Then we must agree to disagree.
I used the example for both AC and DC where the load does not change. If one increases the V to increase the I then R must change. That is a law that cannot be broken.
But we are getting OT.
LED anchor light AWG requirement
- Thread starter Richard19068
- Start date
I used the example for both AC and DC where the load does not change. If one increases the V to increase the I then R must change. That is a law that cannot be broken.
But we are getting OT.
I will readily admit that I am treading at the edges of my knowledge and understanding.
The answer to your question lies in the way LEDs work as compared to incandescent bulbs, where Ohm's law reigns.
In really simplistic terms as I understand the LED, it takes a certain amount of power to cause the LED to light up. The electronics that govern the LED then convert the available power into a current level the LED can use. Power = volts x current, so as the voltage drops the current must increase to yield the same power (watts). As voltage rises the amount of current needed is less, but the power required remains constant.
Works the same with solar panels, the panels have a high open circuit voltage, somewhere in the 20+v range but produce very little current. The controller drops the voltage down to a useable level which causes the amount of current produced to increase.
If you are protecting a wire with a 15 amp fuse or breaker, I would not use anything less than 16 ga. For a run into the mast, 14 ga.
This should give you some ballpark figures if interested
Weight of copper wire
http://colonialwire.com/wp-content/uploads/2013/09/WIRE-WEIGHTS1.pdf
Breaking strength
https://www.calmont.com/wp-content/uploads/calmont-eng-wire-gauge.pdf
Weight of copper wire
http://colonialwire.com/wp-content/uploads/2013/09/WIRE-WEIGHTS1.pdf
Breaking strength
https://www.calmont.com/wp-content/uploads/calmont-eng-wire-gauge.pdf
First off the LED is not a "linear device" like a resistor. Ohms law doesn't exactly apply. Put 2.5V on a white LED and no current flows, put 2.9V on it an some current flows, put 3.5V on it an a lot of current flows. The curve is ex
Ohm's law
V=IR
Where V= voltage, I = Current and R = resistance.
How does an LED "draw more current" if the voltage is regulated by the battery? I don't know everything about LEDs so I'm happy to be schooled but if what you suggest is true, the LEDs must have the ability to vary their resistance as a function of applied voltage. I do know a little about band-gap theory and I don't ever recall any discussion about variable resistance in a diode.
jssailem
SBO Weather and Forecasting Forum Jim & John
- Oct 22, 2014
- 23,080
This appears to be an exercise in Minimum/Maximum.
What is the minimum wire to run in a 50 length? What are the parameters that I am concerned with:
And I did use 14awg to wire the lights on my boat.
What is the minimum wire to run in a 50 length? What are the parameters that I am concerned with:
- Weight of wire per 120 ft - 14AWG = 1.53 lbs 16 AWG = .94 lbs I would venture to say 0.59 lbs is not a significant difference for a 31 foot cruising boat aloft.
- Voltage drop of wire over 120 ft - Favors 14AWG at 3%
- Can the wire handle the current - 14AWG again.
And I did use 14awg to wire the lights on my boat.
Wow, I didn't expect so many responses. Thanks.
Looks like 16 AWG duplex would be a reasonable choice.
Looks like 16 AWG duplex would be a reasonable choice.
The wire gauge calculators only consider voltage drop but, as mentioned previously, you also need to consider the strength of the wire. If you had an LED on your mast that only draws .01 A you could get away with a 28 AWG wire, but it wouldn't last long. I'd go with 14 AWG for mechanical strength in a mast.
The discussions on power/voltage/current/resistance and Ohm's law are valid for resistive loads but are not accurate for constant power or inductive loads such as motors. With a resistive load, if you lower the voltage, the current drops following V=I*R. With a constant power load if you lower the voltage the current goes up P=V*I.
LEDs are low resistance and operate at a prescribed current. Some lower power LEDs use a resistor to limit the current but higher power LEDs use a constant current driver. Since the LED has a given power dissipation the driver input current will vary inversely to the voltage.
The discussions on power/voltage/current/resistance and Ohm's law are valid for resistive loads but are not accurate for constant power or inductive loads such as motors. With a resistive load, if you lower the voltage, the current drops following V=I*R. With a constant power load if you lower the voltage the current goes up P=V*I.
LEDs are low resistance and operate at a prescribed current. Some lower power LEDs use a resistor to limit the current but higher power LEDs use a constant current driver. Since the LED has a given power dissipation the driver input current will vary inversely to the voltage.
