Aux light circuit over rated?

GreensvilleJay

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Equipment
BX23-S,57 A-C D-14,58 A-C D-14, 57 A-C D-14,tiller,cults,Millcreek 25G spreader,
Apr 2, 2019
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this..... presumed 8.8A load.

The keyword is 'presumed', you should really put ammeter in the wiring and see exactly what the current is.
Some manufactures lie all the time in their specs, especially in the audio field. Lik eyou CAN get 100 watts of power from a single USB port !

BTW no company manufactures LEDS in the USA or Canada. Yes, several 'assemble' but not MAKE the actual LED die
 
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mikedlee

New member

Equipment
M6060
Feb 8, 2021
12
2
3
Blountsville, Alabama
2019 B2650 with the 120W circuit wiring tucked up under the rear fender. Just installed 4 new LED lights under my canopy and wired 'em up. Using a relay, I took power from the OEM circuit and switch it with the tail light hot wire. My LEDs come on when the head lights are on.

Each light is 2.2A, for a total of 8.8A load. Ran for ~.56 seconds and popped the 10A fuse. Replaced fuse, and same thing again. For grins, I tried a 15A fuse for a few seconds and it ran great. I pulled that out and put a new 10A in again. I will run a new fused circuit to the battery for power that is more robust. Easy fix.

Not griping, just letting folks know that the circuit is pretty limited. More so than I would have figured. I'm using high quality lights, and am surprised that they pull 15% more than advertised.
My math does not match these numbers 120w at 8.8A is 13.63V.
13.63 is the alternator output to charge the battery. The battery is rated as a 12-volt battery and wiring should be rated for 12-volts plus current. 120 watts divided by 12-volts is 10-amps. Calculations should be based on 12 volts. Just my opinion.
 

Mark_BX25D

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Bx25D
Jul 19, 2020
1,753
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Virginia
My math does not match these numbers 120w at 8.8A is 13.63V.
13.63 is the alternator output to charge the battery. The battery is rated as a 12-volt battery and wiring should be rated for 12-volts plus current. 120 watts divided by 12-volts is 10-amps. Calculations should be based on 12 volts. Just my opinion.

If you are measuring 12v at your battery it's failing.
 

mikedlee

New member

Equipment
M6060
Feb 8, 2021
12
2
3
Blountsville, Alabama
If you are measuring 12v at your battery it's failing.
You are correct about the battery showing signs of failure at 12 volts. But, is the battery always 13.63 volts which is what the calculations were based on? For example; if the battery voltage is 13 volts (still a good battery) that would be 9.23 amps on the 10 amp fuse being talked about. Hopefully the wire is rated for a lot more than 10 amps. I am just saying to use the numbers that something is rated for. from the factory. When I go to the parts store to purchase a battery I don't find a 13.63 volt rating label on the battery. It may very well measure at 13.63 volts or close to it if new or fully charged. I'm suggesting that one should use the rating numbers to be safe rather than risk problems.
 

GreensvilleJay

Well-known member

Equipment
BX23-S,57 A-C D-14,58 A-C D-14, 57 A-C D-14,tiller,cults,Millcreek 25G spreader,
Apr 2, 2019
11,401
4,899
113
Greensville,Ontario,Canada
What's unknown is how the lights get their power. Are they voltage driven, current driven or power driven ? You'd have to open them up and see what 'electronics' is in there. Ideally current driven is best as that's how LEDS actually operate. Most though go for a voltage design and some go for power. 3 white LEDs in series will nicely run on a '12 volt battery', so very cheap to make. A 'power' designed module will pull more current as the voltage goes down to maintain the '100Watt' output. That can easily pop the inline fuse.
Battery voltage depends upon type of battery and of course the alternator or dynamo has to put out a higher voltage in order to charge the battery. Typically 2-3 volts more and it has to compensate to line loss through the battery cables, hopefully 'remote sense' AT the battery, and temperature changes. The ideal alternator would be a '4 wire' unit,slightly higher cost to make, but far better to recharge the battery.
 

Mark_BX25D

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Bx25D
Jul 19, 2020
1,753
1,275
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Virginia
You are correct about the battery showing signs of failure at 12 volts. But, is the battery always 13.63 volts which is what the calculations were based on?
That's the normal operating voltage for automotive circuits.

For example; if the battery voltage is 13 volts (still a good battery) that would be 9.23 amps on the 10 amp fuse being talked about.
No, you are going backwards. You are assuming the wattage remains constant. It does not.

In a DC circuit, to find the amperage, you start with the actual voltage of the circuit and the resistance of the load. You can't find the amperage by starting with the rated wattage, then the voltage, then finding the amperage.

You have to start with the normal operating voltage, then use the rated wattage (which is always calculated at normal operating voltage) to find the resistance of the load.

THEN you use the resistance of the load, and the actual voltage, to find the current. So, lower voltage will mean LESS current and yes, lower wattage.
 

6feettogo

Member

Equipment
B2650
Apr 22, 2020
81
50
18
Newman Lake, Wa
What I would do is put a 15 amp fuse in temporarily and check the amp draw with a clamp meter. This would let you know for sure what it is. I would check it with the tractor not running and with the tractor running since the voltage would probably be a little higher when running which should lower the amp draw.
 

Chad D.

Active member
Sep 21, 2019
242
115
43
Eugene
this..... presumed 8.8A load.

The keyword is 'presumed', you should really put ammeter in the wiring and see exactly what the current is.
Some manufactures lie all the time in their specs, especially in the audio field. Lik eyou CAN get 100 watts of power from a single USB port !

