Okay so My car came with aftermarket mirrors when I bought it... I always get the turn signal notifications because the canbus thinks a bulb is out. How do I go about fixing this I've
read tons of forms and can't find a good answer for this! Can I put a resistor inline with the wire to the leds or change settings for voltage in the canbus or disable that function altogether?
Thanks in advance!

2006 C280 4MATIC

 

Are you sure the car thinks it's the mirror bulbs that are out and not one of the actual turn signal bulbs in the headlights or tail lights?

 

I 'm pretty sure it is the mirrors...




My mirrors are not stock!


Anyone?

 

Ooooo. Nice! No idea on resistors etc. But removing the covers is pretty easy.
There's clips that hold them on.
And the lights are like a plug in circuit board, so theoretically you could
try a resistor and see if it does anything.Those lights do look stock.
Check the connections. I get an occasional error on the mirrors.

 

I've actually pulled them off they cut the plugs and directly wired them to the mirrors.

My problem is I have no idea what resistor to put in there if that would even work, I was hoping I could just disable the canbus option for the mirror warnings!

 

How about a small potentiometer inline, turn it till the error goes away?
Start at zero ohms, and turn slowly....
You can then measure the resistance and buy a small resistor.
Changing the canbus, well you'd need SDS, if it was even possible.

 

btw, don't put the resistor inline, put it parallel.

 

Thanks for the new insite... problem is now I have even more questions. What kind of potentiometer? (1K, 10K, 100K, 1M, 2K, 20K, 200K, 500, 5K, 50K Ω) Also how do you put a resistor in parallel?

Thanks for the help so far!

 

All of this is dependent on the idea that the signals are drawing too little, ie probably LED's not pulling enough.
If it was pulling too much, it would register as a short, and lights wouldn't work at all. Computer would kill the circuit.

Parallel goes across the pos and neg, not inline.
That would be series.

In the 1st image on this page, power source is on the left, just imagine R1 is your resistor, and r3 is the lights.
There's no R2.
https://www.allaboutcircuits.com/tex...llel-circuits/

So connect some wires to a pot like this one, connect across the pos and neg before it gets to the signal light.
https://www.digikey.com/product-deta...iABEgLu2vD_BwE

say maybe a 0-500K. WAIT A MINUTE! Edit:
I decided to do some actual math. Measure the voltage. I'm assuming it's 12V, but just make sure, measure it when it's blinking.
Also lets assume that it's supposed to draw somewhere between 5- 10 watts? Maybe 20 tops?
That means the total current of the circuit would be somewhere between .4 and 1.6 amps when it's on.
E/I =R so, 12v/.4= 30ohm and 12v/1.6= 7.5 ohms TOTAL resistance.
So, get a smaller pot say 0-1000 ohms.
Remember you only need to ADD a small amount of resistance, not the total.
Say you need to add 1 watt to the circuit that would be 150 ohms.

Turn it ever so slowly until the errors go away. AND the lights still work. Too much resistance, it will think there's a short and cut power. Damn car is too smart.
Disconnect, measure with an ohmmeter, and buy a couple like say 1 watt resistors that size, solder 'em in, and you're done.
Don't have a meter? Harbor freight has one that's adequate, digital for like 10 bucks, correction 6.
https://www.harborfreight.com/7-func...ter-98025.html
sometimes they are free, I have a few of them,
along with my trusty Fluke.

You can get all this stuff at a Radio Shack.

 

BTW- I think I forgot to mention, the price of my advise is your carbon fiber mirror covers...

 

Btw, this would be a great question for an electrical/etc engineer interviewing at a very small company.
At very small companies, it's not uncommon for the "Electrical Engineer" to do everything for a product, except for the mechanical design. Now, with Solidworks/etc, a Mechanical Engineer may not be needed for simpler enclosures/designs.

It's simple.
Yet, it's involved, because of the need to get rid of the heat from the resistor.


Some basics:
Equations:
V = I times R (Voltage equals Current times Resistance)
Wattage = Volts times Current.

Metrics:
Battery Voltage: From ~11v to 15volts.
Bulb Current: Turn Signal ~2+amps @12v.

