Title: **Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 28, 2016, 05:36 am**

Post by:

First let me start off by saying I am just trying to understand a concept.

This is not for a specific project or problem. There is no xy-problem involved. The schematic below is just for illustration of what I am trying to ask.

Second I apologize if it takes me too long to explain the questions I am having.

I am going to try to be as through as possible so hopefully I will not be accused of wasting people's time for omitting important information or being unclear as to how I arrive at certain values.

Formulas used in my example can be found here. (http://www.physicsclassroom.com/Class/circuits/U9L4d.cfm)

Let us assume for the example that the power supply is a 5v 1A switch-mode power supply.

The LEDs are 3mm red LEDs with a voltage drop of 1.8 volts.

The resistors are all 1/4 watt 5% metal film resistors.

There are 3 circuits in the following schematic. From top to bottom they are labeled in red 1,2 and 3.

(http://s5.postimg.org/x6e1jnj3r/led2.png)

Now circuit 1 is how I normally would approach constructing a circuit of this type.

Each branch of the circuit, due to the individual LED current limiting resistors is using 21.33mA.

V/R=I

( 5-1.8 )/150=I

.02133 = I

Total circuit current is the sum of the branches of the circuit or 106.67mA.

I_{tot}=I_{1} + I_{2} + I_{3} + I_{4} + I_{5}

I_{tot}=21.33 + 21.33 + 21.33 + 21.33 + 21.33

I_{tot}=106.67

Total circuit resistance R_{eq} is 30 ohms.

1/R_{eq}=1/R_{1} + 1/R_{2} + 1/R_{3} + 1/R_{4} + 1/R_{5}

1/R_{eq}=1/150 + 1/150 + 1/150 +1/150 + 1/150

1/R_{eq}= .00667 + .00667 + .00667 + .00667 + .00667

1/R_{eq}= .03335

R_{eq} = 29.985 rounded to 30 ohms.

Total circuit power is 341mW.

P=I^{2}R

P=( .10667)^{2}*30

P=.34135W

Question 1a: Do I calculate the total circuit and branch values correctly for circuit 1 or if not where am I making errors?

Now circuit 2 is where I attempt to create a circuit with similar total circuit values to circuit 1.

This is based on the idea that in circuit 1 with the resistance of each branch being equal the current through each branch was also equal.

The current through each branch was 1/5th of the total current.

So total circuit resistance is 33 ohms.

1/R_{tot} = 1/R_{1} + 1/R_{2} + 1/R_{3}

1/R_{tot} = 1/100 + 1/100 + 1/100

1/R_{tot} = .03

R_{tot} = 33.333

Total circuit current is 96.97mA.

I=V/R

I= ( 5 - 1.8 )/33

I= .09697

Power of the circuit is 310.3mW.

P=I^{2}R

P=( .09697)^{2}*33

P= .3103

Given the near 1/3rd watt of power is why I used 3 resistors to dissipate the current instead of a single 30 ohm resistor. Otherwise I felt I might be accused of overloading my imaginary resistor.

I have seen it suggested several times both here on this forum and other sources on the internet that this is an acceptable practice.

But this concept of the power dissipation being spread out equally over multiple paths is why I wondered if current might follow a similar logic.

Question 2: Would the current through each branch in circuit 2 be equal to 1/5 of the total current or 19.39mA?

Now on to the final circuit.

As can be seen the circuit is incomplete with an unknown resistor ready to be placed into the circuit to complete the circuit.

So say it was desired to lower the current through each of the LEDs to 15mA and just simply lowering the supply voltage was not an option. Also replacing each of the 150 ohm resistors with 220 ohm resistors was not an attractive option either.

If we pretend that we cut the circuit so that everything to the right of LED 12 and R11 is gone we are left with a simple series circuit.

A 68 ohm resistor could be put in place of the unknown resistor giving a total resistance of 218 ohms and 14.8mA of current for the LED.

But this is a parallel circuit with 5 LEDs and it does not seem to work out even though each LED branch looked at individually is a series circuit.

