Disconnecting one headlight

[wheatonFJR]

...I recommend watching Apollo 13 for energy saving ideas.

Watch yer Gimbles!!!

 

[gixxerjasen]

Ok, I know he's already posted his reasons (and there's folks here running WAY more stuff than he intends to run) but I can't help myself.

The reason he wants to do this is that he enjoys the daily conversations at stop lights explaining to people that he does not in fact have a burned out bulb, that it's supposed to be that way on purpose.

Yep, my 2001 GSX-R600 had the one for low and one for high setup. Yep, VERY experienced with these conversations.

 

[Justin]

I've been mulling over the idea of disconnecting the stock highbeams since the LED aux lights go to full with the highbeam switch. With 10,200 lumens of LED lighting, the stock headlights are useless anyway. That extra 110 watts of power would be nice when it's cold

- might even add a 3rd set of LEDs lol

 

[900gc]

Ok, I know he's already posted his reasons (and there's folks here running WAY more stuff than he intends to run) but I can't help myself.
The reason he wants to do this is that he enjoys the daily conversations at stop lights explaining to people that he does not in fact have a burned out bulb, that it's supposed to be that way on purpose.

Yep, my 2001 GSX-R600 had the one for low and one for high setup. Yep, VERY experienced with these conversations.

LOL. My Cagiva Gran Canyon had this headlight setup. I had a guy get out of his car one time at a stop light and start walking back towards me practically triggering a fight or flight reaction as I had no idea what his intentions were. He says; "You got a headlight out". I say; " Thanks, it's supposed to be that way. One is low beam and the other high." Boy scout.

 

[bramfrank]

My second lamp died almost 8K miles after the first one. When a lamp dies its current goes to zero and the voltage increases in the headlight circuit -- reduced voltage drop. The remaining lamp receives a bit higher voltage and its wattage increases by a squared relationship -- V squared.

during operation the electrical system carries a maximum voltage of 14.2 VDC regardless of how many lamps run. Having one burn out does not increase the voltage in the circuit unless the bike is somehow running below stator/RR output voltage.

We have disconnected lights when a stator fails or there is some reason to conserve power, i.e. to divert loads elsewhere. In my opinion, it's better to convert to HID and consume 35 watts per light than unplug one. More light, less power use. Works great for me.

Constant Mesh is quite correct.

While the output of the regulator will be held to 14.2 volts (or whatever your particular regulator happens to deliver - there is some variation), given the anemic wiring and ridiculous grounding system there is significant voltage drop between the regulator output and the headlight lamp sockets - just look at the side marker lights of any FJR that has a headlight modulator to get an idea of just how much voltage is being dropped.

When a bulb decides to go, since the second bulb will likely have the same amount of aging, the remaining filament is as fragile and worn as the first - and when the voltage pops up because of the reduced load imposed by the blown bulb, the second is very often not far behind, quickly pushed over the edge by that higher voltage at the base of the socket.

It is good practice to replace both bulbs when one goes.

 

[Justin]

35 watt HIDs consume more than 35 watts each due to losses in the ignitor/ballast to generate the high voltage needed for the Xenon gas discharge. Using them will save you a few watts over the 50/55W halogens, but it is hardly worth doing that for that reason. The reason I have them installed is because it makes you stand out more in the daylight, and gives more light down the road at night. It also takes the entire load current of the headlights off of the ignition switch, which (as all 2nd gen owners know) is a good idea.

+1

I tested my 35 watt HIDs on the Busa with my ammeter and they were pulling over 4 amps. That's 48+ watts at 12 volts, and over 56 watts at 14 volts which is typical running voltage. This was the DDM tuning slim ballast 35w kits.

 

[bramfrank]

35 watt HIDs consume more than 35 watts each due to losses in the ignitor/ballast to generate the high voltage needed for the Xenon gas discharge. Using them will save you a few watts over the 50/55W halogens, but it is hardly worth doing that for that reason. The reason I have them installed is because it makes you stand out more in the daylight, and gives more light down the road at night. It also takes the entire load current of the headlights off of the ignition switch, which (as all 2nd gen owners know) is a good idea.

+1

I tested my 35 watt HIDs on the Busa with my ammeter and they were pulling over 4 amps. That's 48+ watts at 12 volts, and over 56 watts at 14 volts which is typical running voltage. This was the DDM tuning slim ballast 35w kits.

