NX - 1st Gen (2015-2021)

Headlight Flicker

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Old 11-03-15, 04:28 PM
  #31  
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Originally Posted by Swacer
Nothing on your car is measured in "power".
Horsepower?
Old 11-03-15, 05:00 PM
  #32  
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Originally Posted by NXOwner
Horsepower?
Touché lol
Old 11-03-15, 05:10 PM
  #33  
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1) I think every new vehicle is known for severe battery drain issues. Its way too many electronics to control everything that basically cannot ever be fully "asleep".

2) In regards to flicker, have you had condensation in the headlight unit? That would be my first Troubleshoot area. Then onto the harness and its connectors.

3) Intermittent problems are a PITA. You as the customer will notice it, but cant do anything to fix it. Warranty/SAs either claim its in your head or magically the car will fix itself and then they cannot do anything LOL
Old 11-03-15, 05:47 PM
  #34  
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I was an electronic technician in the Air Force. It was a long time ago, but electricity hasn't changed. Let me address some of the below. Please note that all of the comments have been cross-checked to either a reliable source of electrical engineering or an auto enthusiast site or LED manufacturer's website.

Originally Posted by Swacer
Nothing on your car is measured in "power". Do you perhaps mean joules? Nothing is measured in joules. It's measured in voltage, which is volts, and current which is measured in amps.
Things are measured in power, which is measured in watts. Joules are a measure of power over time, also called "energy", similar to measures like watt-hours. A joule is also the energy used when an electric current of one ampere passes through a resistance of one ohm for one second. A formula for power is P=IR, meaning power (in watts) = current (in amperes) times resistance (in ohms). Light bulbs and LEDs and almost everything else is rated in watts, as this is how much power they use when operating. One watt is the rate at which work is done when one ampere of current flows through an electrical potential difference of one volt.

Originally Posted by Swacer
Voltage = current * resistance
Correct, but not the only formula applicable here. We also need Current (amps) = Voltage / Resistance (ohms) and Power (watts) = Voltage^2 (^2 = squared) / Resistance (ohms).

Originally Posted by Swacer
So here is your classic example of why your LED lights won't work, but your starter can. Assume that your battery is a 12V car battery. Let's look at a parallel circuit with just the starter and led headlights.
Listening...

Originally Posted by Swacer
Your starter is a high resistance component. V=IR says that means the starter only needs very little current to operate.
Incorrect. A typical starter motor's resistance is about 0.04 ohms. As such, it requires 300 amps (12 volts / .04 ohms = 300 amps) to run (the starter in question was a Ford. Sorry Chevy people!). This, by the way, uses 3600 watts of energy (12^2 / .04).

Originally Posted by Swacer
Your led headlights on the other side are very low resistance, which means they require a very high current, as I had originally said.
(Note: some of the following is a bit oversimplified for those who are not electrical engineers or the like).
LEDs do not have typical "resistance" since they and other diodes are not "Ohmic resistors" (whereas a regular incandescent light bulb is an ohmic resistor). Rather than having a resistance, they have an "I-V characteristic curve". To avoid having to understand this curve, think of it as a "turn-on voltage" (call it TOV). If the actual voltage is less than the TOV, no current will flow through the LED. If the actual voltage is more than the TOV, resistors inserted in series will control the current such that the voltage drop across the diode is exactly the TOV. Each diode's operating voltage will be a particular range of volts; this information can be found on an LED's "data sheet". In auto applications they often start well below 12V. A typical headlight I found was labeled as "Input Voltage: 9V-32V" and uses 15 watts of power.

This 7-minute video is a good learning tool about the I-V (current-voltage) curve:

Originally Posted by Swacer
As your battery dies, the voltage decreases. Thus V=IR changes. You go for your starter, now your battery has 6V, well if the low current is enough to operate the starter (as R is constant) the starter will turn. However, with a lower current (as R is constant for the LEDs as well) you may not be able to support the current required to operate the lights.
No. As your battery dies it loses the ability to push current (in amperes) at its rated voltage (as its own internal resistance increases). Yes, eventually the battery's voltage will decline, especially when measured under load, but a typical battery (of almost any kind) will deliver current at the rated voltage across upwards of 85% of its charge.

Originally Posted by Swacer
Thus, as the batteries die, their ability to operate the "super efficient" LEDs is impacted by inconsistent current and potential loss of lights. They are efficient for long term on the alternator (alternator use is what impacts gas mileage), but their dependence on the initial battery is high.
Bottom line: if the battery won't turn over the starter the LEDs probably will still work. If the LEDs don't work the starter hasn't a chance.

Last edited by 15RC350F; 11-04-15 at 04:51 AM.
Old 11-03-15, 07:35 PM
  #35  
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Originally Posted by 15LexNX2t
I was an electronic technician in the Air Force. It was a long time ago, but electricity hasn't changed. Let me address some of the below. Please note that all of the comments have been cross-checked to either a reliable source of electrical engineering or an auto enthusiast site or LED manufacturer's website.



