Yeah, that could be my fault because I work with other engineers on a daily basis, and we don't talk in individual parts resistor/diode language, we say things like, "Why don't you try using a latch circuit or a peak detector circuit?", we all know what the function of those circuits are and how they work and it tells you a lot more information about the circuit than saying, "why don't you use a resistor/diode?", because those 2 components can we wired in several different configurations and tells you nothing about the intended function of the circuit. It's then up to the engineer to decide which components he wants to use to build that circuit. Go google, "Peak Detector Circuit diagram" and see how many different circuits, with different components you find, where they all have the same intended function but have different components where they can be adjusted to best meet the application the circuit is being used in.

You don't have the "full circuit" diagram shown. Correct me if I'm wrong. Wasn't the original intention of this circuit, to turn on auxiliary lights when the bright lights were turned ON. So that voltage source that is triggering your aux circuit lights, is also powering the JKs main bright head lights at the same time correct? Your circuit diagram doesn't show the existing wiring, which is powering the OE head bright lights. You might be ok there when the diode was added in your circuit, but there is a potential for the Cap to be discharging through the OE head lights....which means your "R" value in that (5xRxC) equation you just learned about could have gotten even smaller and now that you've done some math, you can see how big of a difference that R value can have on the discharge rate of the capacitor value "C".

Now hold up right there. This is all great and very useful information. You measured 104 ohms across the coil on 1 of your relays. For simplicity sake, let's round that off to an even 100 ohms. Now doesn't your circuit diagram actually show 2 relays? So you're using 1 relay for each aux light, and that same circuit is controlling both relays correct? So, what that means is that the (2) 100 ohm coils of each relay are wired in parallel with each other and what happens to the total resistance "R" when you wire (2) 100 ohm resistors to a circuit? The total resistance drops in half. So now your "R" value is actually 50 ohms and not 104 ohms like you measured, which means your (5*R*C) is actually .25 seconds and not .5 seconds. Don't believe me, then wire the 2 relay coils together like they would be in your circuit, then take a measurement of the resistance, that the circuit is actually seeing.




No it wasn't. However, we're now traveling on the path we need to be going to get you closer to the correct cap size....."IF", there is actually one, where I have my doubts. The (5xRxC) equation is not really useful to us for what you are trying to do, but I'm glad the video showed it, because it got you to thinking of what is going on in this circuit in regards to the Resistance and the Cap value. You see, the (5*R*C) equation, tells you how long it will take for the cap to fully discharge, where we're really not interested in that. A more useful equation for what you are trying to achieve is the (RxC) equation, which is named the RC time constant.

What T(sec)=RxC tells you, is that the Initial fully charged voltage will have dropped by 63% of its initial charged voltage at time "T", in seconds.

I made that initial bigger, because that's a key word of where I'm going.

Here's a graph of a typical Capacitor discharging, just to give you some kind of visual of what is going on, and why this isn't as complicated as it sounds..


Voltage level(V) is on the Y axis and Time (T) is on the X-axis.

See how T=5xRxC the voltage from the cap is fully discharged, and at T=(1xRxC) it is discharged by 63% from its initial charged value?


That wasn't really my intent for asking the minimum voltage value across the coil for knowing what voltage closes the relay. More so to know the minimum voltage level to "keeps it closed". This is a very important piece of information. Thank you for looking it up. We want to try to make sure that relay does not open back up, during the time our capacitor is discharging. So we need to chose the value of the cap "C" so that when it's near its 65% discharge point, the voltage does not drop below 8.4 volts.

Hopefully, by now you can start to understand how your additional diode is actually hurting you. You're not allowing the cap to charge up to a higher "initial" voltage than it could be without out.

Now imagine putting a 6 volt regulator in that circuit, and limiting the cap so the initial fully charged voltage can only charge up to 6 volts. You might as well just go shoot yourself in the foot, but I'll wait to hear back from Sahara_Maverick on how well that experiment is going.

 

MY BAD. I think i hi jacked this tread. SORRY for the confusion.

MY intention is to upgrade my IPF H4 headlights bulbs to a higher watt bulb without frying the factory wiring and the Tipm. So i made this relay harness to isolate tipm and the H4 bulbs (it goes in between the factory H13 plug and the H4 bulb).

This is the new diagram. I remove the diode on the ground from the factory H13 plug and put in a jumper.

One relay is powering up both low beams and the other is powering up both hi beams on the IPF headlights.

I found that info on the tech sheet. Drop-out min. voltage @23*C is 10% of rated voltage ,so the math come out to 1.4volts @14vdc, 35amp 14vdc relay times .1=1.4.
So as long as these caps voltage doesnt drop below 1.4 volts this relay should stay closed, as far as the tech sheet says. But we all know there some wiggle room in there too.
Well you could aways go back to the 1940's and find some 6 volt headlight bulbs to use.
Im going to try some smaller caps to see what happens.
I see what you were saying about the resistance in the circuit if the headlight were hooked up to the harness. It would be much less then 104 ohms. Thanks and Sorry for the confusion.

 

Hear some pics of the finished relay harness. All pics are a little fuzzy.

The cap made 2.

Closer pic of the harness with everything mounted.

Some of the parts in it.

With the cover on.

And the bulbs.

Installed

 

Here is a link to my write-up as promised. https://www.jk-forum.com/forums/jk-w...6/#post3509690

Keep in mind that this is not only to fix this problem but I integrated it into my own version of an SPOD. my solution is worked into the schematics (which are tough to see so PM me if you want a better detailed .pdf)

 

Holy Sh*tballs. I just read through 9 pages of this to realize that I'm going to just install a switch and forget about the high beam relay for my light bar.

Thanks for saving me the time of trying to figure this all out guys!