Electrical schoolwork...

Ti Kan ti at amb.org
Tue Feb 10 02:24:41 EST 2004


Basic electronics 101 here.  I thought you'd know that well,
but apparently not.

An LED is not a resistive device, thus you cannot calculate
based on that assumption.  An LED is a light emitting *DIODE*.
It has a relatively FIXED forward bias voltage.  That 150K ohm
"LED resistance" is an irrelevant spec.  What *is* important
is the forward bias voltage and current requirement of the LED,
and the total supply voltage.  Then the required resistor value
can be calculated from that.

2003 A4 1.8T multitronic
2001 S4 biturbo 6-sp
1984 5000S turbo
1980 4000 2.0 5-sp
 R 1 3 5  Ti Kan
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Huw Powell writes:
> > V = I*R
> > (13.6V - 2.0V) = (.015A)*(R ohms)
> Where is that 0.015A figure coming from?
> I suspect there is an error in the data we are working with (garbage in, 
> garbage out...).
> 150,000 ohms at 3 volts yields 0.00002 amps, according to my trusty 
> desktop calculator.  Which is only, what, .02 mA?  Something seems fishy.
> You and Ti have both made assumptions about the current requirement of 
> the LED not specified in the question.  This problem should be solvable 
> with any of the iterations of Ohm's Law - you can chuckle at "syljay"s 
> results all you want, but they are absolutely correct - given the data 
> we have to work with.
> So, Louis-Alain, which number is wrong?  Is the stated resistance of the 
> LED *really* 150k ohms?
> > If I were to install some red 3V LEDs to backlight the numerous switches
> > around the instrument panel of my 1983 urQuattro, what is the resistance
> > value I must introduce to this circuit to lower the 12V to something more
> > acceptable?
> > 
> > Resistance value of the LEDs: 150 000 ohms Ideal voltage at the LEDs: 
> > between 2.5V and 3V.
> > Electrical voltage at the switches: 13.6V, permanent, switched by the
> > ignition, not the IP rheostat.

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