You'll find that Chinese ballasts tend to be less efficient than branded European ones - then again the Chinese ones cost a lot less. The Philips ballasts on my driving lights pull about 3.1 amps at 13.2 volts measured - about 41 watts. The power difference between my Hellas and Justin's units is dissipated as heat.

 

[Nanahanman]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.

A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

 

[Justin]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.

A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

Hmm, maybe I've been doing things wrong :lol:

I always thought that P(W) = I(A) × V(V) - So if you have a device that draws 1 amp, at 12 volts it will be 12 watts. Then, it would stand to reason that at 14 volts, it would be 14 volts. Please correct me if I am wrong

 

[mcatrophy]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

Problem is you are assuming if (for instance) a headlamp bulb says "48 watts" that it always takes 48 watts. It doesn't.
If we suppose its 48 watt rating is for a 12 volt supply, its resistance is 3 ohms, it draws 4 amps, 12x4= 48 watts.

But ohms law applies. If we present the bulb with 14 volts, assuming its resistance doesn't change, it will draw (14/3 = 4.67) amps, it will consume 65.33 watts.

Just to confuse the issue, though, as it gets hotter, its resistance goes up, so while it will be consuming more than the 48 watts, it won't be as much as 65. This is a normal property of incandescent bulbs.

To confuse the issue a little further, modern power supplies for laptops etc don't behave as an ohm's law load at all, they consume virtually constant power over a range of voltage, so as the voltage goes up, their current draw goes down.

I don't know about HID headlamps, I would hope they are more like the constant power load.

Hope that clarifies things a little.

 

[Justin]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.

A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

Hmm, maybe I've been doing things wrong

I always thought that P(W) = I(A) × V(V) - So if you have a device that draws 1 amp, at 12 volts it will be 12 watts. Then, it would stand to reason that at 14 volts, it would be 14 volts. Please correct me if I am wrong

Err - should read "Then, it would stand to reason that at 14 volts, it would be 14 watts."

 

[mcatrophy]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

Hmm, maybe I've been doing things wrong

I always thought that P(W) = I(A) × V(V) - So if you have a device that draws 1 amp, at 12 volts it will be 12 watts. Then, it would stand to reason that at 14 volts, it would be 14 volts. Please correct me if I am wrong

Err - should read "Then, it would stand to reason that at 14 volts, it would be 14 watts."

As per my above post, a simple load will draw more current at 14 volts than it did at 12, twelve fourteenths fourteen twelfths to be exact. [edit] corrected thanks to bamfrank [/edit]

Do the algebra, you find W=V2/R, power proportional to the square of voltage.

But, as before, headlamp bulbs aren't simple ohmic loads.

 

[Justin]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

Hmm, maybe I've been doing things wrong

I always thought that P(W) = I(A) × V(V) - So if you have a device that draws 1 amp, at 12 volts it will be 12 watts. Then, it would stand to reason that at 14 volts, it would be 14 volts. Please correct me if I am wrong

Err - should read "Then, it would stand to reason that at 14 volts, it would be 14 watts."

As per my above post, a simple load will draw more current at 14 volts than it did at 12, twelve fourteenths to be exact.
Do the algebra, you find W=V2/R, power proportional to the square of voltage.

But, as before, headlamp bulbs aren't simple ohmic loads.

OK, help me put the above into something simple I can understand

Let's use the LEDs I have on my bike as an example, as LEDs are fairly simple. They draw 2.4 amps at 12 volts. Am I not correct in thinking they will draw the same amperage at 14 volts? And based on that thinking, at 12 volts they will be using 28 watts, and 14 they will be using 33.6? I know this is very simplified as it does not factor in real life changes, such as resistance changes due to heat etc.

 

[Old Guy]

I'm sure I'll get flamed for this big time so I won't bother with my reasoning. The bottom line is I'd like to do it, but I've read here that when one burns out, the other follows but I can't understand why. Mechanically speaking, is this a bad idea?

Maybe this is too logical, but if both bulbs are about the same age and the expected life is about the same, wouldn't they be expected to go out about the same time?

.

 

[mcatrophy]

...OK, help me put the above into something simple I can understand

Let's use the LEDs I have on my bike as an example, as LEDs are fairly simple. They draw 2.4 amps at 12 volts. Am I not correct in thinking they will draw the same amperage at 14 volts? And based on that thinking, at 12 volts they will be using 28 watts, and 14 they will be using 33.6? I know this is very simplified as it does not factor in real life changes, such as resistance changes due to heat etc.

