Where can I get the 9005/9006 male/female connectors?

Where can I get the 9005/9006 male/female connectors?

I was gonna to D.I.Y the headlight upgrade kit, so I spent the a hour figuring out how it works and finally got it.

However, I still don't know where I can purchase those connectors cheap. The only thing I found was the heavy duty connector and they cost nearly $20 U.S.D!!



Also, about the power cable, is 16AWG ok? or 14AWG is better? I have plenty of 16AWG so if it's ok, I'm gonna use them first.

Thanks

Darren

thats what they cost. I only used the plugs that go into the headlights. Then I just tapped into the power wire from the regular connectors to switch the new relays. It ends up being much cheaper that way.

Also you wont want only 16G wire. If you have lots of it then run 2 16G wires per wire that you need. So you will double them up. You want the most power you can have going to your bulbs so they are brighter.

you mean you are using the existing stock connectors, cut the stock wire, and put new 14awg on the connectors, right?

I just checked the stock connectors and they have two wires. One is red and another is red/black. The red should be 12V power source and the red/black should be ground.

Also a noob question, when you ground the wires, how do you make it look clean and neat yet easy to install? Do you drill a hole on the chassis and secure the terminal with screws? I found the chassis was quite thick, and my $30 driller couldn't do nothing about it... :?

Thanks again, Stinky!

Hey Stinky, I don't really understand your logic behind doubling up the wires for max power. Power is the quotient of Voltage times Current. Since Voltage is constant and set at +12 the only other variable would be current. So if you were to increase power you'd have to up either the current or the voltage. Now... from what I know doubling up the wires wouldn't do you any good, in fact it would probably hurt the overall objective of increasing power. The reason it would adversely affect your power is because since you're adding another wire for the current to run through, you're also adding more surface area for heat loss and copper loss (resistance that exists naturally from the wire), granted over short distances it's not going to be a terribly large amount of loss, but it's a loss nonetheless. Your best bet is to use a larger gauge wire if anything... then you increase the cross-sectional area of the conductor with minimal increase in surface area. I'm not trying to call you out on this to make you look bad at all because from the 8 million posts that I've seen you have a lot of good information to share... but I will have to disagree with you on this one. Maybe you could better explain to me why you'd want to double up the wires?

hmm, that makes sense too. Using a bigger gauge is definnately better than using 2 small gauge cable. I have tons of unused 16AWG remote wire from amp installation kits, so I just wonder if I could just use them instead of buying new cables.

ok, now get to the topic.

In the manual pg 236, battery's open voltage fulled charged is 12.6 - 12.8 Volts ( I remember someone said that when running, it will be 13.8v instead of 12v? could anyway confirm it?)

Anyway, let's assume the power source is 12V, and I put two 100w high beam bulbs and two 100w low beam bulbs. So the maximum wattage when all the lights on is 400w. The power cable from light base to battery will be 7 feet top.

How do I calculate the current, the amp, and the suitable guage size?

Ok, I got the exactly numbers now.

Current * Power = Wattge

Here we should wire parellely so two high beams get 12v, and two low beams get 12v too.

16.7A * 12V (battery power) = 200W ( total of two bulbs)

For 16 amp, the gauge size must be at least "12" to avoid any signal lose...

Hmmm......I guess 16 is way too thin.

Here is the way I see it, if you want to upgrade the wiring you can't simply put better connectors on the stock harness. You could put the stock connectors on a new harness but you will still have a region where the power will be lost (think of it like a pipe getting too small). Your best bet would be to make your own wiring harness to replace the stock one. You will also need relays to switch your new headlights on and off. The relays would be wired with the coil going to the stock harness so that when the light switch is on the coil is energized and the circuit closes and your new bulbs light up. You will need a wire comming from the battery going into a fuse, to the contacts of the relay, then to the bulb. If your bulbs are 100W (I=P/V) they will draw 8-9A each so if you have both connected it should be less than 20A, sooo.. you will need a 20A fuse in your fuse holder, and you will need 12 Gauge wire to handle the current. If you want to use your 16 Gauge wire you should use three pieces in parallel. As for grounding, use another 12 Gauge wire going to the chassis. You should solder and put heat shrink tubing on all connections to ensure they last. Also you should put all the wires in some wire loom to make it look nice, and to keep it protected. Make sure your relays can handle the right current (most automotive relays can handle 20A but make sure). you will need two of these setups, one for low beams on for high.

