Need electrical help!

[vyper144]

So I hate halogen bulbs... And GM has odd ways of wiring and designing electrical systems. I have aftermarket projector headlights for my Impala that take H1 bulbs for both hi-and-low-beams(stock took 9006/9005). The headlights are independent. The DRLs and High-Beams are combined. My DRLs run at 6volts and the high-beams at 12volts. I tried buying the DRL harness for the HIDs but I couldn't seem to understand how they were supposed to be wired, AND they took too long to warm up (they're not very effective if I need high-beams in a short amount of time)... SO! I'm looking at the option of replacing the halogens with some LEDs. I had to use resistors for the turn signals because I replaced them with LEDs and got the hyper-flashing problem, so I'm going to make the safe assumption that I will need resistors to keep the DIC from saying "High-Beam Out". I know NOTHING about resistance, or electrical circuits for that matter...


Do I need the resistors?

If yes, what resistors?

I've seen online 50w6ohm and 50w8ohm; Whats the difference?

Does it matter what type of LEDs I get?




P.S. This is keeping me busy and is inexpensive enough while I search for the parts I need for my top swap, so RICH17 can stfu!

Click to read more...

 

[SleeperLS]

I can't help with wireing...but I can help with some heavily ported L67 heads for your top swap

 

[BrianG]

Do you need resistors? Yes. What resistors? No idea. You'd have to measure the load across the halogens first to be able to choose the right resistor.

The fact that you have to use a resistor in the turn signals is NOT just to suppress the "turn signal out" message, it's to put a load on the circuit. LEDs essentially add no load to a circuit which pretty much means that you're applying 12v across a short = bad. Adding the resistor in series with the LED bulbs creates the load that a conventional bulb would have. The same applies to the headlights/DRLs/brights. We'll have to take a look at the harness you're working with, take some measurements, do some math, and then verify with James that we figured everything right.

 

[02BlueGT]

Do you need resistors? Yes. What resistors? No idea. You'd have to measure the load across the halogens first to be able to choose the right resistor.

The fact that you have to use a resistor in the turn signals is NOT just to suppress the "turn signal out" message, it's to put a load on the circuit. LEDs essentially add no load to a circuit which pretty much means that you're applying 12v across a short = bad. Adding the resistor in series with the LED bulbs creates the load that a conventional bulb would have. The same applies to the headlights/DRLs/brights. We'll have to take a look at the harness you're working with, take some measurements, do some math, and then verify with James that we figured everything right.

LED's have a load what your talking about. It is under the threshold of the signal out setup in most cars. If LED did not have an electrical load, they would not work.

 

[02BlueGT]

Google search nets that a H1 bulb has a resistance of 2.6 Ohms and operates at 55 watts. Adding a 6 Ohm bulb shouldn't need a resistor as long as it is in a similar wattage range.

H1 bulb would pull 5.4 amps @13.5v
6 ohm Led would pull 2.25amp @ 13.5v

 

[BrianG]

If LED did not have an electrical load, they would not work.

That's not correct.

 

[02BlueGT]

That's not correct.

Any resistance is a load. If It was not putting any load on the circut it would either be a superconductor (even those have a miniscule "load") or an open circuit. Wires are technically a load. The "load" is the part/item/place/material that the electricity is doing work including just passing through a medium.

 

[BrianG]

Any resistance is a load. If It was not putting any load on the circut it would either be a superconductor (even those have a miniscule "load") or an open circuit. Wires are technically a load. The "load" is the part the electricity is doing work or passing through a medium.

A diode has ~0 load, as are the wires assumed to be ~0 load for all intensive purposes. Current -> infinity when voltage is applied to a diode essentially equating to a ~0 load/resistance/impedence.

 

[02BlueGT]

A diode has ~0 load, as are the wires assumed to be ~0 load for all intensive purposes. Current -> infinity when voltage is applied to a diode essentially equating to a ~0 load/resistance/impedence.