Firstly, the actual LED lights themselves are not ohmic loads at all. To a first approximation, the voltage across them is constant over a wide range of current, typically around 2 volts, but depends on the exact make-up of the semiconductor used to make the thing.

When small ones are used as indicator lights in (say) domestic equipment, they are often driven from something like 5 volts through a resistor, the resistor is chosen to allow the required current, let's say 10mA, with the 2 volts across the diode, so 3 volts across the resistor. Ohm's law gives us R = V / I, 3/0.01 = 300 ohms.

For automotive work, I'm a bit out of my knowledge zone, however I presume tUhe "ballast" is used to drive a constant regulated current through the LEDs over a range of input voltage. They probably run the LEDs in a series/parallel arrangement so that the voltage across a few LEDs in series would be nearer the lower end of the battery voltage, let's say 8 volts, so a string of 4 LEDs in series (if my 2-volt voltage per LED is valid).

Then parallel two or more strings (care needed to match the voltage of the strings, standard practice).

Now use a switching regulator to drive a constant current through the LEDs.

This is where it may get difficult to understand. An efficient switching regulator has an inductor in series with the load. It switches the inductor to the battery. Inductors have the property that when a voltage is applied, the current builds up exponentially. When the current reaches the required value, it switches the inductor to ground. The inductor, having stored magnetic energy in its magnetic path, continues to drive current through the load, but now decreasing exponentially.

When the current reaches a lower value, the inductor is switched back to the battery, and the cycle repeats, the current rising and falling a small amount around the mean value.

The mark/space ratio is adjusted according to the supply voltage, switched to the battery for a relatively longer time if the battery voltage is lower, or shorter if the battery is higher.

If all the electronic gubbins is lossless, then the current into the LEDs will be constant, so the power will be constant. As the battery voltage goes down, the current from the battery will go up, and vica versa.

This would be the "best" way of driving the LEDs.

A simpler, lossy, but cheaper solution would be to use a resister in series with the LEDs. The resistor would be chosen to allow a higher than the mean current required. This is then switched to to battery or left open, again altering the mark/space ratio to make the average current the required current. Here, the current would be constant from the battery regardless of voltage (until the battery voltage got too low). But there would be heat dissipated in the resistor, the higher the battery voltage the more the waste heat.

 

[Fred W]

hoo boy...

Thank goodness I missed most of this discussion. But for the record, (incandescent) headlight bulbs are most certainly purely resistive loads. To figure power just measure any two parameters (I, E, R) during actual use and do the math.

 

[bramfrank]

That's 48+ watts at 12 volts, and over 56 watts at 14 volts

Kinda curious about the math that got you to those numbers.A fixed load will draw LESS current (amps) as the voltage applied rises.

watts divided by volts =amps

eg...10 watts at 10 volts is 1 amp ...10 watts at 20 volts is .5 amp

If your "35 watt" device is drawing 4 amps at 12 volts it is actually a 48 watt device (as your test results confirm) so at 14 volts it should draw about 3.5 amps (less current load)at the same 48 watts.

Please correct me if I am off base here.

I didn't create Ohms Law I just have to obey it

Hmm, maybe I've been doing things wrong

I always thought that P(W) = I(A) × V(V) - So if you have a device that draws 1 amp, at 12 volts it will be 12 watts. Then, it would stand to reason that at 14 volts, it would be 14 volts. Please correct me if I am wrong

Err - should read "Then, it would stand to reason that at 14 volts, it would be 14 watts."

As per my above post, a simple load will draw more current at 14 volts than it did at 12, twelve fourteenths to be exact.
Do the algebra, you find W=V2/R, power proportional to the square of voltage.

But, as before, headlamp bulbs aren't simple ohmic loads.

Wouldn't that properly be; fourteen twelfths?

 

[Nanahanman]

I stand corrected and am checking my coffee to see if I was drugged. Not thinking straight...

For the hard core math types, see this link

https://www.allaboutcircuits.com/vol_1/chpt_2/4.html

 

[mcatrophy]

...

As per my above post, a simple load will draw more current at 14 volts than it did at 12, twelve fourteenths to be exact.

Do the algebra, you find W=V2/R, power proportional to the square of voltage.

But, as before, headlamp bulbs aren't simple ohmic loads.

Wouldn't that properly be; fourteen twelfths?

Erm, yes, you're quite right. Thanks. Brain fade. I'll edit my original.

At least somebody reads my ramblings

.

 

[Justin]

At least somebody reads my ramblings

.

If it makes you feel better, I did as well :lol: