Minimum wire diameter for certain currents?
- Thread starter Erasmus
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JWP_EE
Enlightened
rizky_p
Flashlight Enthusiast
try google and type "awg" or "american wire gauge".
or maybe this.
http://www.powerstream.com/Wire_Size.htm
or maybe this.
http://www.powerstream.com/Wire_Size.htm
MrAl
Flashlight Enthusiast
Hi there,
The allowable current in a wire depends a lot on what it is
to be used for, ie the application. Wires that are short can
handle more current than wires that are long, and sometimes
voltage drop is the more important thing anyway.
For the application of 120vac house wiring, here are a few
examples:
#12 is good for 20 amps
#14 is good for 15 amps
A little math tells us that these ratings are calculated roughly
by allowing 300 circular mils per amp of current.
Other applications however, such as transformer coil design,
may require 600 circular mils per amp of current, which would
mean #12 wire would only be good up to 10 amps.
Even with the above numbers it is possible to require a heavier
wire if the wire is very long and the voltage drop is critical.
If so, the diameter of the wire has to be increased. For these
applications usually the resistance of the wire is calculated and
the voltage drop calculated from that and the normall running
current. The wire size is increased until an acceptable voltage
drop is obtained.
For a rule of thumb, calculate the wire size using 300 circular mils
per amp (assuming wire is in free air, or 600 circular mils if not)
and use that wire in the application. Run the application for
several hours and at elevated ambient and see if the wire gets
hot. If it gets too hot increase the wire size or run a second
length of wire parallel to it (one for each wire in a pair).
If you are worried about voltage drop, calculate the resistance
of the wire and use Ohms Law to calculate the voltage drop
at the expected current:
V=I*R
where
V is the voltage drop
I is the current
R is the resistance the wire
Note that if there are two wires (a pair) then R will be the sum of the resistances
of both wires.
A convenient formula for voltage drop is as follows:
V=22*Lenght_In_Feet*Current_In_Amps/Area_In_CMils
or in words:
The voltage drop is equal to twenty two times the length of the wire in feet
times the current in amps divided by the area in circular mils.
Note that in the above formula if there are two wires from point A to point B
and the distance from point A to point B is 10 feet, then the total length of
the wire is 20 feet (2 times 10).
The allowable current in a wire depends a lot on what it is
to be used for, ie the application. Wires that are short can
handle more current than wires that are long, and sometimes
voltage drop is the more important thing anyway.
For the application of 120vac house wiring, here are a few
examples:
#12 is good for 20 amps
#14 is good for 15 amps
A little math tells us that these ratings are calculated roughly
by allowing 300 circular mils per amp of current.
Other applications however, such as transformer coil design,
may require 600 circular mils per amp of current, which would
mean #12 wire would only be good up to 10 amps.
Even with the above numbers it is possible to require a heavier
wire if the wire is very long and the voltage drop is critical.
If so, the diameter of the wire has to be increased. For these
applications usually the resistance of the wire is calculated and
the voltage drop calculated from that and the normall running
current. The wire size is increased until an acceptable voltage
drop is obtained.
For a rule of thumb, calculate the wire size using 300 circular mils
per amp (assuming wire is in free air, or 600 circular mils if not)
and use that wire in the application. Run the application for
several hours and at elevated ambient and see if the wire gets
hot. If it gets too hot increase the wire size or run a second
length of wire parallel to it (one for each wire in a pair).
If you are worried about voltage drop, calculate the resistance
of the wire and use Ohms Law to calculate the voltage drop
at the expected current:
V=I*R
where
V is the voltage drop
I is the current
R is the resistance the wire
Note that if there are two wires (a pair) then R will be the sum of the resistances
of both wires.
A convenient formula for voltage drop is as follows:
V=22*Lenght_In_Feet*Current_In_Amps/Area_In_CMils
or in words:
The voltage drop is equal to twenty two times the length of the wire in feet
times the current in amps divided by the area in circular mils.
Note that in the above formula if there are two wires from point A to point B
and the distance from point A to point B is 10 feet, then the total length of
the wire is 20 feet (2 times 10).
Last edited:
lctorana
Flashlight Enthusiast
This should help:
http://www.dse.com.au/cgi-bin/dse.filereader?48619a2800254dfa273fc0a87f9c06bc+EN/catalogs/SUP1000032
It's just the product range from one shop, but it gives you a great ready reckoner, IMHO.
http://www.dse.com.au/cgi-bin/dse.filereader?48619a2800254dfa273fc0a87f9c06bc+EN/catalogs/SUP1000032
It's just the product range from one shop, but it gives you a great ready reckoner, IMHO.
TorchBoy
Flashlight Enthusiast
I've found the same list on the NZ site very handy at times.
Darkpower
Newly Enlightened
- Joined
- Dec 11, 2007
- Messages
- 185
True, but much more importantly, if the wiring is part of a permanent house or building installation, it needs to be done in acordance to the local building code. Most county and municipal electrical codes follow National Electrical Code (NFPA 70). A good reference, is Ugly's reference which is well worth buying at a local Home Depot or Lowes.
As far as hobby type wiring like much of the projects found here, there is much more leeway, but using the heaviest gauge that the project can accomodate is the best policy especially if the project is going to have high current drain.
