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Electro-philes: crit a 280Z HL diag.

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    So, here's a new 280Z headlight relay wiring diag. that needs some vetting by experienced minds.

    As will become apparent, this is no quickie solution, as it entails reversible modifications at a couple connectors and a few hard-to-source parts.

    On the plus side, it addresses issues like low-amp switching at the rotary combination, using the (E) power source for relay actuation, and allows keeping the relays inside the cabin. Plus, it includes the Hi-Beam indicator on the Speedo and it frees two spots at the fuse panel (fog lights?).

    Note: the fuses are sized for stock incandescents, but the wire sizes will accommodate hotter bulbs.

    So look it over and, if you will, offer corrections/adjustments as needed.

    If elements look familiar, it's because the diag. was built on the bones of a previous effort for 240s (with a special thanks to it's Author).

    Your attention, help, and support are appreciated.

     

    HLrelays diag_2a_s.jpg

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    Just saw this, so my comments may be moot. Interesting design. Nice graphics.  What year is your car?

    It looks like your shunt diodes are across the coil, while they should be across the contacts.  The arc you are trying to suppress will occur at the contacts.  The 20A fuses are probably too high.  OEM are 10A, IIRC.  If you're planning to use LEDs, then the current will be even less.  What are you trying to protect: relays, wiring, bulbs?  While a 20A relay will run to 20A, it shouldn't ever.  Ideally, you'd like max load <80% of rating so you can have some margin of protection - call it safety factor.  A 55W halogen would run just over 4A at low battery charging voltage (13.5V), so a 10A fuse should be adequate.

    As for wire gauges, as a general guideline, 350CMA (circular mil area) per amp will provide good service with acceptable IR loss.  A 12AWG wire will certainly easily handle 4A, and if you are into the wire harness anyway, maybe it's no big deal.  It's just not necessary for your intended purpose.  That is as long as you're not wiring against what we used to call the gorilla factor - you're trying to prevent some monkey who might be working on your car from damaging a small wire through lack of care.  In such cases, then 12AWG might be just right.  As an example, the 12AWG wire is about 6500CM which should be good for more than 18A.

    Also, it looks like you trying to energize the coil of your HB relay through the indicator lamp?  Not sure if that's actually the plan, but you may find the relay is jumpy without a good, solid connection to ground.  Some relays don't need quite as much coil current to throw as others might. I'd suggest grounding one side of the coil, and using your stalk switch HB signal on the other.  Tie the HB indicator to pin 87 on your HB relay.

    Have fun.

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    16 hours ago, ETI4K said:

    It looks like your shunt diodes are across the coil, while they should be across the contacts.  The arc you are trying to suppress will occur at the contacts. 

    The 20A fuses are probably too high.  OEM are 10A, IIRC. 

     

    Also, it looks like you trying to energize the coil of your HB relay through the indicator lamp? 

    As you said, all of this is probably moot, but thought I have some comments.

    First, the shunt diodes are in the correct place. You're trying to squelch the flyback kick from the relay coil. The relay contacts are on the secondary side and have absolutely nothing to do with that. Many relay manufacturers include diodes inside the relays already (in the correct circuit location), and that's what it looks like the OP was planning to use.

    Second, the original 10A fuses each powered just one headlight side. But in this new scheme, each 20 fuse powers BOTH sides at the same time. Not getting into whether it's completely appropriate or not, doing the math simply doubles up the original fuse size from 10A to 20A. You did the math and came up with 4A for each bulb... I wouldn't want to run 8A continuous (from both sides at the same time) through a 10A fuse.

    Lastly. the connection to the high beam indicator is fine. The HB indicator bulb is in parallel with the relay, not series. They both get "drug to ground" through the switch.

    OK... I'm back into my hole now.  Haha!!

     

    • Like 1

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    Posted (edited)

    Not being familiar with the original 280Z wiring, but on a 240Z with the NLA Dave Irwin headlight wiring mod the ground is changed to a single ground for each headlight rather than the way Datsun originally did the wiring. The original 240Z wiring would not work with LED bulbs because the post normally used for ground was power and high/low beam was controlled by switching grounds. See this thread;

     

    Edited by w3wilkes

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    Well if that won't teach me to stop thinking while on pain meds (recovering from shoulder surgery) - It certainly seemed okay yesterday. 🙂

    CO: I was looking at the diodes to suppress arcing from the load when the relay opens which would reduce contact damage.  Not sure I ever used one across a coil, but I certainly agree coil inductance can cause trouble in certain circuits.  My oversight on the 20A feeding both - I agree with you.

    Thanks for the corrections!

    • Like 2

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    9 hours ago, ETI4K said:

    CO: I was looking at the diodes to suppress arcing from the load when the relay opens which would reduce contact damage.  Not sure I ever used one across a coil, but I certainly agree coil inductance can cause trouble in certain circuits. 

    LOL. Glad to help. Here's one more.

