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Resistor to add to LED indicators


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Probably a quick one for you electronics type peeps, but I've bought some LED indicators(fog light conversion investigation), and when I tried them out, the flashing was quicker.

 

I presume they are less of a load on the flasher unit and I'll need to wire a resistor in-line with them.

 

Does anyone know what resistance I need?

 

Cheers

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Need to know the forward voltage (Vf) and the forward current (If) to tell you the proper value, if it is indeed a current limiting type resistor you need.

 

However I would have thought that repacement LED indicators would have had one of these build in...

 

Odd. Are they unusually bright too?

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Actually I made a cockup. That's a 650 ohm resistor. That seems awfully high, especially for 15 LEDs that are (I'm assuming) in series).

 

Let me have a think...

 

*EDIT* No it isn't, it is a 65 ohm one. Get with the programme, Darren!

:stupid:

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Shit. This is complicated. :(

 

Firstly, all the LEDs can't be in series because even if the volts dropped across each LED was only 1V, then there would not be any volts left over from the supply to require a current limiting resistor.

 

Therefore I reckon that the whole bulb is a mixture of LEDs in series and parallel, with the resistor compensating for a small remainder. Trouble is that withoug the Vf and If fgures for each individual LED I can't work out either the arrangement or the total load. :(

 

I can get close by assuming that the whole bulb consists of 3 sets of 4 LEDs in series, plus the three remaining LEDs in series with the resistor. If each resistor has a Vf of 3V and an If of 50mA then this gives a resistor value of 60ohms. The other odd thing is that I don't think that a 65 ohm resistor is a standard value, but according to the colour banding that's what it is.

 

Anyway, assuming that I'm close, then the total load will be 0.25A. The equivalent load for a 5W bulb would be ten times that.

 

It makes sense that the flashing may be quicker but there's too many unknowns at the moment to be sure how to fix it.

 

Soz, dude!

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Daz, done some investigation -

 

Found out the company that supply the LED's (as well as other companies), can also supply a 'Load Equalizer' (sic). Having done a bit of searching, this I think is just a resistor in a box.

 

There's some discussion about it here :

 

http://www.grandamgt.com/forum/showthread.php?s=&threadid=2448&highlight=load+equalizer

 

Does this mean I just need to find a couple of resistors that electrically 'look like' a standard front indicator bulb and wire each in parallel with each one of the new LED indicators?

 

Some more stuff :

I've received a *LOT* of emails about LED bulbs and their applications with the ATS system lamp controllers. The actual controller modules do not care if they switch LED, incandescent or pure resistive loads. So the use of LED or related bulbs with the ATS lamp controllers is not an issue.

 

Now, the stock FORD flasher module (electronic) design does care as it has an internal current "sensing shunt" that detects "low" load and will cause the unit to step into a 2X flash (double) rate mode. This is to indicate "bulbs out" and is part of the safety design. In figuring up the actual load for the turn signal flasher, you can count each bulb and note that standard 3057 wedge based bulbs use 2.25 A each.

 

So, three (2 in back, 1 in front) would be 6.75A total. The trigger point for bulbs out in the flasher is >4A, which would be in effect if you were to use LED bulbs in the back. You can install a device called a "LOAD EQUALIZER" sold be several LED/Automotive bulb dealers or distributors that will keep the flasher load above this point. This is nothing more that a power type wire-wound resistor wired in "parallel" (across) the power feed to the rear bulbs at the trunk wiring harness connector(s).

 

Here is a sample of the type of products out there ready to install.

 

http://store.cyrilhuze.com/CyrilHuze/accessories/details.asp?ProductID=10096

 

Now, you can build you own by purchasing some 10 OHM/15-20W "wire-wound" power resistors and soldering connection wires onto them and use one on each rear side for dummy load (wired across the brake/turn signal feed and ground). Use heat shrink tubing on the solder connections and silicone adhesive can be used to mount them on the rear trunk bulkhead, as they will get warn to the touch when in continuous use (extended periods holding brakes on).

 

The dummy load in addition to the three LED's would add the extra bulb "load" to the rear to keep the flasher in normal mode. If the flasher is on the edge of it low current trip point, you would add one more in parallel to each

side to simulate a standard 3057 bulb load. For most folks, one per side is all that's needed.

 

In this way, you can return to stock type light bulbs by simply removing the load equalizer and LED's. Everything is back to "normal".

 

What wattage are the front indicators? are they 15W?

 

Cheers

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I think that post gves you all the info you need. :thumbs:

 

Whatever the wattage of the standard indicator, you can work out the equivalent resistor by using this formula:

 

R = V squared / P

 

P = equivalent bulb power in Watts

V = supply voltage (14 with engine runing or 12 without).

R = resistance in Ohms.

 

So to mimic a 15W bulb would take a 9.6 (call it ten) Ohm resistor in parallel with the LED unit. As the post says, the resistor will need to have the same power handling capability as the bulb it replaces.

 

Unfortunately, Maplins do not stock anything above 10W wirewounds, so you will need to spread the power between two 10W ones. To get the proper resistance you will need to double the value of each resistor.

 

So to simulate a 15W, 10 Ohm resistor you can use two 22 Ohm 10 Watt ones in parallel. This will give am actual resistance of 11 Ohms with a power handling capability of 20 Watts.

 

Order code for these parts is H22R (19p each).

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Darren - You're a star mate!:thumbs:

 

But I've just checked the bulbs and they're 21W :eek:

 

If I recalculate, I get :

 

6.9 Ohms for 12V and 9.3 Ohms for 14V.

 

So if I use 3x33 Ohm resistors, it should bring it back to 11 Ohms but with 30W power handling?

 

Cheers

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I'd try to get as close as you can but err on the high side. If you are aiming for around about 10Ohms, then even 1 ohm out is a 10% error.

 

Also, although you can calculate the R value for either 12V or 14V, which every you finally go for should be checked against both supply voltages. For example, if you base your calcs on 12V, when the engine is running you will push more current through the resistor and therefore require more power handling from it.

 

Mind you, if you are forced to go to three resistors in parallel, you will probably have enough headroom power-wise for this not to be a problem.

 

Incidentally, you can get 7W wirewounds, so three of those would handle 21W exactly, but there's no sense in limiting yourself - especially since if you use dissimilar values the lowest resistance will actually dissipate more power than the other two. Best to stick to 10W ones.

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OK - Got round to adding the resistors today. All work fine. Light flashing is at the standard speed.

 

Used 3x 33Ohm 10W resistors. Left them flashing for about 2 minutes. The resistors got hot, but I was still able to hold them in my hand so should have no problems.

dscf0115.jpg

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