Darkpower
Newly Enlightened
- Joined
- Dec 11, 2007
- Messages
- 185
24 awg should do it. That is very little current.Hmmm this is going much further than my needsJust need to know the minimum wire dia for 700mA, 1000mA and 1200mA for an LED flashlight mod
filibuster
Enlightened
- Joined
- Dec 27, 2005
- Messages
- 205
This wire sizing calculator won't help with voltages not standard with solar power (12, 24, 48) or amps outside of 1-100A, but it might be useful for some applications:
http://www.freesunpower.com/wire_calc.php
Sorry, I don't think it will be too much value for your LED flashlight project unless you take it up a notch or two.
http://www.freesunpower.com/wire_calc.php
Sorry, I don't think it will be too much value for your LED flashlight project unless you take it up a notch or two.
Last edited:
mdocod
Flashaholic
Personally, I wouldn't go smaller than 20 if possible. Technically speaking, 28 gage would carry it, but with more voltage drop. If you have room for 18-20 gage, use it
TorchBoy
Flashlight Enthusiast
If it'll only be a few centimeters (the odd inch or three) then you can use pretty thin stuff because the resistance won't cause much heating of the wire, or voltage drop. To choose a wire I start by deciding on the amount of volts I'm happy losing across it, and going from there.Hmmm this is going much further than my needs
blackdragonx1186
Newly Enlightened
Personally, I wouldn't go smaller than 20 if possible. Technically speaking, 28 gage would carry it, but with more voltage drop. If you have room for 18-20 gage, use it
agreed.
most things i work on i use 1/0....except torches that is, lol. wee bit hard with the wire's larger than most cells.:devil:
MrAl
Flashlight Enthusiast
Hi again,
Way back when, i worked on power supplies that used 0.25 inch
thick copper buss bars for the input power. These units where
30,000 watt power supplies that sometimes were 5 feet high,
3 feet wide, and 2 feet deep. I guess you could power 30,000
1 watt Luxeons with one of those things.
Working on small flashlights you can get by with a bit smaller
wire
Number 22 gauge is a good size to keep around. Number
24 would work fine for currents in the 1 amp range too with
short distances.
Way back when, i worked on power supplies that used 0.25 inch
thick copper buss bars for the input power. These units where
30,000 watt power supplies that sometimes were 5 feet high,
3 feet wide, and 2 feet deep. I guess you could power 30,000
1 watt Luxeons with one of those things.
Working on small flashlights you can get by with a bit smaller
wire
Number 22 gauge is a good size to keep around. Number
24 would work fine for currents in the 1 amp range too with
short distances.
Here is a good site with the wire gauges, uses, current capacities.
Once you get to the main page, scroll down to see the tables with the current capacities.
http://www.powerstream.com/Wire_Size.htm
Has calculators for voltage drops, etc. Works for DC or AC voltages.
Based off of the needed current capacities that were stated above, if he wanted to go with a single wire size for all the applications, he would need to go with at least 20ga. which is good for approximately 1.5A (1500ma) when used for power transmission applications. He could get away with 21ga. but that would not allow any safety overhead.
If he wanted to always use the smallest gauge and allow for some safety overhead, then for his application he should use;
Using Power Transmission column
700ma = 22ga.
1000ma = 21ga.
1200ma = 20ga.
Note - wire sized 1 size larger for each current to allow for some safety overhead.
Once you get to the main page, scroll down to see the tables with the current capacities.
http://www.powerstream.com/Wire_Size.htm
Has calculators for voltage drops, etc. Works for DC or AC voltages.
Based off of the needed current capacities that were stated above, if he wanted to go with a single wire size for all the applications, he would need to go with at least 20ga. which is good for approximately 1.5A (1500ma) when used for power transmission applications. He could get away with 21ga. but that would not allow any safety overhead.
If he wanted to always use the smallest gauge and allow for some safety overhead, then for his application he should use;
Using Power Transmission column
700ma = 22ga.
1000ma = 21ga.
1200ma = 20ga.
Note - wire sized 1 size larger for each current to allow for some safety overhead.
Last edited:
lightforce2
Enlightened
Why not conect a DMM between the led & the driver or battery, measure the current in amps, then place the wire into the circuit and see if there's a drop in current, no drop in amps = no resistance
MrAl
Flashlight Enthusiast
Hello again,
I have developed a simple formula to determine the wire size needed for a given
current. This is good when working with wiring inside of flashlights where the
length is not exceptionally long.
The formula is:
n=30-2*ln(I^2)
where
n is the wire gauge and
I is the current.
More:
I have developed a simple formula to determine the wire size needed for a given
current. This is good when working with wiring inside of flashlights where the
length is not exceptionally long.
The formula is:
n=30-2*ln(I^2)
where
n is the wire gauge and
I is the current.
More:
Last edited:
HKJ
Flashaholic
It is also possible to download MiscEl and use the wire pages for these calculations. It will show voltage drop in the wire (and a lot of other stuff).
http://www.miscel.dk/MiscEl/miscelWires.html
http://www.miscel.dk/MiscEl/miscelWires.html
MrAl
Flashlight Enthusiast
Hi again,
Of course a slight simplification to my formula is this:
n=30-ln(I^4)
It doesnt get any simpler than this
In words:
The wire gauge number is equal to 30 minus the natural log of the current taken up
to the fourth power.
BTW, this formula comes from the physical properties of wire heating due to current
flow in the wire.
Of course a slight simplification to my formula is this:
n=30-ln(I^4)
It doesnt get any simpler than this
In words:
The wire gauge number is equal to 30 minus the natural log of the current taken up
to the fourth power.
BTW, this formula comes from the physical properties of wire heating due to current
flow in the wire.
Last edited:
