Electro-philes: crit a 280Z HL diag.

[ETI4K]

Wow, that's weird.  My wiring diagram ('76) has the fusible link feeding pin 7 and pin 6 feeding headlights.  Something's screwy here.

We must have been trained to different conventions as to illustrating relays.  The way I've always shown it is as though the coil would draw the contact arm towards it when energized.  So, the de-energized state would look like below.   Gee, this is like when two NASA locations were working in different units (imperial vs metric) and did not tell each other. ?

The suspenders and belt idea sounds smart.  Make them all easy to get to and document your additions for posterity (and troubleshooting!).

 

 

 

[Captain Obvious]

What makes your approach so novel or better than the other hashed out options that have been out there for years? What makes you think that your mousetrap is better?

I think it won't work. The relay you're trying to energize with the current coming back from the filaments? I think that relay is going to sit there and buzz when you switch to high beams.

 

Edited April 5, 2020 by Captain Obvious

[ensys]

Mr.K:

Well, there we are; I wasn't working with a '76 wiring diag. and you don't have one for '77. I don't know what to tell you about the differences between them, but I used the familiar "Classic Zcar" version, which I have found to be quite accurate for '77s like my own.

Meanwhile, while my relay diagrams may not be mechanically precise, but they are diagrammatically correct, taken from those shown for the respective relays. Still, I appreciate a continuing education; for example, I have never seen a 2-pole relay that is both Normally Open and Normally Closed at the same time. Live and learn.

Mr.C.O.:

What makes you think it isn't? You tell me. After all, isn't that the point of such discussions among learned gentlemen?

I must admit Sir, that it's hard to find anything constructive in baseless scoffing at my temerity for trying a clean-sheet solution. I'm not selling anything, nor am I here shouting "Eureka, ain't I great?"; I'm looking for someone to tell me where I go wrong.

What would be constructive would be some clear representations of a few of those time-tested solutions of which you speak. Especially those that preserve all the functions of the (e) controls and indicators without materially altering the stock harness. I'll look forward to that.

Again, all constructive criticism is welcome. We're all here to learn from each other, are we not?

 

Edited April 5, 2020 by ensys

[SteveJ]

19 hours ago, ensys said:

 

For your consideration; Plan B

I believe this configuration will preserve all the original switch and indication functions with the least alteration of the (e) wiring.

I invite your comments.

The short answer is that it won't work.

The longer answer is that you will never have sufficient current flowing through the second relay coil. You have a parallel path to ground. I've made a simplified drawing of your circuit that I hope allows you to see the issue. If you removed the high/low beam switch from the car and had the headlight switch on, you would find you have 12VDC to ground on both the red/black wire and red/white wire. The headlight would not be shining, and the filament just acts like an ordinary piece of wire.

In your drawing, you have the low beam circuit on a normally closed contact. That gives it a straight path to ground.

This actually renders your other design flaw moot. The second flaw is that the headlight and second relay coil are in series. The relay coil has a resistance of around 80 ohms. The high beam filament has a resistance of less than 4 ohms. Since they are in series, that means the voltage drop across the coil is going to be at least 20 times more than the voltage drop across the headlight filament. At less than 5% of its designed operating voltage, the headlight would not even have a dim glow.

All working examples of headlight relays isolate the positive and negative for the relay coils from the positive and negative for the headlights. Also, it's smarter to have two positive wires and one negative wire for the headlights in case you ever want to use LED headlights.

[ensys]

 

Now that's what I'm talkin' about. I thank you Mr.J for your thoughtful contribution.

Since my forte runs more to structural than electrical engineering, I must beg some patience with my slow grip. I will have to chew on this a bit before I can offer a proper rely.

More later, and again, my thanks for helping me out.

[ETI4K]

ensys,

I applaud your effort for trying to make this improvement without substantially altering the harness.  There's never been only one way to solve a problem.  I look forward to seeing your ideas.

[ensys]

Mr.J:

With all due respect for your obvious acumen,  I believe that your diagram has simplified some germane events out of the circuits. And while the narrative may simply be beyond my grasp, some elements puzzle me. With your indulgence, I would hope some dialog might illuminate my errors.

