Trying to understand current dispersal control across a parallel circuit

The resistance of the LED changes with the voltage across it.
At 5V, across the LED, is might drop to 5 ohms. The current then
would surely burn it up.
This is what is meant by being non-linear.
A resistor keep almost the same resistance, regardless of the voltage
across it ( within reason ). It is called a linear device.
Diode in forward conduction are non-linear. As the voltage goes up the
resistance goes down.
To understand better, you might look up "load line".
If you had an exact plot of your LED, voltage against current,
you could plot a load line to hit the exact, say 20ma.
Do also remember, everything has a tolerance. When it is said
that the LED has X.X volts at 20ma, it will never be exact.
Again, you can use the load line plot to see how that effects
the amount of current in the LED.
Include the tolerance of the resistor that you are using for
more effects.
Drawing a load line drawing you can also see that the regulation
of current to the LED is better, the more voltage you are dropping
across the resistor.
If say you had a LED that ran at 3.3V and you had a supply at 3.6V,
the tolerance of a 10% resistor would make a much greater than
10% change in current.
Dwight

I think what is making this so difficult for me to grasp is there is no mention of resistance really on LED datasheets.
If it (the LED) had resistance characteristics why is there no graph of resistance vs voltages beyond Vf? Or at least a mention of of it somewhere?
Like this datasheet goes so far as to mention the capacitance of the LED. But no ohm values to be found.
There is a power value 60mW, but no mention of what values were used to arrive at that other than ambient temperature(which does not seem rather helpful).

It seems like the consensus is to just force Ohm's Law to work. Since the resistors can be measured for resistance and the current and voltage can be measured any discrepancy is just shrugged off as equivalent resistance so the Ohm's Law formula is happy.

Since LarryD has been quite sporting in trying to answer my repetitive non-understanding of all this I will attempt to try and answer the four questions posed to me.

  1. How much resistance does a 1N4007 diode have when it is forward biased?
  2. When it is reversed biased?
  3. How much resistance does a fresh 9 volt carbon battery have?
  4. What is the total circuit resistance in the schematic below. Itot = 106.67ma
  1. Using this datasheet the diode has a forward voltage of 1V @ 1A.
    R=V/I so 1 ohm
    Though I figure that is incorrect. Since the same logic would not work in circuit 3 example from my last set of circuits. Using values from that example of Vf of 2V and a circuit current of 13.6mA I get 147 ohms for the LED in the circuit, not the 226 you said.

  2. DS says reverse bias voltage is 700V RMS(not sure what that means) and only reverse current value I see is at peak of 5uA. The 700V 70% of the peak reverse bias voltage listed at 1000V, no idea if that relationship also works for current but maybe it does so non-peak current of .0035A. So 700/.0035 = 200k ohms.

  3. No ideas how to even begin to calculate that. Assume manufacturing discrepancies play a huge role.
    Reading though this I am going to say 35Ω

  4. I am going to say it cannot be determined. What if there was an 8.2V battery under the box?

The questions were loaded.

  1. How much resistance does a 1N4007 diode have when it is forward biased?
    Like a LED, you really cannot say for sure what the resistance is.
    If you were told what the current flow was then you might use ohms law to calculate an equivalent resistance.
    But you weren’t, the best answer is as you did, make some assumptions.
    Full points for using the data sheet and basing your answer on quoted voltage to current characteristics.

  2. When it is reversed biased?
    This is a bit tricky.
    A reversed biased diode has all the applied voltage dropped across it.
    As you pointed out there is a reversed biased leakage value on the data sheet.
    This is 5uA at the DC Blocking Voltage, for a 1N4007 this is 1000VDC.
    You could say the reversed biased resistance was 1000V/5uA = 200,000,000 ohms = 200Meg
    That is, an equivalent resistance of 200Meg at the rated PRV (peak reverse voltage).
    In practice, at 5V, a reversed biased 1N4007 has a leakage current of zero amps.
    (I know this from experience)
    5V/0A = infinite (for all intents and purposes)

  3. How much resistance does a fresh 9 volt carbon battery have?
    The point here is, we sometimes miss the fact that an electronic device may have resistance.
    Ex: a capacitor has an ESR (effective series resistance) .
    Note: as a battery is discharged the internal resistance goes up leaving less voltage for the load.
    Full points.

