Trying to understand current dispersal control across a parallel circuit

[polymorph]

It is a trap to calculate "resistance" of a nonlinear device.

[dwightthinker]

This is not a particularly good quality graph.
The lines should have more of a knee but it is about
right at 20ma.
with a proper graph, on can see what size resistor to
use, with a non-linear device.
If one draws a straight line from the supply voltage at 0 current
through the desired current point on the selected diode line,
to the current line, at 0 volts, you can use that current
divided into the supply voltage to tell you the desired resistance.
This is called a load line( it still uses ohm's law ).
It is useful for determining not only the size of resistance
to use as a current limiter but also it can be useful
to see what say a 10% change is resistor might do
to the actual current in the LED.
One can draw a new line for each resistor size and see
where it crosses the LED's line.
If one had a graph showing the variation of voltages
the LEDs might have, you could see what the tolerances
of the different devices would have.
This graphical method still applies Ohm's law to non-linear
devices.
It is useful to show how poorly a LED running at 4V can
be current regulated with a 10% resistor size.
Dwight

[polymorph]

Alright, a more accurate graph can be seen on page 4:

http://www.phys.uconn.edu/~hamilton/phys258/N/led.pdf

The knee is certainly not at 20mA.

Another more accurate graph:

[Frostline]

with a proper graph, on can see what size resistor to
use, with a non-linear device.

What defines a proper graph?

I ask because look at the current vs forward voltage graph for this red LED.

On the graph it shows 20mA and 2V matching pretty well.

But above in the tables of the datasheet it states the forward voltage can be as high as 2.5V at 20mA.
So to me that graph may not show my particular LED well at all.  It may have a forward voltage of 1.8 or 2.3V at 20mA.  So how can that be factored in creating a load line?

If one draws a straight line from the supply voltage at 0 current

Going to call that point A.

through the desired current point on the selected diode line,

Going to call that point B

to the current line, at 0 volts,

Point C

you can use that current
divided into the supply voltage to tell you the desired resistance.
This is called a load line( it still uses ohm's law ).

So since point C is about 15mA it would be 5/.015 or 333.3?

But what is that value telling me?

Say we have this circuit


Using the LED from the datasheet V_f = 2V
And I want to limit the current to 12.5mA
So formula is V-V_f / .0125 = R
R=240
So I would need a 240Ω resistor in the circuit above.
According to LarryD my total circuit resistance is V_tot/I_tot=R_tot or 5/.0125 = 400Ω
So 400 minus the value of the resistor of 240 leaves an equivalent resistance of 160Ω for the LED.

None of those numbers is 333.3
Is the load line telling me the R_tot =333?
How does that show what current limiting resistor should be should be?
Or is it showing what the current limiting resistor should be so my R_tot is still 400Ω?

One can draw a new line for each resistor size and see
where it crosses the LED's line.

How? Where did resistor value have influence where I drew the load line?

Or is this all not making sense to me due to the poor quality of my drawing?

[Paul__B]

Sorry, you drew your graph wrong.

You have put point B at 13 mA, not 15, and you have not drawn zero volts.

[dlloyd]

So since point C is about 15mA it would be 5/.015 or 333.3?

(5-1.5)V/0.15mA = 233Ω

[Frostline]

Sorry, you drew your graph wrong.

You have put point B at 13 mA, not 15, and you have not drawn zero volts.

You are half right.
Point B according to instructions is the desired current which is 12.5mA.

But yes point C is wrong.  I need 14 more divisions along the X axis since 1.5V =\= 0V.

So new graph



So then that gives me 5/.021 or 238.1Ω which must be what my current limiting resistor needs to be since that is very close to calculated value of
 (V - V_f ) / I = R
( 5 - 2 ) / .0125 = R
R=240Ω

But I don't yet see the point of the load line.

Because if I just look at the original graph from the datasheet a current value of 12.5mA corresponds to a forward voltage of about 1.96V
So (5- 1.96) / .0125 = R
or 243.2Ω which is about as close as trying to draw a load line (at least as poorly as I draw).

[Frostline]

(5-1.5)V/0.15mA = 233Ω

This is because I failed to scale the X axis  to 0 (the subtracting of the 1.5 from source voltage) ?

[polymorph]

Stop calculating the resistance of the LED. "Resistance" is only meaningful when the device has a linear voltage and current relationship.

To get the load line to tell you what value of resistor to use, place a point on Vcc at 0A of current.

Now place a point on the LED curve that is at the desired current.

Draw a line through those two points. Note where it crosses the zero voltage line. Divide Vcc by that current.

That is the resistor value.

However, do you really need it to be that close? Easier to just note what the approximate voltage is on the LED at the desired current, then (V_CC-V_LED)/I_LED = R

[dwightthinker]

I was hoping the load line of the resistor would show him how
the LED responds to different amounts of current.
Polymorph made it clearer than I could.
The voltage of his second chart doesn't go to 5 volts. I'm sure you can
print the chart out and draw a line to the 5v or even 9v is needed.
One point to note. The larger the voltage drop, the smaller the difference
in current is.
Draw some load lines, starting at different voltages, say 3,3v instead of 5v
for the source.
Dwight

[Frostline]

The voltage of his second chart doesn't go to 5 volts.

Could be due to a browser issue or something not showing the entire image but my second chart goes from 0 to 5.5V along the x-axis.

The larger the voltage drop, the smaller the difference
in current is.

Right, I get that part.  The slope of the load line will change relative to the supply voltage at a given current.

I misunderstood what you were stating the load line was going to show.  Or maybe I am now misunderstanding it. 
What it seems now is that where the load line crosses the y-axis is the value needed for the current limiting resistor.
What I thought it was going to show was the equivalent resistance of the LED at a specified current.

There seems to be a catch-22 here.  Take this LED as an example.

Say you want the LED to be at 1/2 intensity and the only physical resistor you have is a 1/2 watt 560 ohm resistor.

What supply voltage is necessary?

Fig 2 in the datasheet shows the I_f which can then be used in Fig 1 to find the V_f.

But R_eq of the LED can't be solved without V. 

[dwightthinker]

The graph already shows the resistance at any particular
current. You just have to do the math. For any particular current,
you can see the voltage. For any particular voltage,
you see the current. Applying Ohms law will show you the resistance
under those conditions. Change the current and the resistance
changes. Change the voltage and the resistance changes.
It is clear from the graph, or should be.
If you don't know either the current or the voltage, you can't calculate the LEDs
resistance.
The load line allows you to select a specific resistance value
or in reverse take a specific resistor and calculate what the current
would be.
The problem becomes much more difficult when you add a
resistor in series with to a string of parallel diodes and resistors.
The load line can't calculate this for you since you don't really know
the voltage or the current at each diode.
You can calculate an approximation based on the fact that once
the LED diode is drawing current above the knee, it will be almost
a constant voltage.
The current voltage plot itself is an approximation for the average
LED diode of that manufacture.
There is a threshold voltage that is characteristic of the material
used and then the current grows exponentially with voltage
plus some series resistance drop. If you knew the coefficients
of these variables, it would be possible to solve the combination
but not something I'd want to tackle.
Dwight