Following the path of your narrative, let us begin with the Secondary relay:

While I acknowledge that the metaphor of hi/low switch removal baffles me (as does the notion that the HL could somehow operate without it), it is the case that the returning hi (wh/red) and low (blk/red) HL circuit wires both have 12v potential on their way to the switch, looking for ground. However, please note that the low bm. circuit is diverted before the connector, leaving only the hi beam circuit to enter the switch to become the coil activator for the Secondary relay via the (e) ground path when hi bms. are selected.  

Thus, low bms. are the relay default when HL are activated, by virtue of no current to the coil, and the N.C. low bm. switch in the relay. My problem is in seeing how this arrangement would leave too little current to power the low-amp coil draw.

And while being wrong is always on the table for me, isn't it the case that when two circuits share a common source and termination, they are in parallel? Wouldn't this obviate the whole "voltage drop" thing?

I understand the "norm" of design you describe, but hey, flow is flow. Dipping into circuits to power their own flow control seems little different than the multitude of taps and diversions that are characteristic of every harness. Besides, in a lighting circuit, what is the actual consequence of the failure of the coil circuit, when the flow it controls fails?  To me, this is the essence of a "moot point".

If I seem stubborn in my ignorance, it's because I am strongly motivated by the potential elegance of a solution that integrates into the function of two complex switches to accomplish their functions in a different way, while providing operation indistinguishable from original, without a single permanent alteration of the orig. harness. Call it a fetish.

Finally, yeah well, if I was smart, I wouldn't fool around with old cars because I am entranced by their time machine qualities. Or; if I wanted the affliction/affectation of hi-tek, I'd drive newer cars.

So, Sir, in full realization that I risk exposure of being electrically inept, I fervently hope my narrative will help my on-going education..

Please, fire away.one and all.

 

Edited April 5, 2020 by ensys

[Captain Obvious]

12 hours ago, ensys said:

I must admit Sir, that it's hard to find anything constructive in baseless scoffing at my temerity for trying a clean-sheet solution. I'm not selling anything, nor am I here shouting "Eureka, ain't I great?"; I'm looking for someone to tell me where I go wrong.

What would be constructive would be some clear representations of a few of those time-tested solutions of which you speak. Especially those that preserve all the functions of the (e) controls and indicators without materially altering the stock harness. I'll look forward to that.

I neither scoffed, nor expressed the sentiment of "temerity" in your work. Nor did I suggest you were selling anything or shouting "ain't I great".

In fact, I myself took a different approach to my headlight relay upgrade design. And if someone were to ask ME "What makes your approach so novel or better than the other hashed out options that have been out there for years? What makes you think that your mousetrap is better? ", I would have a sound technical answer for them that would not include accusing the asking party of scoffing, expressing temerity, or the rest. But whatever.

As for previous solutions?
https://www.google.com/search?q=headlight relay site:www.classiczcars.com

https://www.classiczcars.com/forums/topic/50416-75-280z-headlight-relay-upgrade/

http://zhome.com/ZCMnL/HeadlightRelays/JudkinsRelay.htm

https://forums.hybridz.org/topic/91938-headlight-relay-for-260-280/
 

http://www.zhome.com/DaveRelay/DaveBuild/DaveInstruction.htm

 

https://www.classiczcars.com/forums/topic/60198-240z-led-headlight-wiring/

 

http://www.danielsternlighting.com/tech/relays/relays.html

 

 

[Captain Obvious]

lmgtfy

Edited April 7, 2020 by Captain Obvious

[SteveJ]

Think of the headlight as two resistors. They are both connected to the battery positive. Since the Normally Closed contact on the second relay is connected to ground, you can see the green and blue connect. That means the low beam circuit has a path to ground until that coil could energize. However, even with the switch in the high beam position, the coil can not energize.

