LED lightbar wiring questions?
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    Question LED lightbar wiring questions?

    Couple questions regarding wiring up a toggle switch to a lightbar.

    Lightbar is: Auxbeam 7200 lumen, 72 watt, 13.5" long.

    Switch is: https://www.superbrightleds.com/more...r-switch/1878/

    https://www.amazon.com/Auxbeam-7200l.../dp/B00T7GYDTC

    Questions are:
    • What gauge wire should I use with this?
    • Do I also need a relay between the toggle switch and the battery?
    • Do I need an inline fuse? If so, what amp? (guy at SuperBriteLights said the switch has a built in 20 amp fuse?? But, I think the lightbar only draws 6 amps... so wouldn't I need a smaller fuse before the switch to make sure the light didn't get fried?)


    Thanks for the assist !! This is all confusing to me. But I think with a little help, I can do it.
    Last edited by JoeFriday1979; 09-23-2016 at 02:11 PM.

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    Ray_PA's Avatar
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    You are correct, the light bar will draw 5.4 amps at 13.8 vdc or 6 amps at 12 vdc. According to the switch specs, there is no integral fuse. The spec does say the switch is rated at 20 amp.

    Concerning wire size, 16 gauge automotive wire is large enough for 6 amps considering the distance that you will be running the wire. If you want to oversize the wire, you can use 14 gauge.

    Concerning using a relay. The light bar only draws 6 amps which is not much and the switch is good for 20 amps. Even allowing for some safety factor, you will be working the switch at less than half of its design load capabilities. No need for a relay.

    Now for the inline fuse. YES you want to install an inline fuse. You can install the fuse holder in the power feed to the switch. Concerning amp rating for the fuse. A 10 amp fuse will protect the circuit. The 16 gauge wire will handle 10 amps in the length of wire that you will be using. The switch is rated for 20 amps. So, the 10 amp fuse will protect the circuit from overload.

    I believe you are confused as to how amperage in a circuit works. The amperage draw in a circuit is totally dependent on what load(s) are connected to the circuit, not the fuse or breaker size. In this circuit, the LED light bar is the only load and it will draw 6 amps, more realistically, probably 5.5 amps. It doesn't matter if you install a 20 amp fuse, the light bar will still only draw 5.5 - 6 amps.
    Now, installing a 20 amp fuse will not protect the circuit from overload if a short circuit occurred, so you want to protect the circuit with a fuse that is based on the maximum allowed amperage of the smallest wire or switch ampacity in the circuit. In your case, that would be the 16 gauge wire. So, a 10 amp fuse would be what you want.

    Be sure to make sure all your wiring connections are tight. Solder connections if possible. Solderless connectors will usually corrode over time which will cause voltage drop and increased amp draw. If I use solderless connectors, I also solder them after I crimp them on the wire.
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    Quote Originally Posted by Ray_PA View Post
    You are correct, the light bar will draw 5.4 amps at 13.8 vdc or 6 amps at 12 vdc. According to the switch specs, there is no integral fuse. The spec does say the switch is rated at 20 amp.

    Concerning wire size, 16 gauge automotive wire is large enough for 6 amps considering the distance that you will be running the wire. If you want to oversize the wire, you can use 14 gauge.

    Concerning using a relay. The light bar only draws 6 amps which is not much and the switch is good for 20 amps. Even allowing for some safety factor, you will be working the switch at less than half of its design load capabilities. No need for a relay.

    Now for the inline fuse. YES you want to install an inline fuse. You can install the fuse holder in the power feed to the switch. Concerning amp rating for the fuse. A 10 amp fuse will protect the circuit. The 16 gauge wire will handle 10 amps in the length of wire that you will be using. The switch is rated for 20 amps. So, the 10 amp fuse will protect the circuit from overload.

    I believe you are confused as to how amperage in a circuit works. The amperage draw in a circuit is totally dependent on what load(s) are connected to the circuit, not the fuse or breaker size. In this circuit, the LED light bar is the only load and it will draw 6 amps, more realistically, probably 5.5 amps. It doesn't matter if you install a 20 amp fuse, the light bar will still only draw 5.5 - 6 amps.
    Now, installing a 20 amp fuse will not protect the circuit from overload if a short circuit occurred, so you want to protect the circuit with a fuse that is based on the maximum allowed amperage of the smallest wire or switch ampacity in the circuit. In your case, that would be the 16 gauge wire. So, a 10 amp fuse would be what you want.

    Be sure to make sure all your wiring connections are tight. Solder connections if possible. Solderless connectors will usually corrode over time which will cause voltage drop and increased amp draw. If I use solderless connectors, I also solder them after I crimp them on the wire.
    Awesome! Thanks Ray! And you are right - when it comes to electrical, I am confused. LOL..... but you did a great job in explaining things.

    If I were to go to a 14 Gauge Wire, would I still use a 10 amp fuse.... or a 15?

    I am good with the 16 gauge wire if you think that will suffice. Thanks again for the response!

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    Quote Originally Posted by JoeFriday1979 View Post
    Awesome! Thanks Ray! And you are right - when it comes to electrical, I am confused. LOL..... but you did a great job in explaining things.


    If I were to go to a 14 Gauge Wire, would I still use a 10 amp fuse.... or a 15?

    I am good with the 16 gauge wire if you think that will suffice. Thanks again for the response!
    Thank You, but actually, it is not what I think that matters. haha!!! Seriously, electrical work is not about someones opinion, it is about sizing the wire, switches and fuses/circuit breakers based on the amperage draw of the circuit load. This is really just a matter of sizing these items based on the wire and switch mfg. ampacity information. Of course, it is also a good idea to design any circuit with a 20% safety factor built into it. It is not a good idea to design a circuit so that all of the components will be working at their maximum rating.

