Re: TI-H: NS Illuminator (okay, I guess I'll go a bit more in depth :)
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Re: TI-H: NS Illuminator (okay, I guess I'll go a bit more in depth :)
>Here's a small explanation of resistance and voltage in series circuits:
>(assuming I remember last year's physics correctly)
>
>First, let's pretend we have a 4V circuit with two 1 ohm resistors.
>Each resistor would eat up 2V. If you had a 3 ohm resistor and a 1 ohm
>resistor, the 3 ohm would use 3V and the 1 ohm 1V. So, the voltage used
>by each load is the load's fraction of the total resistance; i.e.
>Vload=V*Rload/(Rload+Rload2+Rload3+...).
Yup.
>Well, find out the resistance of your LED (I'll call it R) and then
>multiply it by 3V (total voltage) and divide it by 2.4V (voltage for the
>LED). Take that quotient and subtract the resistance of your LED, and
>you have the resistance of the resistor that you need. Example, if your
>LED used 4 ohms, you take (4*3/2.4)-4=1 ohm and you use a 1 ohm
>resistor.
Generally it's too cumbersome to think of "LED resistance", mainly
because it's a ficticious quantity... Most of the time you think of
forward current, or the current at which it reaches it's optimal
brightness. This current is usually specified in an LEDs datasheet
(check the mouser/digikey catalogs, they usually give you a "light
output at current XXX.X" rating). An LED's actual resistance will
vary exponentially with the current, so it's safer to work with the
current draw instead.
So, to follow ohms law more precisely, you would take the battery
voltage (3V) and subtract the known forward voltage of the LED
(generally between 1.6 and 2.1 volts). This will give you the
approximate voltage across the resistor. You know the current the LED
requires (from the specification), and you know the voltage across the
resistor, and you can solve for the required resistance (R=V/I).
So the equation becomes:
Current through the LED = (Battery voltage - LED voltage)/ Resitance
>Even if the diode works, a resistor has a couple advantages. First, a
>constant .6V drop yields too little voltage if your batteries drop below
>1.35V each. With resistors, when the batteries lose voltage, the
>resistor takes less voltage too, so your batteries can drop to 1.31V
>before the LED gets less power than it needs.
True, but there are some clarifications needed on this...
What's actually happening with the diode/LED combination is that the
diode is "simulating" a resistor, but a nonlinear one! The two
devices are very similar, and they share the drop across them. It's
actually a very risky process. Here's the graph of what's happening:
C * *
u * *
r * *
r *
e ~2.0v * ~0.7v
n =Vf * * =Vf
t | * * |
| * *
| ** **
| LED ***** ***** Diode
|****** *********
-----------------|------------------- Voltage
Equilibrium
The diode's graph is flipped... on the "real" voltage scale it should
be in the same direction as the LED voltage graph, but this is much
easier to visualize.
The LED takes up a certain amount of voltage, and so does the diode,
but they are "fighting" for the voltage so they eventually settle into
an equilibrium point. The current is where the two graphs cross.
Solving for the current isn't possible directly, you must solve
iteratively solve it using a calculator. Since the total voltage (3v)
isn't enough to overtake the two voltages required by the diodes, the
equilibrium point settles into a reasonable current level. However,
if you gave it enough voltage, this would result:
C * *
u * *
r * *
r * *
e ~2.0v * * ~0.7v
n =Vf * * =Vf
t | * * |
| * *
| ** **
| LED ***** ***** Diode
|****** *********
---------------------|----------------------- Voltage
Equilibrium
The two lines cross near infinity! This would result in a blown
circuit (current->infinity), and this is also why you shouldn't ever
use two diodes in this configuration without a series resistor.
When you use a resistor instead, this is what results:
C *
u** *
r ** *
r *** *
e ** *
n *** *
t ***
| * **
| ** *** Resistor
| LED ***** **
|****** ***
--------------------------------- Voltage
Being a linear device, the resistor's graph (called the "Load Line" in
most textbooks) will intersect the LED somewhere along the line. As
you can see, it's nearly impossible to allow the current to go to
infinity because of the way a resistor works in the circuit. You see
these types of graphs when you get into second semester electronic
circuits, etc. (so if you don't understand them, don't worry, even
college EEs don't sometimes :) The slope/position of the line is
derivable from the circuit but I won't go into that.
The actual place these two lines intersect can be found by the
equation I derived above,
I=(Battery Voltage - Vf)/R
Also, as you can see on the graph, the "LED Voltage" or "Vf" is just
merely a *guideline* for how much voltage the LED will drop when it's
at a reasonable level. It's *not* some fixed quantity that always
holds true when you use it in a circuit! You can see that the LED is
still passing current even when the voltage across it is less than Vf.
Same with the diode. So, if there's only a very small current flowing
through a diode, the voltage across it is no longer ~0.7 volts, but
nearly 0.0v! This is why you can't count on "diode drops" in a
circuit when the current can dip below a few milliamps. If the
current you're working with is a few nanoamps, the diode drop will be
*zero* volts, not 0.6 volts. So, the voltage across a diode can swing
anywhere from zero to a little bit past Vf (far past Vf results in
infinite current).
I know, I know, I should probably write some sort of FAQ on this, but
oh well no time :).
-Mel
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-The TI-Memory Expansion Homepage
-http://www.egr.msu.edu/~tsaimelv/expander.htm
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