Things are measured in power (which in the Metric system is joules, but in the English system is watts). P=IR, meaning power (in watts) = current (in amperes) times resistance (in ohms). Light bulbs and LEDs and almost everything else is rated in watts, as this is how much energy (power) they require to operate.



Correct, but not the only formula applicable here. We also need Current (amps) = Voltage / Resistance (ohms) and Power (watts) = Voltage (squared) / Resistance (ohms).



Listening...



Incorrect. A typical starter motor's resistance is about 0.04 ohms. As such, it requires 300 amps (12 volts / .04 ohms = 300 amps) to run (the starter in question was a Ford. Sorry Chevy people!). This, by the way, uses 3600 watts of energy (12^2 / .04).



(Note: some of the following is a bit oversimplified for those who are not electrical engineers or the like).
LEDs do not have typical "resistance" since they and other diodes are not "Ohmic resistors" (whereas a regular incandescent light bulb is an ohmic resistor). Rather than having a resistance, they have an "I-V characteristic curve". To avoid having to understand this curve, think of it as a "turn-on voltage" (call it TOV). If the actual voltage is less than the TOV, no current will flow through the LED. If the actual voltage is more than the TOV, resistors inserted in series will control the current such that the voltage drop across the diode is exactly the TOV. Each diode's operating voltage will be a particular range of volts; this information can be found on an LED's "data sheet". In auto applications they often start well below 12V. A typical headlight I found was labeled as "Input Voltage: 9V-32V" and uses 15 watts of power.

This 7-minute video is a good learning tool about the I-V (current-voltage) curve: LED Tutorial: Plotting the IV Curve for a 10W LED - YouTube



No. As your battery dies it loses the ability to push current (in amperes) at its rated voltage (as its own internal resistance increases). Yes, eventually the battery's voltage will decline, especially when measured under load, but a typical battery (of almost any kind) will deliver current at the rated voltage across upwards of 85% of its charge.



Bottom line: if the battery won't turn over the starter the LEDs probably will still work. If the LEDs don't work the starter hasn't a chance.
Boom.......
Old 11-03-15, 07:44 PM
  #36  
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Originally Posted by Swacer
Boom.......
Swacie, it wasn't my intent to tear you down, only to make sure that what people take away is what is correct. Have a beer on me.

Also, thanks for making me look at crap I haven't dealt with in thirty years!

Last edited by 15RC350F; 11-03-15 at 07:48 PM.
Old 11-04-15, 03:54 AM
  #37  
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I'm not bothered lol My elementary explanation/understanding was invalid. That's all. Nothing wrong with being told when you're wrong lol
Old 11-05-15, 03:58 AM
  #38  
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Another morning, no noticeable flicker.
Old 11-05-15, 11:58 AM
  #39  
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Originally Posted by 15LexNX2t
Things are measured in power, which is measured in watts. Joules are a measure of power over time, also called "energy", similar to measures like watt-hours. A joule is also the energy used when an electric current of one ampere passes through a resistance of one ohm for one second. A formula for power is P=IR, meaning power (in watts) = current (in amperes) times resistance (in ohms). Light bulbs and LEDs and almost everything else is rated in watts, as this is how much power they use when operating. One watt is the rate at which work is done when one ampere of current flows through an electrical potential difference of one volt.
Since we're getting all technical, I will throw in my thoughts... There is a confusion of units and/or terms here.

Power is the rate of energy consumption. In metric units, power is measured in watts (W) and energy is measured in joules. A watt is equal to one joule per second (1 W = 1 J/s). Therefore, P=IV (where P = power, I = current in amperes, and V = voltage in volts). Alternatively, using Ohm's law, P=I²R.

@15LexNX2 seems to have a good understanding of electrical principles so I think this was just a typo or mix-up. :P

Originally Posted by 15LexNX2t
No. As your battery dies it loses the ability to push current (in amperes) at its rated voltage (as its own internal resistance increases). Yes, eventually the battery's voltage will decline, especially when measured under load, but a typical battery (of almost any kind) will deliver current at the rated voltage across upwards of 85% of its charge.
This is why batteries are rated in terms of CCA (cold cracking amps) and why high CCA is important during winter months. Cold temperature increases the internal resistance which reduces the current (amperes) that the battery can deliver.

Last edited by computerwi; 11-05-15 at 12:03 PM.
Old 11-05-15, 12:50 PM
  #40  
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Originally Posted by computerwi
@15LexNX2 seems to have a good understanding of electrical principles so I think this was just a typo or mix-up. :P
I did mix up the terms and I thank you for the correction.
Old 11-06-15, 03:50 AM
  #41  
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Nothing again today, so I'm going to probably put this thread to rest. I can't help but think that my driver size headlight is not as bright as my passenger, but I think its a placebo effect in which I'm "looking" for something to be wrong as opposed to it really being wrong. The car has 7001 miles on it now. Barring anything else going wrong, I'll make sure to mention that I want them to look at it when I go in for 10,000.
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