[V8] Headlight question
urq at pacbell.net
Wed Aug 6 12:41:05 PDT 2008
... I have opened up one of these modules ... and they are basically as you imagined. The thing is that they are not calibrated in any way ... they just measure the voltage drop across the sense resistor in both circuits and flag when the difference in voltage drop is over a threshold. I forget the manufacturer of the chip that does the comparison, but it was a standard part, designed for use in measuring current differences in this way. This characteristic is used to advantage by those who implement individual relays for the left and right headlamps ... as long as both relay coils draw similar currents the auto check function is satisfied. The other interesting characteristic is that if both headlamps are burned out the autocheck system will not complain.
The "sense resistors" are reasonably high current carrying bars, so I doubt there'd be any significant difference in the voltage drop from the switches to the unrelayed bulbs with the module in place as compared to a solid wire. I'd be willing to bet the voltage drop across the in line fuses is greater. There's a single wire that goes from the module to the auto check system, so what I do is to relocate the bulb out monitor to the relay box and send a single wire back to complete the circuit to retain proper function of the bulb out warning system.
San Jose, CA (USA)
Dave Saad <dsaad at icehouse.net> wrote:
Well - yes and no -
First off, I have not looked inside this particular module (one of
the only things on my car I have not opened up btw :-)
but to measure current (or amps) the easiest and cheapest way is to
measure a voltage drop across a very small resistor. I am pretty
confidant this is how the Audi system works.
Inside there has to be a current sense resistor somewhere around 0.1
ohms. When you know the wattage of the bulbs in the circuit, you can
easily calculate what the voltage drop across the sense resistor
should be. If you have too little or too much current, the voltage
drop across the resistor will be out of range and the system turns on
the warning light. An open circuit means no voltage drop, and too
much current means too much voltage drop. I doubt the system checks
for over current, but who knows. This all means that the warning
system acts as a current limiting device that will dim your total
possible light output somewhat. When you relay the system, you are
eliminating as much of those tiny current limiting resistances as
To fake out the system, you either need to fool it into thinking the
current draw is in range (not easy), or disable the warning (probably
easy). I don't know what the wiring mod does, but I suspect it
simply removes the signal wire to the autocheck display - right?
Can anyone tell that school is out, and I have very little to do?
On Aug 5, 2008, at 5:18 PM, Tony and Lillie wrote:
> The headlight warning system works on resistance, I believe. So, if
> resistance of the bulbs is significantly different than the
> resistance of
> the relays (about 200ohms each) then it will trigger the warning. I
> wired past it on my car. I had installed relays for the stock DOT's
> on the
> V8, and installed 100W high beams also. Didn't want to burn up that
> (expensive) headlight switch.
> Tony Hoffman
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