Tangerined
TJ Enthusiast
Strange, current in the loop should be the same at all points in the circuit.
TJ LED front turn signals
- Thread starter bcdawson27
- Start date
Strange, current in the loop should be the same at all points in the circuit.
To be sure, I’ll do that check again today after I check for voltage drop.
My understanding is that that the resistor will indeed limit current output.
Yeah that's about all I've ever checked was amperage. My bulbs are 0.4A if I remember right. When I measured total A it was always way down like 0.03-0.05A for the bulbs and resistors I use.
I think what Tangerined is getting at is that the wattage I calculated is correct, but the resistor is not seeing it all. So your resistor is probably seeing the 0.1A but also at a lower voltage than the 12.8, leading to lower wattage ratings at the resistor than calculated.
As soon as I get home I'm going to start testing because I'm really curious too. would make me feel even better about my oversized resistors.
And as for the current being different at certain points of the circuit, I would think that should be the same no matter what too. But I will look into that as well when I test mine.
Power is I^2 * R so your power is about 1/10 watt if 0.03A is correct and 1W if 0.1A is correct. The current in a series circuit should be the same everywhere like @Tangerined says so I don't know what's going on with your measurements.
Power is E*I (basic formula)
R= E/I (so 12.8/.13= roughly 100 ohms like I’m at so I believe this checks out).
I wanted to add this in case it helps since it discusses current limiting resistors.
Edit: I’m no math whiz and my brain is melted from tax class but I’ll do my best to hang with everyone’s formulas/equations.
Last edited:
Alright so I made some interesting findings:
A few notes:
-Amperage was always the same no matter where measured
-220 ohm 2W resistor used
-Battery was performing at 12.47-12.48V
Amber 194 bulb from VLEDS (very bright):
-Rated to be 0.4A in normal operation - true
-amperage found at start and end of circuit with resistor in place: 0.04A
-voltage across resistor: 8.52V
-voltage across bulb: 3.96V (these V’s check out with the 12.48 battery voltage)
P = V^2 / R
P = 8.52(8.52) / 220
P resistor = 0.33W
2W resistor so that will live forever.
Considering I had 0.04A in the circuit @ 12.48V, this means the total wattage was 0.49W, so this means the bulb was operating at 0.15-0.16W compared to its normal 5ish W.
Generic white 194 bulb from parts store (dim, cheap)
-0.06A normal operation
-amperage found at start and end of circuit with resistor in place: 0.015A
-voltage across resistor: 3.81V
-voltage across bulb: 8.66V (these check out)
P = V^2 / R
P = 3.81(3.81) / 220
P resistor = 0.066W. Would live forever also
220 ohm resistor hooked straight to battery +/-:
-0.05A draw
-voltage across resistor: 12.47V (duh)
-voltage from resistor leg to battery: 0V (duh)
P = V^2 / R
P = 12.47(12.47) / R
P resistor = 0.70W (ran pretty hot for a 2W resistor but still in an operable range)
Overall findings:
-The resistor definitely ran hottest when hooked straight to battery at full voltage (still within safe rating though)
-resistor still ran warm with bright powerful amber bulb hooked up (but still safe)
-clearly the brighter/more powerful led bulb hooked up, the higher the voltage will be at the resistor. Not sure why it’s inverse like this
At the end of the day, I think it’s safe to say we are all well within the limits of the resistors being used. Definitely interesting to learn the info though. I don’t understand how the voltage biases towards either the bulb or resistor based on what brightness of bulb is being used, but at least I can measure and do the math to verify it’s safe.
A few notes:
-Amperage was always the same no matter where measured
-220 ohm 2W resistor used
-Battery was performing at 12.47-12.48V
Amber 194 bulb from VLEDS (very bright):
-Rated to be 0.4A in normal operation - true
-amperage found at start and end of circuit with resistor in place: 0.04A
-voltage across resistor: 8.52V
-voltage across bulb: 3.96V (these V’s check out with the 12.48 battery voltage)
P = V^2 / R
P = 8.52(8.52) / 220
P resistor = 0.33W
2W resistor so that will live forever.
Considering I had 0.04A in the circuit @ 12.48V, this means the total wattage was 0.49W, so this means the bulb was operating at 0.15-0.16W compared to its normal 5ish W.
Generic white 194 bulb from parts store (dim, cheap)
-0.06A normal operation
-amperage found at start and end of circuit with resistor in place: 0.015A
-voltage across resistor: 3.81V
-voltage across bulb: 8.66V (these check out)
P = V^2 / R
P = 3.81(3.81) / 220
P resistor = 0.066W. Would live forever also
220 ohm resistor hooked straight to battery +/-:
-0.05A draw
-voltage across resistor: 12.47V (duh)
-voltage from resistor leg to battery: 0V (duh)
P = V^2 / R
P = 12.47(12.47) / R
P resistor = 0.70W (ran pretty hot for a 2W resistor but still in an operable range)
Overall findings:
-The resistor definitely ran hottest when hooked straight to battery at full voltage (still within safe rating though)
-resistor still ran warm with bright powerful amber bulb hooked up (but still safe)
-clearly the brighter/more powerful led bulb hooked up, the higher the voltage will be at the resistor. Not sure why it’s inverse like this
At the end of the day, I think it’s safe to say we are all well within the limits of the resistors being used. Definitely interesting to learn the info though. I don’t understand how the voltage biases towards either the bulb or resistor based on what brightness of bulb is being used, but at least I can measure and do the math to verify it’s safe.
The following test results should make more sense.