Stinkys logic as far as putting the wires in parallel was to minimize the power lost in the wiring, with one 16 Gauge wire there would be excessive power lost as heat in the wire. By putting the wires in parallel the net resistance decresses thereby minimizing the power lost in the wire and ensuring the lightbulbs get the majority of the voltage from the battery. As Hirolla said you are increasing the cross-sectional area, but this is a good thing as resistance is related inversly to cross-sectional surface area so a large cross-sectional area lowers the resistance. (R=pL./A ).


Anyways let me know if you need a wiring diagram for the wiring harness.

Scott

Sorry Scott, gotta disagree with YOU now. First of all, Current does NOT vary Inversely to The Cross-sectional area of the conductor... Resistance varies inversely to the (A)... just as you illustrated in your perfectly accurate formula. The thing is, you got it backwards... here's how:

R= Resistance
L= Length of Conductor
A=Cross-sectional Area of Conductor

The formula is R= (L/A)

This means that as A goes up, R goes down (because they vary INVERSELY to each other)... as L goes up, R goes up (because they vary DIRECTLY to each other). For example if we're using a conductor that is 6 feet long with a CS area of say 3 then: R=(6/3) so R=2. Let's increase the size of the CS area to 6 (A.K.A using a larger gauged wire) so R=(6/6) so R=1... hmmm, the larger the CS area, the smaller amount of resistance.

SO... I'm right, the larger the gauge of wire, the less resistance offered by it. I hope I explained this well enough for everybody to understand. If anyone has questions about why Current varies DIRECTLY to CS area... I'd be happy to explain.

Ok let me sum it up...

larger gauge wire = less resistance

longer wire = more resistance

more wire (the parallel theory) = more SURFACE area = more heat loss = less current = less power

Besides, putting two wires in parallel with no load in between is absolutely pointless, no matter how you look at it if you take any kind of reading on any part of those wires, they are going to be IDENTICAL electrically speaking. That would be POINTLESS. Sorry guys, I'm not here to make enemies, only trying to shed some light to some who may not know so much about electrical engineering (which I do in the Navy).

lol, opps if you replace the word current with resistance in the last sentance of my post iit all makes sense. As well I was using surface area and cross-sectional area interchangably. Damn me and my fast typing and slow brain

I meant to say

One thing that does need to be cleared up from your initial post is where you where saying that a second wire in parallel would not help, in actual fact it would greatly help because it doubles the crossectional area. since the wires are in parallel the current is shared between them, and hence the power lost in the wire is cut in half. In actual fact you would need three 16Awg wires in parallel to handle the necessary current, but it will work well

Scott

Scott

three 16awg? screw it I just go buy some 12 awg then.

I wonder what gauge they put on those light upgrading kit that they sell online?

I believe what stinky was saying is, if you ALREADY HAVE lots of 16GA, and want to actually use it, then use two runs per lamp. He's not saying it's better than one big cable. He's trying to help you not waste a bunch of wire you already may have.

BTW...two 16GA will handle far more current than a single 14GA. Since current, (more-so with AC but still does in DC) tends to flow along the surface of the conductor (skin effect), two 16GA wires will provide more surface than a single 14GA. I'm not sure how you came up with two wires equaling more heat loss, if you're heating the wire at all you're wasting current that was intended to heat the light filament, not the wire. Kind of a moot point, if the wiring is getting warm at all, something is very wrong. But using P=I/E, a 65 W light bulb is only drawing just over 5 amps. 12GA is rated for 30. Overkill.

OK point taken, if you wanna save the $ on wire, then you can double the wiring. There's an easy way to almost eliminate "skin effect" though... use braided or multistrand wire. Plus I'm really not sure if skin effect is even an issue here since that mostly has to do with magnetic fields and antennas (RF distribution). At any rate, in MY opinion, not to step on anyone else's toes anymore... I would just cough up the money to buy a slightly larger gauge wire, minimize any soldering and dressing work, and not have to think about it one more minute. The choice is yours DarrenWang I tend to choose quality work for quality results... but it has its price of course.

BTW, I just have to say that I'm extremely impressed by the amount of knowledge a lot of you posess... it's good to have an actual conversation about this sort of stuff. Feels like we should all be congregating around a big table discussing electrical design... getting paid the big bucks mind you!

And one last thing... I honestly don't think that this is even as big of an issue as we made it out to be. For a seven foot span, you could use telephone wire and it wouldn't make a lick of difference. Ok Ok, maybe not telephone wire but you get the idea...

Oh yeah and 03ToyMan, more wire does mean more heat loss... not because I'm sitting there cooking it with a butane torch either... it's because heat is a natural byproduct of current running through resistance... hence the need for heat sinks. The more surface area, the more heat loss. Do you get what I mean now?