If the 20 Watt incandescent bulb was doing the job when it burned out then I would just replace it with an LED bulb and not worry about it if the existing 16 gauge wiring looks good. The original bulb would have been using 12 volts if the battery voltage had dropped to 11.5 you probably would not have even noticed the drop in the light output at first. Of course if you had tried to start the engine at that moment it would have been obvious. The bulb found below requires only 10 volts to operate so you would have to experience a drop of 2 volts before the lamp had low voltage problems since it uses one quarter of an amp current at the lamp.
tower LED Replacement Bulb Warm White 10 to 30V DC 3 Watts Omni-Directional B15d Socket
Name Value
Amperage 0.25 Amp at 12 Volts
0.125 Amp at 24 Volts
Best Use Tower Navigation Bayonet
Base Type DC-Bay
Color White
Voltage 10 to 30 Volts DC
Wattage 3 Watts
Product Overview
At 210 lumens, this is the highest output LED tower cluster available for marine use. It combines 18 surface-mount (SMD) LED bulbs arrayed in 6 planes for 360 Degrees visibility. Proprietary PWM constant-current chipset results in high performance across the entire 10 to 30V DC range.
tower LED Replacement Bulb Warm White 10 to 30V DC 3 Watts Omni-Directional B15d Socket
Name Value
Amperage 0.25 Amp at 12 Volts
0.125 Amp at 24 Volts
Best Use Tower Navigation Bayonet
Base Type DC-Bay
Color White
Voltage 10 to 30 Volts DC
Wattage 3 Watts
Product Overview
At 210 lumens, this is the highest output LED tower cluster available for marine use. It combines 18 surface-mount (SMD) LED bulbs arrayed in 6 planes for 360 Degrees visibility. Proprietary PWM constant-current chipset results in high performance across the entire 10 to 30V DC range.
Hi Richard. Yeah sooo many responses but you never told us what anchor light you are installing. It will have specs including a voltage RANGE it will operate at and a draw (could be in milli-amps or watts). Most all LED lights now use a built in DC to DC converter as their FIXED power source to the LED group. A cheap and efficient way to regulate the current and voltage to the LED group. This never changes and shouldn't for constant brightness and better LED life. As stated above current + voltage = power. On the fed side of the DC - DC converter the POWER draw will change based on the voltage applied But not by much. Simply put (leaving a few details out) power in almost equals power out. These fixtures however are different than simple loads but close enough to assume some of the simple math above. That said check the specs of the anchor light you are going to install and the resistance per/ft of the wire you buy (usually per 100' or 1000' and divide the run length you estimate). Then you can use ohms law to get your answers (or we can do it for you
The Hella Marine I did was spec'ed at 9-33V at less than 1W. So worrying about 16Ga voltage drop is more or less moot.
The Hella Marine I did was spec'ed at 9-33V at less than 1W. So worrying about 16Ga voltage drop is more or less moot.
I understand the breakdown voltage (or band gap voltage) but I assumed that once you exceed the band gap, the diode followed Ohm's law.
Someone already mentioned this..
Current in a diode has an exponential increase with voltage once "on" ( equation in this link Current vs. voltage properties of a diode ). A diode never looks like a resistor.
If you look at that equation, there is also a fairly strong temperature dependence (ie, T in the equation) so you cant just apply a fixed voltage to a diode as the current (and brightness) would vary a huge amount with temperature. LED would be too dim at low temp and have excessive current (and damage) at high ambient temp.
Adding a resistor in series with the diode reduces the temperature dependence but the more the temperature dependence is reduced, more power is lost across the series resistor. Generally only low power applications (like a display LED) would use a resistor in series because the power loss in the resistor is small.
But for higher power LED applications (like Nav lights), a more efficient and more complicated method is typically used. Typically this would be a very simple buck/ boost magnetic switching converter that simply monitors the current through the LED diode and by using feedback, adjust the LED drive voltage to hold the current constant. If the temperature changes, the circuit will adjust the LED drive voltage in order to keep the LED current constant.
If this magnetic switching circuit was 100 percent efficient, the power used for a given temperature would be constant regardless of the input voltage changing. Ie, if you doubled the input voltage, the input current would go to 1/2 (constant power). They are never 100 percent efficient.. but way better than than the simple resistor "ballast".
You can usually tell if a LED light has the magnetic buck/ boost more efficient drive method because they typically have a very large input voltage range.. like even 5 to 30 volts. Care must be taken in the design to insure that harmonics of the switching frequency dont radiate and cause other problems.
Current in a diode has an exponential increase with voltage once "on" ( equation in this link Current vs. voltage properties of a diode ). A diode never looks like a resistor.
