melting and cause a fire!
The
video gear will draw no more than 1 amp total, directly off the car's
12V battery. My question is, what is the maximum length of cable i
can
use to remain safe? Thnks for your advice.

---
22AWG has a resistance of about 2 ohms per thousand feet at 20C,

---
Sorry about that, that's completely off the wall. It's actually got a
resistance of about 16 ohms per thousand feet so a 1000 foot cable would
exhibit about a 32 ohm round-trip resistance.
---

means a 2 conductor 22AWG two conductor cable will have a resistance of
about 2 ohms per 500 feet of length, since one 500 foot, one ohm
conductor will go from the battery to the video gear, while the other
one will return to the battery from the video gear.

With the video gear drawing one ampere, a 500' cable will drop:


E = IR = 1A * 2R = 2 volts

And will dissipate:

P = IE = 1A *2V = 2 watts, so there'll be no fire hazard since the
heat from that dissipation will occur along the total length of the
cable and will barely heat it.

More to the point, since the 500' cable will waste 2 volts, then with
12V at the battery the video gear will only see 10 volts if it's drawing
1 ampere, which will shorten the time you can work until your battery
gets exhausted.

View in Courier:

12V |------ 500'-------| 11V
/ | | /
+-----------[1R]----------+
|+ |
[BAT] [VIDEO GEAR]
| |
+-----------[1R]----------+
\ | | \
0V |------ 500'-------| 1V

Since the resistance of the conductors equates to 2 ohms per thousand
feet, that's 0.004 ohms per foot of cable and you can figure the voltage
available for the video gear like this:


El = Vbat - (Il * 0.004 * L),

Where Ev is the voltage into the video equipment with a 1 A draw.
Vbat is the voltage at the battery terminals
0.004 is the resistance of the cable, in ohms per foot, and
L is the length of the cable in feet.


Just trying out a 100' cable gives us:


El = Vbat - (Il * 0.004 * L)

= 12V - (1A * 0.004R * 100')

= 12V - 0.4V = 11.6 volts


Since you'll be recording off a car battery it'll probably have enough
capacity to let you record without worrying about how much time 'til it
goes flat, but what you will need to know is what's the longest cable
you can use before the input voltage to the video gear is too low for it
to work properly.

Let's just arbitrarily pick a number, 9V say, and let's use that as the
absolute lowest input to the video gear which will allow it to work, OK?

Then we can draw:

12V
It-- /
+---------+----Vb
| |
| [Rc]
|+ |
[BAT] 9V--+----Vl
| |
| [Rl]
| |
+---------+----GND


Assuming that the video equipment is drawing 1 ampere with 9V across it,
then the current in the circuit will be the same In Rc (the resistance
of the cable) and Rl (the resistance of the load) since they're in
series.

The drop across Rc will be:

Vrc = Vb - Vl = 12V - 9V = 3V

and, since it has 1 ampere through it, it's resistance must be:

E(rc) 3V
Rc = ------ = ---- = 3 ohms
It 1A

since that represents the resistance of the cable, the length of the
cable will be:


Rc ohms 3 ohms * foot
L = ------------ = --------------- = 750 feet.
0.004 ohms 0.004 ohms
---------
foot

The low-voltage threshold for your video gear will most likely be
different than 9V, but you can use the foregoing to figure out just what
you need.

---
If we redo do the immediately above with 22AWG at 0.016 ohms per foot
per conductor (0.032 ohms per foot cable), then:


Rc ohms 3 ohms * foot
L = ------------ = --------------- ~ 94 feet.
0.032 ohms 0.032 ohms
---------
foot



JF