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post #1 of 27 (permalink) Old 11-21-2012, 11:03 AM Thread Starter
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Floor Shifter Light Black Box

The shifter lights connect to a small black box. My lights are out and I thought that may be the problem, but it isn't. I did notice that, after disconnecting the power, it is still able to keep the lights on for a few seconds. Exactly, what is it? Can I use this black box to keep a light on, a little longer, than courtesy lights or is there something more proper?
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post #2 of 27 (permalink) Old 11-21-2012, 11:08 AM
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That black box is a power inverter, taking the 12V car power and ramping it up to something in the 100v AC range.
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post #3 of 27 (permalink) Old 11-21-2012, 11:14 AM Thread Starter
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Originally Posted by cdmccul View Post
That black box is a power inverter, taking the 12V car power and ramping it up to something in the 100v AC range.
So Dodge decided the shifter lights needed AC, while the other lights were DC? Makes no sense....to me. Thanks for the explanation. So what would I use to keep a light on, say 15 seconds longer, than the courtesy lights?
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post #4 of 27 (permalink) Old 11-21-2012, 11:56 AM
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Time delay relay or equivalent circuit.
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post #5 of 27 (permalink) Old 11-21-2012, 12:25 PM
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Originally Posted by BatCaveTrep View Post
So Dodge decided the shifter lights needed AC, while the other lights were DC? Makes no sense....to me. Thanks for the explanation. So what would I use to keep a light on, say 15 seconds longer, than the courtesy lights?
The lights in the shifter, it is my understanding, are EL lit - Electro-Luminescent... Like the backlighting in the special gauges. It is like the backlighting on an Indiglo watch - and it runs on very high voltage (well, 100 volts or so).
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post #6 of 27 (permalink) Old 11-21-2012, 02:03 PM Thread Starter
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Quote:
Originally Posted by peva View Post
Time delay relay or equivalent circuit.
Gotcha.

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Originally Posted by cdmccul View Post
The lights in the shifter, it is my understanding, are EL lit - Electro-Luminescent... Like the backlighting in the special gauges. It is like the backlighting on an Indiglo watch - and it runs on very high voltage (well, 100 volts or so).
I knew the type of lighting, but didn't know it requires that much voltage. Well don't need it anymore, since the acutal lighting is bad.
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post #7 of 27 (permalink) Old 11-23-2012, 01:35 PM Thread Starter
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A few questions about LEDs. Looking at making my own, instead of buying a strip.
  1. With the viewing angle, if it is 15 degrees, is that measured from looking at the top of the LED and then angling to 15 degrees, down all the sides or only one side?
  2. What is reverse current? reverse voltage?
  3. What is continuous forward current?

I am looking at these, but not sure on resistors, brightness, viewing angle and such. I don't want the lights too bright where they illuminate the inside but also don't need to much of a viewing angle, since you're not looking under the bezel. I'm going to attempt to use the colors for selection, when driving and when parked. I may be jumping in too far for a novice. We will see.
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post #8 of 27 (permalink) Old 11-23-2012, 02:05 PM
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I would strongly suggest starting out with strip LED...

The LEDs you picked out there are not going to work, as they are 2 volt - you may be able to run them in series to get 6 of them to take 12 volts (were 7 or 8 would be safer as car voltage can often be 14-15 volts), but then you have to do the math to find what resister you need to not over-current them... The math isn't hard, but I just don't have it in front of me.

If you start with strip LEDs, then you can put one under each shift indicator, and you can cut and move the next LED to the next indicator opening.

The 15 degrees means that the the light spread is fairly tight - it is a 15 degree viewing total angle...

shaped like this -> V not a 180 degree patter like this -> -- or a 270 degree pattern like this -> ^

If you get my drift... if not, I'll try to draw a picture.
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post #9 of 27 (permalink) Old 11-23-2012, 02:32 PM Thread Starter
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Quote:
Originally Posted by cdmccul View Post
I would strongly suggest starting out with strip LED...

The LEDs you picked out there are not going to work, as they are 2 volt - you may be able to run them in series to get 6 of them to take 12 volts (were 7 or 8 would be safer as car voltage can often be 14-15 volts), but then you have to do the math to find what resister you need to not over-current them... The math isn't hard, but I just don't have it in front of me.
Wouldn't that work though; 6 LEDs (PRND <- ->)? They would all be on, at the same time, just one would have a different color when selected.