You're absolutely right about the stranded wiring providing more surface area than solid. The skin effect is far more a concern with AC up in the RF range, than it is with DC, that's very true.

I also agree, just empty the wallet a little and get some good, 14GA, teflon jacketed stranded wire. With the teflon jacketing even if the wire did manage to heat up before the fuse opened, you wouldn't have to worry about it catching fire.

You're doing this in the Navy now...I did this in the Air Force in 1979. Good to know the military is still training it's people well!

I do get what you mean. But look at it like this...two 16GA provides more copper (silver, whatever) for the current to use. Which means if more current can get thru, there is obviously less resistance to current flow. Resistance to current flow is what causes heat.

Yeah, that's what I meant when I said "... it's because heat is a natural byproduct of current running through resistance..." I guess I nuked it out, you definitely make it clear with the way you put it. As far as the whole doubling the wire thing... I think we should just agree to disagree on it, he he he. Anyway, you said it best when you said that this is totally moot anyway, overkill is probably the exact thing that's happening here, after all it's not like we're designing an electrical system for an F-18, is just some headlights for cryin' out loud. The fact of the matter is that all ya really need to do is find out how much current the 16AWG wire is designed for and go with it from there... after that it's only an issue of preference.

Oh and 03ToyMan... you said that you're glad to know that the military is still training people well... I just want to say that it's good to know that a military man from 1979 still has his full deck of cards... I'm getting worried breathing in all these chemicals... J/K... but you never know!

Too funny! While I was waitinng for my security clearance they put me in the NAV-AIDES shop. We'd visit the end of the flight line once in a while, and a friggin B-52 would rotate and go over us so low we could feel the blast and heat! About the only thing I got disability wise out of the military was a bad case of deafness to jet-engine frequency sound!

Oh and that whole attitude thing from combat in Grenada...

Yeup, you hit the nail on the head...these are headlights. Two 65W bulbs is 130 watts...just over 10 amps. Like you said phone wire would handle it, but what the heck, if you're gonna do it, do it with the good stuff. Truthfully, using two 16GA in parallel would work, but if say one of the crimps was weak, then everything would be passing thru the other 16GA...not the idea we want. I took ISO classes, a example they used was on one of the Apollo moon shot training missions three astronauts burned to death on the ground. It was due to a bad crimp in a pure oxygen atmosphere. Makes me solder EVERYTHING now.

Getting back on topic (another hijacked thread), for the 9005 / 9006 connectors, hit the junkyard and find a totalled whatever. Cut the connectors off the front end harness.

These are one of the best/deepest discussion in the forum. Thanks for all your help. I know what to do now and will post a picture after it's done.

Yes, I must agree I enjoy a good electronics/physics discussion We gotta get our brains together to work on a big project, Mine for right now is a OBD-II performance meter, I have some basic stuff working on a PC but I need a higher data rate system so I can do performance analysis (0-60, 1/4 mile, fuel economy etc), Right now I only get 3 samples / second and I have to use matlab to interpolate the data to get reasonably useful data. I would like to do a embedded system with a text based LCD.

Need more time, shrug...

Scott

Are yo doing that for your Corolla? If so I read the modern ECUs keep the previous (guesing two seconds) of relevant data in them. Maybe you could hit the port on it and pull that out, poke it in a serial port and record it on the fly. Not sure what it records, if it's used for accident investigation it would at least contain speed, throttle / brake position. Maybe jsut use it as a couple more lines on the graphs.

I'm using the real time data logging from the OBD-II standard. There are many companies who have done it already I just don't want to pay the hundreds of dollars for a system. There is lots of people with the same idea (opendiag, freediag) some are even open source.

Scott

Guys I am in too big of a hurry to read everything you all said.

Just to further back up my claim about the 2 wires.

Voltage drop is a result of the resistance of the wire and the amount of current going through it. So if I have 2 A goig through 2 - 16G wires I will have less voltage drop than if I only had 1 - 16G wire. I want as close to 12V at my light bulb so I can get the full rated power. SO if you have 100' of 16G wire then run 2 of them for each run to each bulb. If you dont have 2 then get a roll of 12G or even 10G (overkill) and run it with ONE of those.

I am an electrician so dont test me on this guys! I can start pulling numbers and formulas out all day. The bottom line is

Bigger wires with shorter runs, or lots of thin wires will get better results over the stock wiring. And thats also why when you ground your amp you want to run 2 ground wires to it that are the same size as the main power wire.

Also if running 2 wires in paralell is a waste then why do you do it when you feed a huge building with 500 KCmil feeders?? because you want to decive up the current evenly. And thats exactly what it does!

