View Full Version : Question for the Electronic Gurus
11-27-2002, 09:17 PM
If replacing the bulbs in the rear of our trucks with LED's is not putting enough load on the system for the flasher to work properly. Why couldn't a relay on the circuit be used? Would this be enough of a load for the flasher to work properly? Use any relay on each light and run the wire from the switch to the coil and run the switched terminal to the LEDs.<br>Just got thinking about this tonight while driving home.<br>Thanks<br>Chuck
11-27-2002, 09:24 PM
I also had a thought about the LED problems.<br><br>Why not just take your trusty ohmeter and measure the resistance of the stock bulbs.<br>Measure the resistance of the new LED setup.<br>Add a resistor in the circuit to bring the LED setup to the resistance of the stock bulbs.<br><br>I only have basic instruction in electronics (3 years high school, half year tech school) but this makes sense.<br>Maybe too much sense.<br><br><br>phox
11-27-2002, 10:03 PM
Phox,<br>I've thought of that too. Would you put it in the power side or ground side?
11-27-2002, 10:30 PM
I've thought of that too. Would you put it in the power side or ground side?
I just talked to the senior version of myself (we work together) and my idea wouldn't work.
He's had a bit more schooling and 30 years of practical usage of electronic theory.
He says we'd have to measure the current the standard setup draws.
Then measure the current the LED setup draws.
Then add a resister "in parallel" to bring the LED setup up to what the standard setup draws.
I'm sure if we knew these numbers, he could tell us exactly what size resister we'd have to add.
Adding it in series would just decrease the voltage, not the current.
The current draw is what makes the blinker blink.
He also says we could put in a different blinker module that doesn't care what the current draw is.
Auto manufacturers use one that cares, so if one blinker bulb goes bad, the blinkers don't blink to alert us to one needing to be replaced.
Or, in the case of our trucks, the blinkers blink fast to alert you.
Isn't this the problem people are experiencing?
I know I've read something about it, either here or other site, but can't remember the details.
I knew deep down that it had to be simpler than what I suggested [undecided].
11-29-2002, 12:44 AM
If I may add my 2c.
If you want to continue the use of the old relay you'll need a resistor in parallel to the LED flashers for each side . The resistor will be dimensioned like this:
Sum up power of all bulbs on one side. = P
Sum up power consumption of all LEDs for one side =Pn
Calculate original resistance: 144/P = R
Calculate LED-set's resistance 144/Pn=Rn
Calculate Resistor: Rz = 1/ (1/R - 1/Rn)
These 2 resistors will be between the + of the turn signal on each side and ground. You need two separate resistors to keep the turn signals separated ( left from right) and the hazard funktion working.
Take a good quality resistor that will be able to dissipate the heat ( found in tech datasheet for the resistor) ...
Downside of this is that you won't know whether your turn signals work or not. The flasher will go on working as long asd it's got the right resistance. Do not use an Ohmmeter on a cold lightbulb to determine anything-except cold resistance. The resistance increases when the filament gets hot. If you measure an H4 headlamp bulb cold you'll find about 0,08 Ohms- this would let you assume 150 Amps , and therefore
1800 Watts .. . I trust the the manufacturer saying 55W .
11-29-2002, 01:08 AM
OK here we go! Using an ohmmeter to the bulb is useless. The tungsten in the bulb has what they call a positive temperature coefficient which is engineering BS to say that as it heats up it becomes more resistive meaning if you measure the resistance cold it will be different while in normal operation.<br> Watts= volts x amps<br> say a turn signal bulb is say 35 watts (actual heat)<br> 35= 12volts x 3 amps<br> volts is like water pressure you know its waiting for you when you turn on the tap<br> amps is like how many gallons per minute<br>the 2 combined = watts.<br>so if the turn signal bulb is 35 watts(you have to find that out) and the new LED light is say 5 watts (about 1/2 amp)you have to make up for 30 watts.<br>35watts -5watts=30 watts<br>OK now you say OHMIGOD how do I figure out what resistor to use.<br>OK I'll give you one more formula<br>watts also = amps x amps x resistance in ohms<br>30watts =2.5 amps x2.5 amps xresistrance in ohms<br>30 = 6.25 x resistance in ohms<br>devide each side by 6.25and you get<br>30/6.25=6.25/6.25 x resistance in ohms<br>6.25/6.25 = 1 so therfore 30/6.25= resistance in ohms<br>is about 4.7ohms which is a standard size resistor BUT YOU HAVE TO GET A RESISTOR GOOD FOR 30 WATTS OR WHATEVER VALUE THE TURN SIGNAL LAMP IS!!!<br><br> hope this helps<br><br>mitch
11-29-2002, 01:17 AM
yomitch : ;D<br><br>AlpineRAM
11-29-2002, 01:42 AM
I will have to reread this tomarrow. My brain is full for today.<br> [eyecrazy] :'(
If you have the old two blade flasher just get a heavy duty electronic flasher as used on bigger trucks, They have an electric motor inside that runs at a a steady rate. It also will give brighter lights when pulling a trailer. If you have the new 5 pin relay there is a new heavy duty coming out but I have been told it has not been released yet, so in that case you have to do the math.<br><br>Cost is about $12.00 each I have one in the 4ways and one in the turn signals.
diesel dave c
12-01-2002, 07:14 PM
Just found an ad for no load flashers in a Street Rodder mag.Check out www.HotronicsProducts.com. Looks like a plug and play install. A bit pricey @ 30 bucks, but it takes away all the hassels<br><br>dave
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