If you look at that equation, there is also a fairly strong temperature dependence (ie, T in the equation) so you cant just apply a fixed voltage to a diode as the current (and brightness) would vary a huge amount with temperature. LED would be too dim at low temp and have excessive current (and damage) at high ambient temp.
Adding a resistor in series with the diode reduces the temperature dependence but the more the temperature dependence is reduced, more power is lost across the series resistor. Generally only low power applications (like a display LED) would use a resistor in series because the power loss in the resistor is small.
But for higher power LED applications (like Nav lights), a more efficient and more complicated method is typically used. Typically this would be a very simple buck/ boost magnetic switching converter that simply monitors the current through the LED diode and by using feedback, adjust the LED drive voltage to hold the current constant. If the temperature changes, the circuit will adjust the LED drive voltage in order to keep the LED current constant.
If this magnetic switching circuit was 100 percent efficient, the power used for a given temperature would be constant regardless of the input voltage changing. Ie, if you doubled the input voltage, the input current would go to 1/2 (constant power). They are never 100 percent efficient.. but way better than than the simple resistor "ballast".
You can usually tell if a LED light has the magnetic buck/ boost more efficient drive method because they typically have a very large input voltage range.. like even 5 to 30 volts. Care must be taken in the design to insure that harmonics of the switching frequency dont radiate and cause other problems.
Last edited:
Most of the tables were written with incandescent and/or other electronics in mind. I replaced all the lights and wiring on my 34’ and used 14 AWG duplex wire for everything. I live aboard and cruise full time and have had zero issues. The LED’s draw so little amperage it kind of re writes the standard in a way. Use quality heat shrink style connections, properly crimp them and you’ll be fine. AA
Off topic from what the OP was looking for (and got lots of good answers) but I wanted to note one more thing about LED drive circuits. Discussed was the resistor "ballast" and the magnetic switching constant current circuit but there is actually another common way for LED constant current drive.
Here is a typical IC used for a linear constant current LED drive method https://www.mouser.com/datasheet/2/256/MAX16815-MAX16828-1515374.pdf
Notice there is no inductor involved in the schematic so no magnetic switching. The circuit senses the LED current by measuring the voltage across a a resistor (Rsense) and uses an active circuit (transistors) to drive the LED so that the measured current is kept constant. The voltage across the string of LED's in series will change with temperature but as long as the highest level this voltage gets to is lower than the supply voltage plus a little for the electronics to work, no problem. And to get the most efficiency out of the circuit, the sum of the LED series voltage would be somewhat close to the minimum power supply voltage so that power losses in the circuit itself are minimized.
The active circuit somewhat looks like a variable "ballast" resistor where the resistance gets adjusted to keep the current constant.
This type of circuit likely is not as efficient as the magnetic switching type.. but probably gets close and its very simple and inexpensive and no switching noise.
This IC has a PWM input that would be used for dimming only.
Anyhow.. might be interesting to someone...
Here is a typical IC used for a linear constant current LED drive method https://www.mouser.com/datasheet/2/256/MAX16815-MAX16828-1515374.pdf
Notice there is no inductor involved in the schematic so no magnetic switching. The circuit senses the LED current by measuring the voltage across a a resistor (Rsense) and uses an active circuit (transistors) to drive the LED so that the measured current is kept constant. The voltage across the string of LED's in series will change with temperature but as long as the highest level this voltage gets to is lower than the supply voltage plus a little for the electronics to work, no problem. And to get the most efficiency out of the circuit, the sum of the LED series voltage would be somewhat close to the minimum power supply voltage so that power losses in the circuit itself are minimized.
The active circuit somewhat looks like a variable "ballast" resistor where the resistance gets adjusted to keep the current constant.
This type of circuit likely is not as efficient as the magnetic switching type.. but probably gets close and its very simple and inexpensive and no switching noise.
This IC has a PWM input that would be used for dimming only.
Anyhow.. might be interesting to someone...
I made some LED fixtures using a buck converter to step down the voltage from 12V and then current limiting resistor. The buck has a constant voltage output I then selected an appropriate resistance to limit current for the LED. This is a dumb system so not truly a current limiting supply but it was cheap and quick that allows for a wide range input.
I am still waiting for @Maine Sail to finish his statement, but I am leaning toward strength for spar wiring as opposed to voltage drop, that wire is hanging free, slapping back and forth over and over and over again something to think about 10 times (gauge) over also always use marine rated wire, none of that apron store stuff.
jssailem
SBO Weather and Forecasting Forum Jim & John
- Oct 22, 2014
- 23,080
That makes sense.
Note: to your benefit, the larger wire also means less voltage drop.