For total circuit resistance I would need to take the equivalent resistance of the 5 resistors already in the circuit plus the new resistor since it is in series.

So that would be 98 ohms.

Total circuit current based on the new total resistance is 32.65mA.

I=V/R

I=3.2/98

I= .03265

Which is a far cry from the desired 74mA total circuit current. So clearly this seems not the way to figure it.

So then I thought of idea two which was simply calculating a needed total circuit resistance to get the desired total circuit current.

I=V/R

.074=3.2/R

R=3.2/.074

R = 43.24

And since 30 ohms is already in the total circuit an additional 13 ohm resistance would be sufficient to replace the unknown resistor and complete the circuit. Thus total circuit current is lowered so available current for each branch is lowered.

Question 3: Am I even close with the second idea and if not how would the unknown resistor be calculated correctly(or is that not possible in this type of circuit)?

Assuming the above calculations were correct in circuit 1, total circuit voltage for circuit 1 would be 3.2

V=IR

V=.10667*30

V=3.2

which is the same as the source voltage minus the voltage drop of a single LED.

So although the voltage is dropped 5 times in the circuit by the 5 LEDs the total voltage effect is equal to a single LED.

Almost as if it is the average voltage drop per branch.

This is why I used 3.2V in my calculation for my second idea before question 3 as well as the calculations for question 2.

Let us then pretend LED5 is replaced with a blue LED that has a 3V drop and R5 is now a 94 ohm (just to keep branch current about the same).

Now the average voltage drop per branch is 2.04V

Question 4: What effect if any would differing branch voltage drops have on total circuit voltage compared to a circuit where the branch voltage drops were all the same? Or asked another way, for the calculation using Ohm's Law for question 3 when trying to calculate R what value would be used for V? 3.2 or 2.96 or something else if LED5 and R5 were replaced as noted above?

Thanks

This is not for a specific project or problem. There is no xy-problem involved. The schematic below is just for illustration of what I am trying to ask.

Second I apologize if it takes me too long to explain the questions I am having.

I am going to try to be as through as possible so hopefully I will not be accused of wasting people's time for omitting important information or being unclear as to how I arrive at certain values.

Formulas used in my example can be found here. (http://www.physicsclassroom.com/Class/circuits/U9L4d.cfm)

Let us assume for the example that the power supply is a 5v 1A switch-mode power supply.

The LEDs are 3mm red LEDs with a voltage drop of 1.8 volts.

The resistors are all 1/4 watt 5% metal film resistors.

There are 3 circuits in the following schematic. From top to bottom they are labeled in red 1,2 and 3.

(http://s5.postimg.org/x6e1jnj3r/led2.png)

Now circuit 1 is how I normally would approach constructing a circuit of this type.

Each branch of the circuit, due to the individual LED current limiting resistors is using 21.33mA.

V/R=I

( 5-1.8 )/150=I

.02133 = I

Total circuit current is the sum of the branches of the circuit or 106.67mA.

I

I

I

Total circuit resistance R

1/R

1/R

1/R

1/R

R

Total circuit power is 341mW.

P=I

P=( .10667)

P=.34135W

Question 1a: Do I calculate the total circuit and branch values correctly for circuit 1 or if not where am I making errors?

Now circuit 2 is where I attempt to create a circuit with similar total circuit values to circuit 1.

This is based on the idea that in circuit 1 with the resistance of each branch being equal the current through each branch was also equal.

The current through each branch was 1/5th of the total current.

So total circuit resistance is 33 ohms.

1/R

1/R

1/R

R

Total circuit current is 96.97mA.

I=V/R

I= ( 5 - 1.8 )/33

I= .09697

Power of the circuit is 310.3mW.

P=I

P=( .09697)

P= .3103

Given the near 1/3rd watt of power is why I used 3 resistors to dissipate the current instead of a single 30 ohm resistor. Otherwise I felt I might be accused of overloading my imaginary resistor.