A electircal circuit Diode and a Light Emitting Diode are not the same electrically:

LED stands for light-emitting diode, so on the surface, it may appear there is any different between the LED and a common diode. Normal diodes, however, are used as resisting semiconductors in electric circuits, while LEDs are designed specifically to produce light as a result of the extra energy caused by their resistance. This leads to several key differences.

Read more: The Difference Between LED & Diode | eHow.com The Difference Between LED & Diode | eHow.com

Read the rest HERE






P.S. Normal Diodes are not ~0 resistance anyway.

When a diode is placed in a circuit and the voltage on the anode is higher than the cathode, it acts like a low value resistor and current will flow.

Read more HERE

 

[BrianG]

Neither of those are very correct, and both are stripped down to such laymans terms that they're neglecting the fundamentals at work.

 

[02BlueGT]

Neither of those are very correct, and both are stripped down to such laymans terms that they're neglecting the fundamentals at work.

So enlighten me then? Also, you are talking assumed and I'm talking actual BTW

 

[BrianG]

 

[02BlueGT]

Using your chart, the white one is 4V(forward) and .02 Amps. Per Ohms law, that gives the white LED a 200 Ohm rating. That is not ~0 Ohms. If I read it wrong and it is .2 amps, it will be 20 ohms, still not ~0 Ohms. I still listening though, tell me more.


P.S. it is also really hard to read the chart in the black background I have, so I may have missed something.

 

[BrianG]

So enlighten me then? Also, you are talking assumed and I'm talking actual BTW

At the voltages at which the circuit the LEDs are being added to, we're working in the functional range of the LEDs. The "simplifed" concepts of current flow through an LED similar to the diagrams above do apply, and for all intensive purposes, we can show/prove that the LEDs do not have a measurable load to sustain themselves and the circuit without blowing a fuse, blowing out themselves, or melting wires.

Diodes and LEDs work pretty much the same, but one emits light due to its chemical composition and construction. An input voltage into any kind of diode has to overcome an internal diode potential (~.7V) in order for current to flow, but once that potential is passed significantly, current -> infinity. Using this simplified operation and current vs resistance relationships:
V = IR
12 = infinity * R
12/infinity = R = 0

That's why we can assume the diode has no load. There is a voltage drop across a diode, but no load.

An LED also does not create light from its resistive properties like a conventional bulb filament does. A diode creates light when an electron jumps the potential gap in a diode and emits a photon.

 

[02BlueGT]

At the voltages at which the circuit the LEDs are being added to, we're working in the functional range of the LEDs. The "simplifed" concepts of current flow through an LED similar to the diagrams above do apply, and for all intensive purposes, we can show/prove that the LEDs do not have a measurable load to sustain themselves and the circuit without blowing a fuse, blowing out themselves, or melting wires.

Diodes and LEDs work pretty much the same, but one emits light due to its chemical composition and construction. An input voltage into any kind of diode has to overcome an internal diode potential (~.7V) in order for current to flow, but once that potential is passed significantly, current -> infinity. Using this simplified operation and current vs resistance relationships:
V = IR
12 = infinity * R
12/infinity = R = 0

That's why we can assume the diode has no load. There is a voltage drop across a diode, but no load.

An LED also does not create light from its resistive properties like a conventional bulb filament does. A diode creates light when an electron jumps the potential gap in a diode and emits a photon.

Yes, I understand how a diode works, but once current is flowing, it is a Load, any resistance is a Load. Even if it is .000000001 ohm and lets 1000000 amps flow through it, it is an electrical load.

IIRC Multiplying something by infinity does not =0, it = infinity. Per standard math theories, Infinity CAN = 0, but it does not as a fact = 0

 

[BrianG]

That's 12 / infinity. That = 0.

And the load that a diode has is so small that it is negligible, so 0. I'm not arguing that scientificlly a diode has no load, I'm arguing that for the circuit in question, the diode adds no (negligible) load.

You are correct about your resistance calculation, but only at 4V. We're at 12 tho, so we can't depend on the diode to produce the load required to keep the current at manageable level, that's why require a resistor in a LED conversion.

 

[02BlueGT]

That's 12 / infinity. That = 0.