My guess at the Current for the OEM lights in the mirror:
Between 0.1amps to 3amps.

Crunch numbers:
Use 11Volts for resistance calculations
R=V/I
High-end R=11/0.1 = 110 Ohms
Low-end R=11/3 = 3.7 Ohms

Use 15Volts for wattage:
Low-end Wattage = 15volts * 0.1amps = 1.5 watts
High-end Wattage = 15volts * 3amps = 45 watts.


My best wild *ss guess is that the bulbs in mirror would take ~between 0.5 - 1.0 amps.
So, that's 7.5 watts to 12watts.
That's a big resistor and a lot of heat to get rid of!

You want the wattage rating at least 50% greater than your calculations.
BUT, wattage ratings on resistor are based on proper air flow and cooling/heat-sinking.

Since you won't have air-flow, forget about the big "sand" power resistors.
Like the following:

Those need air flow.


So, you'd need a "thick film" resistor in a package that allows for bolting directly to a heat-sink (metal).

Manufacturer Part Number : PWR221T-30-8R20F
Description: RES 8.2 OHM 30W 1% TO220
Detailed Description:
8.2 Ohms ±1% 30W Through Hole Resistor TO-220-2 Automotive AEC-Q200, Current Sense, Pulse Withstanding Thick Film

Spec sheet: https://www.bourns.com/docs/Product-...pwr221t-30.pdf
Electrical & Thermal Characteristics
Power Rating @ 25 ˚C Case Temperature 30 W**
Tolerance ±1 %***, ±5 %
Operating Temperature -55 ˚C to +150 ˚C
** Power rating of 2.25 W when mounted free to air (no heat sink).


Good Luck.

 

Summary - Hail Mary ( https://en.wikipedia.org/wiki/Hail_Mary_pass ) pick:
===========
91 Ohms 10watt resistor:
Genuine Dale CW Series 10 Watt 5% Power Resistor, 91 ohms
Price: $1.55 + $3.59 shipping
~$8 for two (including shipping)

From ebay:
https://www.ebay.com/itm/NEW-Genuine...s/391792050930
Dale CW Series 10 Watt 5% Power Resistor, 91 ohms
~$14 for two (including shipping).
===========


More Detail:
Also, because of the current/wattage involved, you can not use a simple cheap trim pot to find the needed resistance.

The C230 door mirrors used 6 LEDs - 3 Horizontally and 3 vertically
If I take a GUESS/ESTIMATE that they use typical ~20ma (milli-amp) LEDs, then....
Volts=I(current) * Resistance
v= i*r
r=v/i
i=v/r

Watts=volts*current(i).
w=v*i
i=w/v

LED = ~20ma
6 LEDs x 20ma = 120ma = 0.12amps
r=11volts/0.12amps
r=91
Use 91 ohms (standard 5% E24 series)

i=15volts/91ohms = 0.17 amps
watts = 15volts * 0.17amps = 2.55 watts

So, with the above calculations, based on the pull-it-from-my-***** guess , I get the suggested
91Ohms and 10watts for a resistor.


Since I leg-hump Pelican Parts like mad, I'll steal their picture.
Here, you can see the 6 LEDs - 3 Horizontally and 3 vertically.
From their web-page: https://www.pelicanparts.com/More_In...-820-01-21-INT




Otherwise,
If you want to do it the "correct way", and use a potentiometer to find the resistance, then I suggest one of the following:

https://www.ebay.com/itm/25W-100-OHM...C/182983705677
25W 100 OHM High Power Wirewound Potentiometer Rheostat Variable Resistor EC
$6.50

100 Ohm Linear Taper Wire-Wound Potentiometer
$11

Uxcell 100W 100 Ohm Ceramic Wirewound Potentiometer Rotary Resistor Rheostat
$14


Good Luck!

 

Forgot to mention, when turning the pot, start at high resistance and work backwards. Starting at 0 ohms might blow the fuse. I don't think heat will be a problem since it's only an intermittent flasher.
I had resistors on my parking lights when I first used LEDs before they had the canbus ones and yeah, it got hot but it was on all the time when the headlights were on.