BTW no company manufactures LEDS in the USA or Canada. Yes, several 'assemble' but not MAKE the actual LED die
Yessir! I intentionally inserted “presumed” in my post. I’m absolutely taking their word for it that the numbers are accurate. Only testing would confirm this, which has not been done.
You’re also right on about manufacturer’s performance claims. Seems there is little regulation when it comes to Off-road lighting, and it certainly shows.
In my opinion, there are three manufacturers of quality lights. They all use different ways to make you think theirs is the best (brightest).
Baja Designs lists output in raw lumen.
Ridgid Industriesuses Lux.
Diode Dynamics uses candela.
They all claim the others are inferior for whatever reason they can “prove”.

Many of the Amazon/EBay/China lights list with numbers that just don’t add up.
 

Chad D.

Active member
Sep 21, 2019
242
115
43
Eugene
What I would do is put a 15 amp fuse in temporarily and check the amp draw with a clamp meter. This would let you know for sure what it is. I would check it with the tractor not running and with the tractor running since the voltage would probably be a little higher when running which should lower the amp draw.
Well, I did all of that except for metering. I know a 15A will hold, but a 10A will blow in a fairly short amount of time. Running, or not.
 

Chad D.

Active member
Sep 21, 2019
242
115
43
Eugene
That's the normal operating voltage for automotive circuits.



No, you are going backwards. You are assuming the wattage remains constant. It does not.

In a DC circuit, to find the amperage, you start with the actual voltage of the circuit and the resistance of the load. You can't find the amperage by starting with the rated wattage, then the voltage, then finding the amperage.

You have to start with the normal operating voltage, then use the rated wattage (which is always calculated at normal operating voltage) to find the resistance of the load.

THEN you use the resistance of the load, and the actual voltage, to find the current. So, lower voltage will mean LESS current and yes, lower wattage.
You lost me a bit here. Now I want to feed my brain…

I was under the impression that the watt demand remained constant, but the amperage required to deliver that wattage would vary based on voltage available.

For an example, use a load with a wide range of acceptable voltages, like a compressor motor that can be wired for 120 or 240. For grins, let’s say it is a 2,400 watt motor. Wire it for 240 volts, it pulls 10 amps. But at 120 volts, it pulls 20 amps.

Could be totally wrong, but that’s how my brain worked it out!

We’re way out in the weeds now, but that’s ok! It’s my thread and the original problem is no longer an issue. We can run down a different road all we want!
 

6feettogo

Member

Equipment
B2650
Apr 22, 2020
81
50
18
Newman Lake, Wa
You lost me a bit here. Now I want to feed my brain…

I was under the impression that the watt demand remained constant, but the amperage required to deliver that wattage would vary based on voltage available.

For an example, use a load with a wide range of acceptable voltages, like a compressor motor that can be wired for 120 or 240. For grins, let’s say it is a 2,400 watt motor. Wire it for 240 volts, it pulls 10 amps. But at 120 volts, it pulls 20 amps.

Could be totally wrong, but that’s how my brain worked it out!

We’re way out in the weeds now, but that’s ok! It’s my thread and the original problem is no longer an issue. We can run down a different road all we want!
Still pulling 20 amps. The difference is that it is only pulling 10 amps out of both sides.
 

GreensvilleJay

Well-known member

Equipment
BX23-S,57 A-C D-14,58 A-C D-14, 57 A-C D-14,tiller,cults,Millcreek 25G spreader,
Apr 2, 2019
11,401
4,899
113
Greensville,Ontario,Canada
NO !!!! it's only pulling 10 amps !
The amps pulled from one side(leg or phase ) goes through the motor then BACK into the 2nd side (leg or phase ). The motor ONLY gets 10 amps of current.
simply measure the current.....with clampon AC ammeter

The 240 v motor is simply connected to a 240v transformer that has an unused center tapped connection
 
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Mark_BX25D

Well-known member

Equipment
Bx25D
Jul 19, 2020
1,753
1,275
113
Virginia
You lost me a bit here. Now I want to feed my brain…

I was under the impression that the watt demand remained constant, but the amperage required to deliver that wattage would vary based on voltage available.

For an example, use a load with a wide range of acceptable voltages, like a compressor motor that can be wired for 120 or 240. For grins, let’s say it is a 2,400 watt motor. Wire it for 240 volts, it pulls 10 amps. But at 120 volts, it pulls 20 amps.

Could be totally wrong, but that’s how my brain worked it out!

We’re way out in the weeds now, but that’s ok! It’s my thread and the original problem is no longer an issue. We can run down a different road all we want!

  1. AC acts very differently from DC. In DC, the wattage is 'driven' by the voltage.
  2. When you are splitting phases, it's another completely different story.
You cannot compare any AC power calculations to DC calculations.


This guy has some great simple videos, for those who want to dig into it. Just remember, AC and DC power calcs are VERY different. These videos are about DC.

 
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Chad D.

Active member
Sep 21, 2019
242
115
43
Eugene
Ok fellas, it’s been a few weeks because of some vacations and a bit of work travel, but it’s all fixed up now. Removed my OEM auxiliary power source from my lighting circuit and ran a (fused) 12ga pair from the battery straight to my relay. Still using the tail light lead to energize the relay coil, as I want all my lights on with a flick of the headlight switch. Also, it prevents accidental dead batteries if I leave the switch on after a few cold ones…

I also took this as my opportunity to install a bigger alternator. 100% bolt on ease and stupid simple. I’ll make another thread for that.
 

Showmedata

Active member

Equipment
LX3310
May 18, 2022
197
157
43
Boulder CO
FWIW, I had the same problem using the OEM accessory circuit on my LX3310. My home-built electric snowblower chute actuators (which shouldn't be pulling more than 5A) would blow that 20A fuse every time. I ran a new home-run circuit to the battery, also with a 20A fuse, and it works perfectly. I can't explain it yet, but somehow that OEM circuit draws a lot more current than just the load you put on the end of it. Maybe the wire is undersized.