    16 gauge automotive wire, at 12 vdc, will handle 10 amps at 15' of length, 18 amps at 10 feet of length. So, as you can see, 16 gauge wire, at the length that you will be using, will easily handle the 6 amp load of the LED light and the 10 amp fuse will quite sufficiently protect the circuit.

    If you go with 14 gauge wire, you could safely go with a 15 amp fuse because the 14 gauge wire will easily handle 15 amps at the length that you will be using and of course the switch is good for 20 amp. Also, if you plan to maybe add another light to this circuit in the future, you may then want to go with the 14 gauge wire.

    So, if I was doing this job, and I knew that I as not going to add to this circuit later, I would use 16 gauge wire, without a doubt and a 10 amp fuse.

    Essentially, the wire size in an automotive circuit is sized to handle the combined amperage draw of all loads that will be on that circuit. The fuse/circuit breaker in a circuit is sized to the ampacity of the lowest ampacity component in the circuit. Most circuits have wires and switches and most times the wire is going to have the lowest ampacity. So, whatever the ampacity of the wire is, that is what the rating of the fuse should be.
    The one thing that needs to be considered when determining the ampacity of automotive wire is the length of the wire. The longer the wire is, the more voltage drop you will have so the lower the ampacity will be.
    I've included a generic automotive wire ampacity chart. This chart must not be used for high voltage wiring!!!!
    Good luck!!
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    Quote Originally Posted by Ray_PA View Post
    I've included a generic automotive wire ampacity chart. This chart must not be used for high voltage wiring!!!!
    Good luck!!
    Ray,
    I like the chart. But I don't understand the wattage column. For example, 6 amps at 12 volts is 6X12= 72 watts. The table says 36 watts. It looks like every entry is half what it should be. What is it that I'm not seeing here?

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    Quote Originally Posted by Ray_PA View Post
    Thank You, but actually, it is not what I think that matters. haha!!! Seriously, electrical work is not about someones opinion, it is about sizing the wire, switches and fuses/circuit breakers based on the amperage draw of the circuit load. This is really just a matter of sizing these items based on the wire and switch mfg. ampacity information. Of course, it is also a good idea to design any circuit with a 20% safety factor built into it. It is not a good idea to design a circuit so that all of the components will be working at their maximum rating.

    16 gauge automotive wire, at 12 vdc, will handle 10 amps at 15' of length, 18 amps at 10 feet of length. So, as you can see, 16 gauge wire, at the length that you will be using, will easily handle the 6 amp load of the LED light and the 10 amp fuse will quite sufficiently protect the circuit.

    If you go with 14 gauge wire, you could safely go with a 15 amp fuse because the 14 gauge wire will easily handle 15 amps at the length that you will be using and of course the switch is good for 20 amp. Also, if you plan to maybe add another light to this circuit in the future, you may then want to go with the 14 gauge wire.

    So, if I was doing this job, and I knew that I as not going to add to this circuit later, I would use 16 gauge wire, without a doubt and a 10 amp fuse.

    Essentially, the wire size in an automotive circuit is sized to handle the combined amperage draw of all loads that will be on that circuit. The fuse/circuit breaker in a circuit is sized to the ampacity of the lowest ampacity component in the circuit. Most circuits have wires and switches and most times the wire is going to have the lowest ampacity. So, whatever the ampacity of the wire is, that is what the rating of the fuse should be.
    The one thing that needs to be considered when determining the ampacity of automotive wire is the length of the wire. The longer the wire is, the more voltage drop you will have so the lower the ampacity will be.
    I've included a generic automotive wire ampacity chart. This chart must not be used for high voltage wiring!!!!
    Good luck!!


    Great work Ray. Excellent explanation and tutorial!!
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    amp = watt / volt

    The title is how to calculate Amps, Watts and volts.

    Amps = Watts divided by Volts

    Watts = Amps times Volts

    Volts = Watts times Amps


    Steve
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    Quote Originally Posted by Deere Ol Man View Post
    The title is how to calculate Amps, Watts and volts.

    Amps = Watts divided by Volts

    Watts = Amps times Volts

    Volts = Watts times Amps


    Steve
    Shouldn't that last example be "divided by"?
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    Quote Originally Posted by keane View Post
    Ray,
    I like the chart. But I don't understand the wattage column. For example, 6 amps at 12 volts is 6X12= 72 watts. The table says 36 watts. It looks like every entry is half what it should be. What is it that I'm not seeing here?

    Keane
    Keane,
    As you know, wattage is the actual electrical power based on the voltage and amperage.
    Based on this, the wattage ratings in this chart are obviously incorrect if you apply ohms law, which is the only thing you can apply.

    That said, in my opinion, these types of charts should be used carefully. Sort of like generic hardware torque charts. Bolt torque is also mfg. specific, just like ampacity charts for automotive wire.

    I actually got this one from Lawson many years ago. It supposedly applies to the cross linked automotive wire that they sold.
    I personally like to use the specific mfg. amp rating for the wire that you are using. The ampacity of a wire, especially automotive wire, has allot to do with the gauge of wire, the insulation and also whether the wires are bundled or run singularly.

    I don't have a good answer as to why the wattage ratings in this chart are obviously incorrect.
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    Quote Originally Posted by jgayman View Post
    Shouldn't that last example be "divided by"?
    Good ol' Ohm's Law

    P=IxE P is power(watts) I is current(amps) E is voltage(volts) R is resistance(ohms)
    I=P/E
    E=P/I

    E=IxR
    R=E/I
    I=E/R
    Last edited by Billy Cravens; 09-25-2016 at 07:23 PM.

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