Quote:
Originally Posted by cdmccul View Post
The 15 degrees means that the the light spread is fairly tight - it is a 15 degree viewing total angle...shaped like this -> V not a 180 degree patter like this -> -- or a 270 degree pattern like this -> ^

If you get my drift... if not, I'll try to draw a picture.
I think I understand the view; Half-circle = 180, which makes sense; so anything greater (179 - 360) would be approaching straight line and anything less (0 - 179) would approaching no light? If that makes any sense at all.
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post #10 of 27 (permalink) Old 11-23-2012, 02:46 PM
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Well, the spread of a laser beam is like 1 or 2 degrees - the spread of a standard light bulb in your house (the classic A21 for example) is like 350 (the base of the bulb casts a shadow)... So, and LED with a 15" viewing angle is a cone spreading out from a small area the size of the top of the LED, upward in a fairly narrow beam. A 180 means that the light cast will be like that from a fire on the ground. A normal light that you'd find on a work light with a shade over it will be around 90 degrees.

Yes, you could with the 6 LEDs do that, and one would change colors based on where the shift indicator wanted to light up - you'd have to switch there so that the different color light is on the same load as the rest of the LEDs (not hard to do, just gotta think about it in the switching system).

I do not know how adding multiple LEDs in series effects what resistor is needed. If you take a standard electronics LED (like what you show there), even rated for 12v, if you don't have some sort of current limiting device in it, it will burn out fairly quickly - see, LEDs have no (or very little) internal resistance. They will draw power as a near dead short till the power available is used up - which in a car is on the order of a hundred amps or so (battery power available) (yes, it is not that high in practical application, I realize that).

PEVA is probably the best to answer this inquiry.
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post #11 of 27 (permalink) Old 11-23-2012, 04:31 PM Thread Starter
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Quote:
Originally Posted by cdmccul View Post
Well, the spread of a laser beam is like 1 or 2 degrees - the spread of a standard light bulb in your house (the classic A21 for example) is like 350 (the base of the bulb casts a shadow)... So, and LED with a 15" viewing angle is a cone spreading out from a small area the size of the top of the LED, upward in a fairly narrow beam. A 180 means that the light cast will be like that from a fire on the ground. A normal light that you'd find on a work light with a shade over it will be around 90 degrees.

Yes, you could with the 6 LEDs do that, and one would change colors based on where the shift indicator wanted to light up - you'd have to switch there so that the different color light is on the same load as the rest of the LEDs (not hard to do, just gotta think about it in the switching system).

I do not know how adding multiple LEDs in series effects what resistor is needed. If you take a standard electronics LED (like what you show there), even rated for 12v, if you don't have some sort of current limiting device in it, it will burn out fairly quickly - see, LEDs have no (or very little) internal resistance. They will draw power as a near dead short till the power available is used up - which in a car is on the order of a hundred amps or so (battery power available) (yes, it is not that high in practical application, I realize that).

PEVA is probably the best to answer this inquiry.
I think pictures,showing the different casts, would help, but I'm going to see if I can find some in the research. Thanks for some answers.
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post #12 of 27 (permalink) Old 11-23-2012, 05:26 PM
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See note 4 of page 2 of this data sheet: http://www.avagotech.com/docs/AV02-0373EN
It says that "θ1/2 is the off-axis angle where the luminous intensity is half the on-axis intensity. The viewing angle is given as 2θ1/2, which would be total angle, θ1/2 being half that, or 7.5 θ1/2 for a 15 stated viewing angle (2θ1/2).

Reverse voltage is the max. reverse voltage that they guarantee current flow not to exceed a stated safe level - beyond that reverse voltage, you could be damaging the LED. They say 10uA max. reverse current (at 5 volts reverse voltage). Normally you wouldn't put a reverse votlage across it, but they're saying you can put up to 5 volts in the reverse bias direction and not risk damaging it. Beyond that, the risk goes up and they offer no guarantees.

Continuous forward current is the max. continuous current that they are designed for. They fail to tell you over how many hours and to what percentage of intensity it would drop to given that continuous forward current rating. With the cut throat competition among the manufacturers, if they are a manufacturer of unkown reputation (i.e., typical ebay sellers), I wouldn'g go much above 1/2 their stated current if you don't want failure or dimming over time. Even with a reputable manufacturer, if the intensity level is important, you stay below the max. rating. If really critical, you find out how the ratings are determined statistically and design accordingly.

For stacking LED's and sizing resistors, below is a past post of mine. It is poor design to put a fixed voltage across an LED or stack of LED's based on a nominal forward voltage spec. because the tolerance variations around that and small power supply variations will create *huge* variations in the current (and thus brightness and life). *Always* design in current regulation or resistors to reduce the current sensitivity to supply voltage variations.

When you read of LED's failing, either poor or no current regulation was designed in, or the manufacturer pushed the current to exceed a reasonable rating so that they could advertise some high brightness spec. so people who didn't know better would purchase their product vs. the competition who may be more honest in their ratings and/or more conservative in the design. No skin off their teeth when the LED's fail after a few months - an outrageously positive ebay rating was given long before that.