    The only time you need to suppress the arcing from a load is when you have a load that creates an arc. Headlights aren't typically one of them.

    The most common load that causes a problem is an inductive load. That's the exact reason they put that diode across the relay coil. That coil is an inductive load. Headlights are not.

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    I agree re headlights not inducing much arcing, especially not a concern with the very low switching frequency - a few times a night, maybe?  Although, I have seen relay contacts weld up on a motorcycle headlamp flasher (a safety add-on that had a small market) for that very reason (but that was about 100 years ago, I think).

    But couldn't the same thing be said of the flyback energy on a coil being controlled, not by a solid-state device, but a mechanical switch?  There's no real switching noise, and the flyback energy from a 200mA(?) coil load couldn't have any negative impact on the 12V power circuit.  Although I suppose it could make its way through to an ECU or solid-state regulator.

    Questions, not assertions.  Comments always welcome

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    4 hours ago, ETI4K said:

    But couldn't the same thing be said of the flyback energy on a coil being controlled, not by a solid-state device, but a mechanical switch?  There's no real switching noise, and the flyback energy from a 200mA(?) coil load couldn't have any negative impact on the 12V power circuit.

    Same thing cannot be said. The bigger the L, the higher the spike, but the same physics still applies.

    inductance.jpg

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    Posted (edited)

     

    Well folks, it's gratifying to see some feedback. Education never ends...

    First, some comments:

    - The relay diagrams were lifted from actual relays, chosen purposely with the integrated backfeed buffers (diodes, etc.). I acknowledge that the buffers may not be necessary in this case, but I like the sense of extra finesse.  Also, my decision process was influenced by a more intimate knowledge of my own driving habits that employ a more frequent use of beam-switching than may the norm for many. Low-to-hi-and-back-again, plus flashing, added to simple on-and-off, adds up, by my reckoning, to quite a few cycles during typical night road work.

    - Same principle at work with the choice of wire ga. As I am not planning production of umteen thousands of units, I like to afford myself the luxury of over design in wiring.

    - I concede that the fuse choices are probably over-kill, again out of caution (15a would likely be a still prudent choice), but easy replacement makes refinement simple.

    Finally, I will note that since posting this postulation, I have made a few alterations in the design, including a change to an 8 pole relay to control the hi-low circuits, and a 4 pole to control on-off. Of course, this flow required several routing changes, all devised to more accurately emulate the operation of the original configuration of inter-related column switches.

     I'll post a revised presentation as soon as time allows.

    I thank you for your interest and valuable critiques.

     

    Edited by ensys

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    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.

     

    A-B WD5_HLsw det.jpg

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    I'm not clear on how the primary relay is controlled.  My understanding is 12V (always hot) comes into the switch on pin 7, and 12V is supplied to the headlamp fuses on pin 6 when the switch is ON.  You've interrupted that connection with the insertion on the primary relay. 

    So if the relay is not energized and the switch is OFF, 30-87 is closed providing power to the headlamps and pin 6, with pin 7 essentially at ground potential. 

    If the switch is ON, pin 7 is hot which energizes the primary coil, opening 30-87.  But as soon as that happens, power is removed from pins 6 and 7 de-energizing the coil. 

    I feel like I'm missing something.

     

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    Mr.K:

    Well, I think you'll find that the hot in is at pin 6, arriving on the red/wh directly from the fusible links.

    The red out of pin 7 goes to the fuse box where it feed the R and L headlights. I interrupt this feed to energize the Power Rely coil when HL are selected from the switch. This event then directly connects the hot feed from the F.L. to the feed to the HL at the fuse block

    Also, the Power Relay is N.O. as shown. Energizing the coil closes the switch, powering the H.L.

     

    This would be a good time to note that despite the fact that the (e) fuse deployment should continue to provide sufficient circuit protection, I am mulling the idea of introducing additional fuse(s) within the added circuits. Call it the spirit of suspenders and a belt....  Any thoughts?

     

     

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    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!).

     

    image.png

     

     

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    Posted (edited)

    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 by Captain Obvious

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    Posted (edited)


    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 by ensys

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    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.

    image.png

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    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.

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    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.

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    Posted (edited)


    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 by ensys

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    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/
    headlight_Starter_relay_upg_1.jpg

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

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

    59d0185d8e2c8_78Zheadlightmod.jpg.a17403

    http://www.zhome.com/DaveRelay/DaveBuild/DaveInstruction.htm
    image018.gif

     

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

     

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

    relaycircuit.gif

     

     

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    Posted (edited)

    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.image.png

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

    image.png

    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 by SteveJ

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    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.image.png

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

    image.png

    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?

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    8 minutes ago, Patcon said:

    Is that the buzz CO was referring to?

    Yes, sir.

    • Like 1

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    So I rigged up a test in my garage. Per my suspicions, the relay coil did not pick up when it was in series with the headlight.

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