Hello everyone,
I know what i’m about to ask is counter intuitive but hey, sometimes we do it just because it involves an LED somewhere, so here it goes:
I want to replace a halogen light bulb with LED one, unfortunately the computer on the car keeps detecting that I have a Burnt bulb obviously that is because the LED is not pulling enough amps, so the solution is to attach that special device with transistors and a pedometer to adjust the current consumption to match what the computer is measuring and full it into thinking I have the right bulb.
how can I achieve this please:
I thought about using one of those heat resistors but those emit a lot of heat and will get the LEDs hot and decrease their lifetime.
btw I measure the require current and it must be above 40w, so something I can adjust to consume between 40w to 50w would be nice 
Recommendations please?
Jay.
Perhaps the computer is only measuring the initial current, at startup? Try connecting the halogen light in parallel and then disconnecting it. If the Led stays on then perhaps it is only measuring it at startup. If so then put a big capacitor 1000µF in parallel across the Led. The initial current rush might be enough. But this is just my guess.
If you have to waster 50W of power as heat just to make your LED work what is the point? IMHO your not coming out ahead, just use the proper bulb.
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@ Jay Jay - I think manipulating the current-sensor would be the better way. No chance to find out how it works and where it is located?
@ Jay Jay -
You need to add a load resistor. The resistor should be connected in parallel with the LED.
[url]http://www.ebay.com/itm/2x-9005-HB3-9145-LED-Bulb-Headlight-Fog-Driving-Light-DRL-No-Error-Load-Resistor-/171785236350?hash=item27ff32af7e:g:cTEAAOSweW5VUBhU&vxp=mtr[/url]
Replaced the halogen with LED on my motorcycle and had the same issue.
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Your only advantage of the LED is longer life as you will need a power resistor in parallel to simulate the load so power wise it will be the same.
Thank you guys for all the suggestions:
I’m trying to avoid the load resistors, because: 1. they have a fixed value and 2. they get pretty darn HOT.
it would be ideal is if I could find/build a load resistor with a pedometer where I can varies it’s load, from a range of values.
I ran into these, can anyone tell me if these can be altered and used for my purpose, at least they have a away to adjust the current load/voltage.
http://www.ebay.com/itm/USB-DC-Electronic-Load-Module-15W-3A-Precise-Adjustable-USB-Discharger-/131821224262?hash=item1eb1284146:g:TgoAAOSwFqJWnYjN
and last but not least can these resistors be used instead of the traditional load resistors, it says dummy load, if yes how would I connect them:
http://www.ebay.com/itm/1pc-RF-Microwave-Power-Resistors-HF-Dummy-Load-60W-Watt-50-Ohms-DC-3GHz-Ham-/311411521233?hash=item488192b6d1:g:UXgAAOSwHnFVtePw
thank you
Jay.
Lacky bands, duct tape and super glue
Dummy load is based on a 50 or 75 ohm load for radio or other testing. I was referring to power resistors of higher wattage. The aluminium (not spelling US guys) that you can get in different resistances.
Have you calculated how much additional load you need? It might not be 50 ohm.
He said in the car. And on the car is usually 50W 6-Ohm load resistor for this purpose.
Jay Jay, if you are dissipating 50W as heat it does not matter what device you use 50W is 50W is 50W.
The driver may not need to sense the total amount of current, you can try different dummy loads to see the current draw and dummy load you will need.
Some cars use an incandescent bulb circuit for creative purposes. On some old VW bugs (and Chevy’s) used a incandescent bulb between the +12V ignition circuit and the alternator ‘start up’ input. This provides +12V to provide a small amount of current to start the alternator charging. Once the alternator is charging there is +12V on both sides of the lamp and it goes out. If the alternator is not charging the lamp stays on, i.e. it is also the idiot light.
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R = V^2 / P (Resistance = Voltage squared over Power)
If you really need between 40 and 50W to trick the thing then the load you need is :-
R = 12V^2 / 50watts = 144 / 50 = 2.8ohms current = 4.2Amps
R = 12V^2 / 40watts = 144 / 40 = 3.6ohms current = 3.3Amps
3.3 ohms is a preferred value that should be easy to find.
This must be rated to at least 50W or possibly even higher. Even then it will still get very hot. Just like the incandescent light bulbs you use might have in your house. This load will be a beefy one like this http://uk.rs-online.com/web/p/panel-mount-fixed-resistors/0136200/ It will also need good ventilation around it.
You can use the RF 50ohm dummy loads if you really want, but you need to put about 14 of them in parallel to drop it down to 3.5ohms.
@ Jay Jay - is the only real problem that the warning light stays on? Just disconnect that light and then your car will be even more efficient
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A pedometer?
Maybe it is my brain damage, but a pedometer measures how many steps you talk, while walking, I think. How would that help here?