Quote:
You always need resistors with LED's to limit current when a constant voltage source is powering them (like when replacing regular light bulbs). Some LED's come with built in resistors, but their packaging or spec. sheet will say that if that's the case. Assume it has no resistor built in if it doesn't say.

Decide on how many milliamps you want going thru the LED at full voltage. (mcd = millicandellas is the measure of brightness at a given current).

R = (V(b) - V(LED)) C
where:
R is in ohms
V(b) is your car's power system voltage (assume 13.8 with it running)
V(LED) is the voltage drop across the LED (assume 1.2 unless the spec. sheet tells you)
C is the current you want in amps. 20 mA = 0.02 amps

So for example: For V(b) = 13.8 volts, V(LED) = 1.2 volts, and desired current is 20 mA (0.020 amps), the resistor would be 630 ohms - find the nearest value to that - 620 is a standard value I think.

Lower resistance will give you more current, more brightness, shorter LED life.

Some of these bulb locations may have lower source voltages, so for a given location, measure the voltage with the brightness turned all the way up with the original bulb installed, and use that number for your V(b).

For multiple LED's in series, use their total voltage for V(LED) in the formula. For example, if one LED voltage drop is 1.2 volts, and you are putting 2 LED's in series, use 2.4 volts for V(LED) in the formula. For 3 LED's, put 3.6 volts in the formula. Do not stack so many LED's that their total voltage gets more than 4 or 5 volts from the source voltage. Because the source voltage varies some, you need to have a resitor in there so that the current (and the LED brightness) is not going to be super-sensitive to those variations. If you use a voltage regulator, you still need a resistor, but you can get a little closer to the source voltage (in that case, the output voltage of the regulator) because the regulator will give less variation in its output voltage than the vehicle's system.

To determine the wattage needed in the resistor, use the formula:
P = C^2 x R
where:
P = power in watts
C = current in amps
R = actual resistor value in ohms

So for our above example of 20 mA (milliamps) and 620 ohms, the resistor would dissipate 0.248 watts. That's right at 1/4W, but you would not use a 1/4W - you want to go up one size which would be 1/2W. Always try to use at least 50% safety margin on the wattage (for one thing, the wattage rating gets de-rated at higher ambient temperatures). You need some safety factor on that so your resistor lasts forever. If your wattage calculation came out to, say, 0.15 watts, you could safely get by with 1/4W. Any more than that, move up to 1/2W. If wattage comes out more than 0.33 or so, use more than 1/2W (or parallel two 1/2 watters of twice the resistance value each for the same net resistance value).


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Last edited by peva; 11-23-2012 at 05:35 PM.
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post #13 of 27 (permalink) Old 11-23-2012, 06:12 PM Thread Starter
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Thanks Peva. So for the sake of an example: 90 = (13.8 - 12*) .02
So I would need a resistor of 90 ohms if I want the normal brightness, but 50 will make the LEDs brighter and 120 will make them dimmer. But you stated not to run too many LEDs, in series, that the voltage reaches 5+, so it would be better of wiring 3 in series, so then the resistor would 390 instead of 90? I am still lost on the viewing angle, but I don't think it really makes a difference due to their location. I could be wrong. Am I on the right path of understanding?
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post #14 of 27 (permalink) Old 11-23-2012, 06:32 PM
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Thanks Bill, for jumping in. He'll take care of you from here.
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post #15 of 27 (permalink) Old 11-23-2012, 08:12 PM
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Quote:
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Thanks Bill, for jumping in. He'll take care of you from here.
Thanks.

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Thanks Peva. So for the sake of an example: 90 = (13.8 - 12*) .02
Umm - no. The '12' needs to be '1.2' in my example, but according to the data sheet on your LED's, should be 2.0 (the forward voltage). So that would come out to 590 ohms (or closest value).
Quote:
...But you stated not to run too many LEDs, in series, that the voltage reaches 5+, so it would be better of wiring 3 in series, so then the resistor would 390 instead of 90?
390 ohms using the 2.0 forward drop if I crunched the numbers correctly. For consistent brightness (i.e., 12 volts engine off, 13.8 engine running), you could put an 8 or 10 (preferred) volt regulator in, use 3 LED's, and use the regulator output voltage in place of the 13.8 volts.

Quote:
I am still lost on the viewing angle, but I don't think it really makes a difference due to their location. I could be wrong. Am I on the right path of understanding?
I could draw it for you, but visualize a cone with its point at the LED die. The total included angle would be the 15, which would be a line at 7.5 from axis (0) swept around to form the 15 included angle" cone.


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Last edited by peva; 11-23-2012 at 08:15 PM.
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