Beware "High Output" Bulbs
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From: Monmouth County NJ
Beware "High Output" Bulbs
I recently posted some information about a headlight question, and I felt like I put a lot of effort into a post that fell upon deaf ears, so... here's my very own thread!
After paying almost $80 for a pair of 9004 PIAA Xtreme White Plus Headlight bulbs, I've had both headlight connectors melt and fuse to the bulbs. I am avoiding these ridiculously overpriced headlights from now on, and sticking to good ol' fashioned $20 Halogen bulbs (no plans for HIDs as of now).
I remember reading about why these manufacturer claimed "high-output / HID-like output" bulbs are basically overpriced bits of rubbish off of the Hella Website, but I can't find that link anymore. Enclosed is my paraphrasing of their findings:
Here's a little taste of some basic electronic fundamentals.
V (Voltage) = I (Current) / R (Resistance)
Let's say the headlight circuit runs on 12V. Headlights create a load, therefore there is a resistance rating. The amount of current flowing in a circuit is based on the total resistance (total load) in the circuit (in this case, headlight bulbs).
Let's say the headlights are on a 10A fused circuit. Fuses are designed as an electric safety device to protect other components and prevent damage to more complex and more expensive parts.
The headlights will draw current slightly less than 10A, leaving room for small voltage spikes and inevitable electric variances that happen in a car's electrical system. Let's say the headlights both create a 0.75 Ohm resistance in the vehicle's headlight circuit.
V = I / R
V * R = I
12V * 0.75 Ohms = I
9 Amperes = I (current)
9A is less than the rated 10A fuse protecting the headlight circuit.
A rather dastardly thing aftermarket bulb manufacturers will do to market brighter light output ("Virtually as bright as an HIDs!") is create bulbs that essentially provide less resistance (see PIAA). Current and Resistance are inversely proportional, so if there is less Resistance (Less Ohms), then more Current (More Amperes or Amps) will be able to flow, in theory providing a brighter light output.
Let's say your aftermarket bulbs create a 0.82 Ohm Resistance in the same headlight circuit.
V = I / R
V * R = I
12V * 0.82 Ohms = I
9.84 Amperes = I (current)
The difference between the stock headlights and the aftermarket headlights in this example is roughly 0.8A, which doesn't sound like much, but to put things into perspective, current flowing as low as 0.1A has been known to stop the human heart, if conditions are just right. Pretty nuts right? So back to our 0.8A increase:
A fuse rated for 10A may never melt and protect the circuit because 9.8A is still less than the 10A fuse rating. Unfortunately, the wiring integral to the fuse box, the connectors for the headlights, or the wires that lead into the wiring into the fuse box may not be designed for a constant flow of current of over 9A. More current = more heat = things melting.
I've had this happen to an uncle's car after trying PIAA bulbs (which cost upwards of $80 dollars for one pair of 9004 Ultra Bright Silverstar something or anothers). I had to replace connectors on both sides because the plastic in the connector had actually fused to the beautiful bright blue PIAAs.
I hope this helps and doesn't throw you off too much. Also, someone please check my Math because I'm CUI (calculating under the influence).
https://www.cobaltss.net/forums/newr...eply&p=3589495
I wish I realized sooner that "EXTREME" marketing translated to "AVOID".
After paying almost $80 for a pair of 9004 PIAA Xtreme White Plus Headlight bulbs, I've had both headlight connectors melt and fuse to the bulbs. I am avoiding these ridiculously overpriced headlights from now on, and sticking to good ol' fashioned $20 Halogen bulbs (no plans for HIDs as of now).
I remember reading about why these manufacturer claimed "high-output / HID-like output" bulbs are basically overpriced bits of rubbish off of the Hella Website, but I can't find that link anymore. Enclosed is my paraphrasing of their findings:
I have a 08 sport w/ the DRL headlights
i baught these 9007 nokya blue bulbs and put them on my car
about 3 days later one of my headlights goes out, so i take it to the shop and there it sits...
they replaced my fuse box , then they found a wireing harness was burned up and melted so now i'm waiting for them to get it becuz somehow it's out of stock or some ****...
my question is, why did these blue bulbs do that?? can the cobalt run nothing but stock bulbs? I put the stock bulbs back in and took it to the shop and said it just stoped working lol but the service guy said sometimes people put differ bulbs in that burn to hot and thats what happens... idk
i baught these 9007 nokya blue bulbs and put them on my car
about 3 days later one of my headlights goes out, so i take it to the shop and there it sits...
they replaced my fuse box , then they found a wireing harness was burned up and melted so now i'm waiting for them to get it becuz somehow it's out of stock or some ****...
my question is, why did these blue bulbs do that?? can the cobalt run nothing but stock bulbs? I put the stock bulbs back in and took it to the shop and said it just stoped working lol but the service guy said sometimes people put differ bulbs in that burn to hot and thats what happens... idk
V (Voltage) = I (Current) / R (Resistance)
Let's say the headlight circuit runs on 12V. Headlights create a load, therefore there is a resistance rating. The amount of current flowing in a circuit is based on the total resistance (total load) in the circuit (in this case, headlight bulbs).
Let's say the headlights are on a 10A fused circuit. Fuses are designed as an electric safety device to protect other components and prevent damage to more complex and more expensive parts.
The headlights will draw current slightly less than 10A, leaving room for small voltage spikes and inevitable electric variances that happen in a car's electrical system. Let's say the headlights both create a 0.75 Ohm resistance in the vehicle's headlight circuit.
V = I / R
V * R = I
12V * 0.75 Ohms = I
9 Amperes = I (current)
9A is less than the rated 10A fuse protecting the headlight circuit.
A rather dastardly thing aftermarket bulb manufacturers will do to market brighter light output ("Virtually as bright as an HIDs!") is create bulbs that essentially provide less resistance (see PIAA). Current and Resistance are inversely proportional, so if there is less Resistance (Less Ohms), then more Current (More Amperes or Amps) will be able to flow, in theory providing a brighter light output.
Let's say your aftermarket bulbs create a 0.82 Ohm Resistance in the same headlight circuit.
V = I / R
V * R = I
12V * 0.82 Ohms = I
9.84 Amperes = I (current)
The difference between the stock headlights and the aftermarket headlights in this example is roughly 0.8A, which doesn't sound like much, but to put things into perspective, current flowing as low as 0.1A has been known to stop the human heart, if conditions are just right. Pretty nuts right? So back to our 0.8A increase:
A fuse rated for 10A may never melt and protect the circuit because 9.8A is still less than the 10A fuse rating. Unfortunately, the wiring integral to the fuse box, the connectors for the headlights, or the wires that lead into the wiring into the fuse box may not be designed for a constant flow of current of over 9A. More current = more heat = things melting.
I've had this happen to an uncle's car after trying PIAA bulbs (which cost upwards of $80 dollars for one pair of 9004 Ultra Bright Silverstar something or anothers). I had to replace connectors on both sides because the plastic in the connector had actually fused to the beautiful bright blue PIAAs.
I hope this helps and doesn't throw you off too much. Also, someone please check my Math because I'm CUI (calculating under the influence).
https://www.cobaltss.net/forums/newr...eply&p=3589495
I wish I realized sooner that "EXTREME" marketing translated to "AVOID".
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