[wek120]

Just wondering if it would be possible to build a small circuit to light up an LED when the 2nd cam/lift timing kicks in?
I would think just tapping into the signal from the ECU that goes to activate the OCV (oil control valve) that triggers the cam change would provide a signal to activate an LED, but any ideas as to how to go about doing it without affecting the signal to the OCV?

 

[wek120]

... well for anyone interested, I hooked up a small relay parallel off the OCV on the VVL (closest to the oil cap with a green/pink and blue/black wires coming out). Wiring diagram says volts between the two wires is 12V above 6,000rpm, though when I measured the output of each side with the oscilloscope, one side generated a consistent square wave output the entire time. Attached a wire off each side and sure enough, it activates a relay (which can be used to light a led)

Anyone know if running a small 12v relay coil between the two would draw enough to interfere with the 12v signal? (I assume an Ideal relay coil resistance =infinity? similar to measuring the volts using a voltmeter)

 

[addertooth]

I would use a dropping resister in series with an LED, total current draw is less than a relay (about 20 mA) AND you do not introduce a voltage spike (counter EMF from the coil in the relay), which may be harmful without adding diode crowbar circuit. My exhaust note changes when the car is on second cam, so unless I lose my hearing, I won't need a light to tell me so.
Note: the resistance/impedance of the coil is lower than an LED, ohms law... P=E X I, E=I X R, as R goes lower (assuming E is constant 12V) then I (current) increases as does power load. I think radio shack sells LEDs which already have a dropping resister, and are built to run at 12v. This saves you from doing the math.

 

[sleepe]

Isn't it obvious?

 

[glb]

and the point of this exercise is?????

 

[malakaone]

You can feel it kick in.

 

[WhatsADSM]

I would use a dropping resister in series with an LED, total current draw is less than a relay (about 20 mA) AND you do not introduce a voltage spike (counter EMF from the coil in the relay), which may be harmful without adding diode crowbar circuit. My exhaust note changes when the car is no second cam, so unless I lose my hearing, I won't need a light to tell me so.
Note: the resistance/impedance of the coil is lower than an LED, ohms law... P=E X I, E=I X R, as R goes lower (assuming E is constant 12V) then I (current) increases as does power load. I think radio shack sells LEDs which already have a dropping resister, and are built to run at 12v. This saves you from doing the math.

Assuming you really want that LED.. ^^ This is correct.

Get the relay out of the circuit. Run an LED (and associated bias resistor) off the solenoid circuit. With the LED there is lower current draw and no induced voltage spike when the ECU tri-states the solenoid ground.

 

[wek120]

Everyone says they can hear or feel it kick in, but mine is pretty hard to notice

As for the LED, I can't hook it up directly like that because it produces about 2.5 volts when it is not above the 6000 mark, so it would be half lit the entire time, and also eliminates the use of a transistor; with a relay, it needs to hit 12v before powering the light. The problem with using a diode to avoid a spike is the waveform (on an oscilloscope) is a square wave, so if the voltage between the two wires goes negative during part, it would cause a short. and adding a resistor to the relay would just drop the volts to the coil and it wouldnt switch

 

[addertooth]

Okay Wek120, you are killin' me. Lets make some basic assumptions to start with.
1. You find an LED which has an operating voltage of 2 volts, and an operating current of 20mA or 0.02 amps.
2. Assume you want it fully illuminated at 12 volts (13.8volts is probably your actual)
This means you have an EXTRA 10 volts you must lose across a resister to keep from burning out the LED. (12 volts available, LED needs 2v, must lose 10v)
10 volts, 20ma (or 0.02 amps) "dissapated" across the resister. R (the value of the resister in ohms equals E (the voltage you want to lose across the resister, assumed 10 volts) divided by I (the current required to fully illuminate the LED 20ma or 0.02 amps)
10/.02 = 500 ohms. The resister will always lose some voltage, so when you apply 2.5 volts across the resister which is in series with the LED you will not have enough voltage and current to illuminate the LED. Another route which provides a knife-edge transition from off to on; place two 5.3 volt zener diodes in series with the LED, the zeners will "lose" a total of 10.6 volts, so the voltage must exceed 10.6 volts before the LED sees even a single volt. (caution: make sure the band on them (cathode) is faced the correct direction or the voltage drop will only be 1.2 volts (frying the LED), instead of the intended 10.6 volts)
But all of this aside, if you cannot hear or feel your second cam kick in, then you may have an actual problem with your car (broken lift bolt comes to mind), OR, have a tune which has lowered the cam changeover to slightly below 5000 rpms, where the high and low lobe on the cam produced almost exactly the same horsepower. My cam has the "low" changeover point programmed in (and I love it, thanks Sector111), even so, my exhaust note changes when the second cam kicks in, even though there isnt a noticable seat of the pants acceleration difference at the changeover point. Cam changeover will be extremely obvious in a stock car (I mean ultra-super-duper-crazy obvious).
Finally, If you cannot fully follow what I have written, you probably should not be tinkering with your wiring. I am not going to fly out to fix your car if you smoke your ECU or wiring.