So trust me and do it!

Resistance values

Here are the resistances (typical) per foot of wire:

16 gage 0.0041 ohms
14 gage 0.0025 ohms
12 gage 0.0016 ohms
10 gage 0.0010 ohms
8 gage 0.00063 ohms
6 gage 0.00040 ohms
4 gage 0.00025 ohms

And for those into serious stereo installs:

2 gage 0.00016 ohms
0 gage 0.000098 ohms
00 gage 0.000078 ohms

The voltage drop along a wire will be the resistance listed above times the length of the wire times the current flowing in the wire.

You're in way over my head. Hell I'm still figuring out a triggered flashing circuit for my city light to follow my alarm system!

Yeah Stinky, maybe you are an electrician, but so am I. Granted I'm in the Navy (the US Navy) and don't do wiring for buildings but I'm pretty sure that the theory is all the same. I play with a lot of power too so that doesn't impress me. My particular ship has 3 Gas Turbine Generators that produce 450 V @ 4009 A (figure out the POWER coming from that!). That's ONE of THREE. My ship could power your entire city alone, hell we could power Las Vegas, so I'm glad about your building. NEway... I am not going to argue your parallel theory until after I sit down, write up a schematic, and I will also ask my professors at my school (these guys are REAL Electrical Engineers). Rest assured that IF by chance you ARE right... I will have no problem admitting that you are. I'm not too proud and if you're right then you're right and I will have come out of this learning something that (in the last 3 years) I've never heard of EVER. Also, could you, and I'm asking this seriously, pull some numbers and formulas that will explain this theory better?

Lets see if I can do this right still.

If he is running only one 16G wire to the bulbs he has a R of .0041ohms per foot. Lets say he only uses 3 feet.

.0041x3 = .0123

Now if he is running a 100W bulb @ 12V it will draw 8.3333A

SO our wire has 8.33A going through it. We need to figure out our voltage drop in the wire.

SOOO .0123x8.33 = .1025V lost in the wire.

So his bulbs will NOT be running at the full 100W because it doesnt have the full 12V!

It will have 11.8975V

So

11.8975x8.33 = 99.1W


BUT if we run 2 runs of 16G wire we will have increased the cross sectional area of it to be a cross between 14 and 12G. AND SINCE r IS ADDEDD inversly. our 2 runs of .0123ohms will now be HALF that.
1/.0123+1/.0123 = 162.601 BUT we need to inverse that again so we get this .00615 ohms.

So our NEW R will be .00615 ohms for 3 feet of wire

.00615x 8.33A = .0512295 volts lost in the wire

So his bulbs now have 11.95V SO the wattage will NOW be 99.54W


I know thats not as impressive so I will show you what his 100W bulb would be with the stock wiring.

we still have 8.33A for the bulb, BUT we now have AT LEAST 10 feet of wire by the time it goes through the stock relays and wiring. AND the wiring is only 18 or even 20G!!

so 20G wire has an R of .0105 ohms per foot.

.0105x10 feet = .105

.105x8.33A = .87465 V drop on the wire

So the bulb will only get 11.13V.

The wattage of that bulb is now

92.67W and that is about 8% less power than what it ias rated at.

And that my freinds is why running relays with bigger wire will make your stock bulbs brighter, and make your aftermarket ones brighter and SAVE your relay from burning out.


Also no offence to anyone here, but if engineers in the US are as good as Engineers in Canada I trust them about as far as I can throw them. I realize that there is A LOT of extra stuff for you guys to think about. But I also find toooo many of them have no actual experience on the tools for the ttype of project they are designing. And on Paper it looks A LOT different than when you actuyally go to hook it up. Electricians may as well be the proof readers for our engineers in Canada, coz we usually have to change 40% or more of it to make it all work.


Now that I have offended everyone, you can all pick apart at my numbers to see where I went wrong. I think I kept it all right though. ANd the points that I made are ALL basic rules of electricity. And they are backed up with formulas. Let me know what your professors say. Im pretty stuborn, so if I am wrong you will need to be sure and post your own formulas and numbers to prove me wrong.

Also I assumed 12V as our pwer. Really you will have about 13 volts to deal with, so our numbers will get a bigger difference between them. Because as the voltage increases or decreases your vol;tage drop will raise and lower as well. So we will assume we froze the clock for a second to get ALL the numbers we collected.

stinky I think you're right on, on every count. And if facts offend someone, well, that's THEIR character flaw.

Saying I heard years ago, still believe it's true. Engineers have to know how things SHOULD work. The rest of us, we have to know how things DO work.

well if all else fails at least I have one guy on my side.