So electrically, when you move the high/low beam switch to high beam, this is what the circuit looks like:

While the coil in the relay is an inductor, for the purposes of this explanation, we will only look at the resistance (impedance) of the coil. At the headlight, there are two branches to the circuit. One goes through the headlight high beam filament and the relay coil and returns to negative. The other branch goes through the low beam filament and returns to negative. Since we know the voltage of the battery (12 VDC) and the resistance of the components of the circuit (4 ohms for the filament and 80 ohms for the coil), we can calculate the current flowing through each branch and the voltage drop across each component.

Total current = Battery voltage divided by Effective resistance

R1 = R(headlight) + R(coil) = 4 + 80 = 84

R2 = R(headlight) = 4

Since this site has limited HTML capabilities, here's a link to see how to calculate effective resistance: https://www.electronics-tutorials.ws/resistor/res_4.html

R (eff) = 3.82 ohms

Total Current = 3.14 A

Indeed, when you apply Ohms Law to each branch of the circuit, it matches our results.

I1 = V/R1 = 12/84 = 0.14 A

I2 = V/R2 = 12/4 = 3.00 A

So now we can calculate the voltage drop across each component for the branch with the high beam coil.

V (high beam filament) = 0.14 * 4 = 0.57 V

V (coil) = 11.43 V

Even if that is enough power for the coil to pull in the contacts, you have another problem. When the relay contacts connect the high beam circuit to ground, the voltage at the coil drops to zero. (In the first diagram, the black wire would have 0 VDC to ground with the relay contact closed. It is all the same wire electrically speaking.) When the coil de-energizes due to no voltage, the contacts will return to their normally closed state. Your headlights would switch to low beam. The coil would energize and swap to high beam. The voltage at the coil drops to zero...ad infinitum until you swap the high/low beam switch back to low beam.

Edited April 5, 2020 by SteveJ

[Patcon]

45 minutes ago, SteveJ said:

Think of the headlight as two resistors. They are both connected to the battery positive. Since the Normally Closed contact on the second relay is connected to ground, you can see the green and blue connect. That means the low beam circuit has a path to ground until that coil could energize. However, even with the switch in the high beam position, the coil can not energize.

So electrically, when you move the high/low beam switch to high beam, this is what the circuit looks like:

While the coil in the relay is an inductor, for the purposes of this explanation, we will only look at the resistance (impedance) of the coil. At the headlight, there are two branches to the circuit. One goes through the headlight high beam filament and the relay coil and returns to negative. The other branch goes through the low beam filament and returns to negative. Since we know the voltage of the battery (12 VDC) and the resistance of the components of the circuit (4 ohms for the filament and 80 ohms for the coil), we can calculate the current flowing through each branch and the voltage drop across each component.

Total current = Battery voltage divided by Effective resistance

R1 = R(headlight) + R(coil) = 4 + 80 = 84

R2 = R(headlight) = 4

Since this site has limited HTML capabilities, here's a link to see how to calculate effective resistance: https://www.electronics-tutorials.ws/resistor/res_4.html

R (eff) = 3.82 ohms

Total Current = 3.14 A

Indeed, when you apply Ohms Law to each branch of the circuit, it matches our results.

I1 = V/R1 = 12/84 = 0.14 A

I2 = V/R2 = 12/4 = 3.00 A

So now we can calculate the voltage drop across each component for the branch with the high beam coil.

V (high beam filament) = 0.14 * 4 = 0.57 V

V (coil) = 11.43 V

Even if that is enough power for the coil to pull in the contacts, you have another problem. When the relay contacts connect the high beam circuit to ground, the voltage at the coil drops to zero. (In the first diagram, the black wire would have 0 VDC to ground with the relay contact closed. It is all the same wire electrically speaking.) When the coil de-energizes due to no voltage, the contacts will return to their normally closed state. Your headlights would switch to low beam. The coil would energize and swap to high beam. The voltage at the coil drops to zero...ad infinitum until you swap the high/low beam switch back to low beam.

Is that the buzz CO was referring to?

[SteveJ]

8 minutes ago, Patcon said:

Is that the buzz CO was referring to?

Yes, sir.