 

[wek120]

Okay Wek120, you are killin' me. Lets make some basic assumptions to start with.
1. You find an LED which has an operating voltage of 2 volts, and an operating current of 20mA or 0.02 amps.
2. Assume you want it fully illuminated at 12 volts (13.8volts is probably your actual)
This means you have an EXTRA 10 volts you must lose across a resister to keep from burning out the LED. (12 volts available, LED needs 2v, must lose 10v)
10 volts, 20ma (or 0.02 amps) "dissapated" across the resister. R (the value of the resister in ohms equals E (the voltage you want to lose across the resister, assumed 10 volts) divided by I (the current required to fully illuminate the LED 20ma or 0.02 amps)
10/.02 = 500 ohms. The resister will always lose some voltage, so when you apply 2.5 volts across the resister which is in series with the LED you will not have enough voltage and current to illuminate the LED. Another route which provides a knife-edge transition from off to on; place two 5.3 volt zener diodes in series with the LED, the zeners will "lose" a total of 10.6 volts, so the voltage must exceed 10.6 volts before the LED sees even a single volt. (caution: make sure the band on them (cathode) is faced the correct direction or the voltage drop will only be 1.2 volts (frying the LED), instead of the intended 10.6 volts)
But all of this aside, if you cannot hear or feel your second cam kick in, then you may have an actual problem with your car (broken lift bolt comes to mind), OR, have a tune which has lowered the cam changeover to slightly below 5000 rpms, where the high and low lobe on the cam produced almost exactly the same horsepower. My cam has the "low" changeover point programmed in (and I love it, thanks Sector111), even so, my exhaust note changes when the second cam kicks in, even though there isnt a noticable seat of the pants acceleration difference at the changeover point. Cam changeover will be extremely obvious in a stock car (I mean ultra-super-duper-crazy obvious).
Finally, If you cannot fully follow what I have written, you probably should not be tinkering with your wiring. I am not going to fly out to fix your car if you smoke your ECU or wiring.

I've actually tested the LED with a dropping resistor and 100K trimpot from the 12v source, and even when the resistance is turned all the way up, the LED still stays dimly lit and keeps the 2.38v to light the LED, even with several different type LED's. So in practicality, that isn't working...

Ahh, completely forgot about using zener diodes, that would definitely make things work as intended, will build up a circuit here and see what I can come up with that. But in answer to your question, yes I do know enough about the electrical aspects to be just fine (which is why I started with an oscilloscope to measure the outputs, and was asking here if someone had a quick opinion what would work Best ex. relay, transistor, zener diode, ect)

As for the lift timing, it is not very pronounced, though you can tell just enough that it does come on at 6k. The MAIN reason I was installing the light was for when you are driving in the 2nd cam, and downshift around a turn, it will let you know if you downshifted slightly out of the 2nd cam. That way it doesn't kick back in and catch you off guard.

 

[kestrel74]

Isn't it obvious?

+9 or 10

 

[wek120]

am I the only one who see's the practicality of this?]

Situation: Keeping it in the second cam while entering a turn. While the turn is just starting to unwind, you downshift to accelerate going out of the turn. You let the clutch back out and it sounds like it is at about 6000 rpms, so you think you never dropped out of the second cam. The rpm's were actually 5950, so it really did drop too low. You start accelerating for a split second and suddenly it kicks on. Since you thought it was in the second cam it catches you off guard and that little kick is enough to break the rear loose, and suddenly you find yourself fishtailing and into a ditch.

Alternate Situation: You let the clutch back out, but the vvtli light went off, so this time you know you dropped a little too low. You back off the turn just enough so when that second cam kicks in, you don't end up in a ditch.

And an LED would be much easier to catch in your peripheral than looking down at the RPM's and trying to tell if it's at 5950 or 6050 in a split second.

 

[ErictheBastard]

Install a Lotus Sport ECU or have a re-flash done by KoldFire or get the CharlieX reflash from BritishRacingGroup. They all lower the cam-switchover point (to the 5200-5750 rpm range) for precisely the reasons you describe. It eliminates the 'kick', yet keeps you on the second cam through gear changes and corner exits.

 

[sturgeongeneral]

I think you are right on track with your thinking. Both electrically and in your idea about an indication of being in the second cam. Please keep us posted!

 

[wek120]

Wish I had know about the ECU flash before doing the mod but on the bright side, it's nice to know how much you actually use the 2nd cam. Tested it out at the autocross last weekend, was surprised how little I was actually using it for autox
Just a quick update though, the waveform is not a steady X volt output, so the despite the zener diodes working from a normal 2.5v to 12v dc switch, they will not work on the actual output (most likely having to do with the square wave). After 12 hours straight of messing with the zeners and led's on it, I finally decided to stick with a very small 12v relay, splicing into the two VVL wires for the relay coil, and 12v power (with a resistor) to the LED. It seems to work just fine with no affect on the VVL itself. You can still feel the kick as it was, and the light comes on when you are driving in the second cam.
For those curious about doing the mod, it was as simple as splicing into the two wires, hook it up to the relay, and running the wires to where you desire the LED.