And the load that a diode has is so small that it is negligible, so 0. I'm not arguing that scientificlly a diode has no load, I'm arguing that for the circuit in question, the diode adds no (negligible) load.

You are correct about your resistance calculation, but only at 4V. We're at 12 tho, so we can't depend on the diode to produce the load required to keep the current at manageable level, that's why require a resistor in a LED conversion.

I'm not arguing that he doesn't need a resistor, I'm arguing your theory. And now I get to argue your math : )

12/∞ may or may not = 0

We are looking at Diodes. Since Diodes do not allow reverse flow of currency, your ∞ it actually a positive infinity. Depending on your school of thought, 0 may or may not be a positive number. And sisnce ∞ is being used as a variable for this argument, it would be the same as saying 12/X = 0 or 12/Y = 0. The problem is that ∞ is not defined. Even it if = .01 to the 100th power of zeros, 12/∞ ≠ 0. But also, if the ∞ does = 0, then yes, 12/∞ = 0. Get my point?

Yes he probably needs a Resistor to run a LED, but if it is rated 6 - 8 ohms, I assume it has a built in Ω as LED's don't have that high of resistance typically. Link to said bulb?

 

[BrianG]

I'm not arguing that he doesn't need a resistor, I'm arguing your theory. And now I get to argue your math : )

12/∞ may or may not = 0

We are looking at Diodes. Since Diodes do not allow reverse flow of currency, your ∞ it actually a positive infinity. Depending on your school of thought, 0 may or may not be a positive number. And sisnce ∞ is being used as a variable for this argument, it would be the same as saying 12/X = 0 or 12/Y = 0. The problem is that ∞ is not defined. Even it if = .01 to the 100th power of zeros, 12/∞ ≠ 0. But also, if the ∞ does = 0, then yes, 12/∞ = 0. Get my point?

No dude, just no.

 

[02BlueGT]

No dude, just no.

Mathematically if 12/∞ = 0 then ∞ = 0, as 12 X 0 = ∞ would also be correct

Sorry if you don't agree,

but I still :hug: you BG

 

[BrianG]

Mathematically if 12/∞ = 0 then ∞ = 0, as 12 X 0 = ∞ would also be correct

You're making it worse man.

 

[Bob Kazamakis]

 

[Euro]

I have no idea what the fuck is going on in this thread.

 

[vyper144]

Mathematically if 12/∞ = 0 then ∞ = 0, as 12 X 0 = ∞ would also be correct

Sorry if you don't agree,

but I still :hug: you BG

Mathematically infinity does not =0. In electricity, you can't have negative load, resistance, or voltage so it cant be less than 0 or negative infinity. Yeah, you're right, the .0000000000000000000001 is a number but is NEGLIGIBLE, as Brian stated. So we'll call it 0.

This may not be a great analogy, but I have exactly $0.01 to my name. $0.01 won't buy me ANYTHING, therefore its negligible and ACCEPTABLE TO SAY I HAVE 0.

 

[02BlueGT]

Mathematically infinity does not =0. In electriity, you can't have negative load, resistance, or voltage so it cant be less than 0 or negative infinity. Yeah, you're right, the .0000000000000000000001 is a number but is NEGLIGIBLE, as Brian stated. So we'll call it 0.

This may not be a great analogy, but I have exactly $0.01 to my name. $0.01 won't buy me ANYTHING, therefore its negligible and ACCEPTABLE TO SAY I HAVE 0.

Flowing electricity in the incorrect direction would be a negative. AC current does this, + to - and back again. You asked about 6 and 8 ohm LEDs so how does that =.0000001?

I am aware that infinity does not equal zero. I was more pointing out that his equations do not produce zero, unless infinity =0.

Being that infity is a reprensentation of the highest number possible,(i stated it was a variable to try and make your equations make more sense) the equations listed would never =0. Sorry if this deffys BrianG logic, but 12/ by any number other than 0 will not = 0.

Please tell me how I'm wrong.

 

[02BlueGT]

P.S. where is the link to the bulbs in question?