  4. What is the total circuit resistance in the schematic below? Itot = 106.67ma
    Back to the box.
    With the applied voltage of 5V and a total current of 106.67mA your equivalent total resistance would be 5/160.67mA = 46.7 ohms.
    At best, if all the 150 ohm resistors were in parallel this would give you 150/5 = 30 ohms leaving 16.7 inside the box.

Click the LINK below the drawing to see what was in the box.

Some conclusions:

  • There is a difference between equivalent resistance and resistance.
  • Equivalent resistance can vary as circuit conditions change where resistance will not.
  • Data sheets are important!
  • Measurements may be necessary to dissect a circuit which quite often is a black box to the technician.
  • It appears, you have quick mind and you will do well in electronics!

Edit:
Let's say there was a LED in the box instead of the 1.8 volt battery and the LED forward voltage drop was 1.8V.
It would be correct to say the power dissipated by the LED would be W=1.8v X 106.67mA = 190mW.
If this was a 16.7 ohm resistor it would also dissipate the same 190mW.
The power calculation is valid maybe you can considered the equivalent resistance calculation as valid :wink:

The reason they don't have voltage to resistance plots is because it
is more useful to have voltage to current plots.
As I said, with such a plot, one can draw a linear resistance
load line to see what resistance is best at some driving voltage.
Knowing the resistance at any particular current is just looking at the
plot and applying ohms law.
Knowing the resistance at any one current or voltage is of little use
at predicting what it will be at any other current or voltage.
Take the time to learn what a "load line" is and it will start to make sense.

A LED diode is not a silicon 1N4007 rectifier diode. Why would you expect its
voltage/current curve to match an LED diode. They are made from
completely different materials.
They do look some what similar but don't match.
Why would you think that 1V at 1A is not 1 ohms. It always is in
my book. A 1N4007 has little that is similar to a LED other than
they are both a type of diode. Silicon diodes do emit light though.
It is a 10.5 microns as I recall ( in the lower infrared band someplace ).
Dwight

Frostline:
I think what is making this so difficult for me to grasp is there is no mention of resistance really on LED datasheets.
If it (the LED) had resistance characteristics why is there no graph of resistance vs voltages beyond Vf?

Because while we are discussing the finer conceptual points of ESR or "effective series resistance" and that only because you mentioned it, that is basically irrelevant to the practical operation of a LED.

There are two ways of operation of an LED. Either you calculate, knowing the approximate voltage drop of the LED, the value of a series resistance corresponding to the specified current and your particular supply voltage, or else you design a circuit to regulate the current over a very wide range of applied voltages, so that the current will be as specified under all conditions. For either of these, it is only required to know the specified (actually, maximum) current for the LED, and its usual voltage drop in operation.

Therefore those are the only (electrical) things specified in the datasheet.

Another reason you can not apply Ohm's law to LED (and other devices) directly, is because Ohm's law is linear and LED are non-linear

It's not about Ohm's law at all, it is about understanding the nature of the component, so Ohm's law is irrelevant.

Ohm's law applies perfectly well to calculating the resistor when you follow the correct procedure.

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

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

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

The knee is certainly not at 20mA.

Another more accurate graph:

dwightthinker:
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?

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

Going to call that point A.

dwightthinker:
through the desired current point on the selected diode line,

Going to call that point B

dwightthinker:
to the current line, at 0 volts,

Point C

dwightthinker:
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 Vf = 2V
And I want to limit the current to 12.5mA
So formula is V-Vf / .0125 = R
R=240
So I would need a 240Ω resistor in the circuit above.
According to LarryD my total circuit resistance is Vtot/Itot=Rtot 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 Rtot =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 Rtot is still 400Ω?

dwightthinker:
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?

Sorry, you drew your graph wrong.

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

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

(5-1.5)V/0.15mA = 233Ω

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.

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 - Vf ) / 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).

dlloyd:
(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) ?

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 (VCC-VLED)/ILED = R

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

dwightthinker:
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.

dwightthinker:
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 If which can then be used in Fig 1 to find the Vf.

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

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