Without a resistor
Voltage reading to the LED = 12.58V
Current = .22A
Power without a resistor = 2.77W (12.58V*.22A)
With a 100-ohm resistor (measured at 96.5 ohms)
Resistor voltage drop = 3.17V
Voltage reading to the LED = 9.4 (math check: 3.17V + 9.4V = 12.57V)
Current = .03A (29.6mA)
Power to the LED with a resistor = .28W (9.4V*.03A)
I used a quick calculator regarding the minimum Resistor Wattage:
This might explain why my resistor is not getting warm (My 1W resistor is more than 11x the minimum rating required for what I'm doing).
Without a resistor
Voltage reading to the LED = 12.58V
Current = .22A
Power without a resistor = 2.77W (12.58V*.22A)
With a 100-ohm resistor (measured at 96.5 ohms)
Resistor voltage drop = 3.17V
Voltage reading to the LED = 9.4 (math check: 3.17V + 9.4V = 12.57V)
Current = .03A (29.6mA)
Power to the LED with a resistor = .28W (9.4V*.03A)
I used a quick calculator regarding the minimum Resistor Wattage:
This might explain why my resistor is not getting warm (My 1W resistor is more than 11x the minimum rating required for what I'm doing).
So the reason 100 ohms gets warm for me is my higher wattage bulb causing more voltage across the resistor I suppose. My resistor is about 5W and is causing my resistor to run at mid 8 volts which would explain it. The little bulb I tried had in the 8’s voltage across it and the remaining 3.9 across the resistor didn’t get the resistor hot.
LED is 5W roughly. It’s like 0.4 or 0.43A at 12ish volts. I use 2W resistors only lately.
My math is showing I’m pulling 0.33W through the resistor. But they’re definitely getting warm and even with 220 ohms, I’ve got 8.X volts across the resistor. I just would have thought 0.33W would run cool.
Me personally I use 470 ohm much of the time. There was no heat from those.
So your LED is roughly 5.36W, your current draw is .43A, and your battery could use a charge (12.47V).
Yes, yes, and yes, but this is just a spare small boat battery I use for projects like this. I charge it once a year or so and leave it alone the rest of the time.So your LED is roughly 5.36W, your current draw is .43A, and your battery could use a charge (12.47V).
This link does a good job explaining voltage drop in more detail which makes it make more sense.
What is your current draw with the resistor in the circuit?
.43A seems like it would be the current reading without a resistor and if that's the case it will be less after you introduce the resistor to the circuit.
What's your voltage reading at the LED (when dimmed by a resistor). It should be battery voltage (12.47V) minus voltage drop (3.9V), which would be around 8.57V.
Then, take your current * voltage to see the LED's power output with the resistor in the circuit.
Edit: For example, my voltage at the LED is 9.4V, and the current is .03A. This means that my LED is producing .28W (9.4V*.03A) of power when dimmed by the resistor (unless I'm wrong which is certainly possible).
Last edited:
I tried to do this test while my jeep was in the shop, but my 12V lantern battery (non-serviceable and barely used) wouldn't even turn on my LED lol.
So I've basically been tearing up while waiting to check on this nonsense.
Hopefully you’ll get it back soon then!I tried to do this test while my jeep was in the shop, but my 12V lantern battery (non-serviceable and barely used) wouldn't even turn on my LED lol.
So I've basically been tearing up while waiting to check on this nonsense.
I got it back yesterday and was doing (failing at) these tests in the dark last night to help me be able to go to sleep.
Wow just skimmed through this.
Macho your setup sounds awesome.
My question is if this is to get the front and side blinkers working were do you actually put all these components. Did you make a box in the engine compartment? Are they just in the wheel well or behind the front grille?
This is just my curiosity kicking in.
Macho your setup sounds awesome.
My question is if this is to get the front and side blinkers working were do you actually put all these components. Did you make a box in the engine compartment? Are they just in the wheel well or behind the front grille?
This is just my curiosity kicking in.
Thanks, I’m happy with it. The wiring goes on the side markers. I don’t value my time very highly so I I pulled the headlight harness out and hid the diodes and resistors inside the big loom between the headlights and then ran the wires out to the side marker plugs. I wanted to make it so the wiring out at the side markers wasn’t bulky and still fit in the small stock loom.
The easier and less time consuming approach would be to just splice on the stuff at the side marker location. Only downside to that is it gets bulky and probably won’t fit in a loom.
Another way it can be done is make a jumper harness that plugs into the side marker plugs and then has a ground wire coming out of it that you bolt to a headlight ground or similar. That way you don’t modify the stock harness. The only real downside to that method is that it will add extra bulk under the fender. Pretty much would have to loop the wiring around under there to waste the slack.
No need for boxes. Macho helped me realize that I needed to locate my running light to the side marker (it's one of the white/yellow wires on each side of my engine bay).
There are two sets of w/y wires on each side. one for the front turn signal and one for the side marker (on each side).
For this part, you simply cut the running light wire on each side and then insert a resistor on each side. Note: If you cut the wrong running light wire (i.e. the side marker running light doesn't shut off but your front marker light turns off) then just splice the wire back together and you will now be 100% sure that you need to cut the other wire (to the side-marker).
Then splice in your resistor, which will dim your side marker's running light output.
The jumper harness method is pretty intriguing really, assuming you can zip tie up the slack and keep the assembly rigid enough to not break resistor legs. Should be fine really. The nice thing about a jumper harness is how quickly it could be reversible. The way I did it is permanent but I don’t expect failures so I’m not worried about it. Would have been nice to just plug in a harness and ground a wire per side though.