Electrical schoolwork...

[Ti Kan]

Huw,

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.

-Ti
2003 A4 1.8T multitronic
2001 S4 biturbo 6-sp
1984 5000S turbo
1980 4000 2.0 5-sp
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 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.