I have seen it suggested several times both here on this forum and other sources on the internet that this is an acceptable practice.

But this concept of the power dissipation being spread out equally over multiple paths is why I wondered if current might follow a similar logic.

Question 2: Would the current through each branch in circuit 2 be equal to 1/5 of the total current or 19.39mA?

Now on to the final circuit.

As can be seen the circuit is incomplete with an unknown resistor ready to be placed into the circuit to complete the circuit.

So say it was desired to lower the current through each of the LEDs to 15mA and just simply lowering the supply voltage was not an option. Also replacing each of the 150 ohm resistors with 220 ohm resistors was not an attractive option either.

If we pretend that we cut the circuit so that everything to the right of LED 12 and R11 is gone we are left with a simple series circuit.

A 68 ohm resistor could be put in place of the unknown resistor giving a total resistance of 218 ohms and 14.8mA of current for the LED.

But this is a parallel circuit with 5 LEDs and it does not seem to work out even though each LED branch looked at individually is a series circuit.

For total circuit resistance I would need to take the equivalent resistance of the 5 resistors already in the circuit plus the new resistor since it is in series.

So that would be 98 ohms.

Total circuit current based on the new total resistance is 32.65mA.

I=V/R

I=3.2/98

I= .03265

Which is a far cry from the desired 74mA total circuit current. So clearly this seems not the way to figure it.

So then I thought of idea two which was simply calculating a needed total circuit resistance to get the desired total circuit current.

I=V/R

.074=3.2/R

R=3.2/.074

R = 43.24

And since 30 ohms is already in the total circuit an additional 13 ohm resistance would be sufficient to replace the unknown resistor and complete the circuit. Thus total circuit current is lowered so available current for each branch is lowered.

Question 3: Am I even close with the second idea and if not how would the unknown resistor be calculated correctly(or is that not possible in this type of circuit)?

Assuming the above calculations were correct in circuit 1, total circuit voltage for circuit 1 would be 3.2

V=IR

V=.10667*30

V=3.2

which is the same as the source voltage minus the voltage drop of a single LED.

So although the voltage is dropped 5 times in the circuit by the 5 LEDs the total voltage effect is equal to a single LED.

Almost as if it is the average voltage drop per branch.

This is why I used 3.2V in my calculation for my second idea before question 3 as well as the calculations for question 2.

Let us then pretend LED5 is replaced with a blue LED that has a 3V drop and R5 is now a 94 ohm (just to keep branch current about the same).

Now the average voltage drop per branch is 2.04V

Question 4: What effect if any would differing branch voltage drops have on total circuit voltage compared to a circuit where the branch voltage drops were all the same? Or asked another way, for the calculation using Ohm's Law for question 3 when trying to calculate R what value would be used for V? 3.2 or 2.96 or something else if LED5 and R5 were replaced as noted above?

Thanks

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 28, 2016, 06:09 am**

Post by:

Circuit 1

5V / .1066A= 46.7 ohms

5V * .106.66 = .533W

5V / .1066A= 46.7 ohms

5V * .106.66 = .533W

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 28, 2016, 06:18 am**

Post by:

Question 2

Total Current is 96ma

In practice you will only know what the current in each leg is by measuring it.

No two LEDs have the same forward voltage drop.

Total Current is 96ma

In practice you will only know what the current in each leg is by measuring it.

No two LEDs have the same forward voltage drop.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 28, 2016, 06:26 am**

Post by:

Question 3

If total resistance is 46.7 use ohms law to calculate the needed resistance.

My eyes are tired, reminds me of school days.

If total resistance is 46.7 use ohms law to calculate the needed resistance.

My eyes are tired, reminds me of school days.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 28, 2016, 06:45 am**

Post by:

Circuit 1So Ohm's Law and not equivalent resistance formula?

5V / .1066A= 46.7 ohms

5V * .106.66 = .533W

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 28, 2016, 06:47 am**

Post by:

Total voltage / total current gives resistance ;)

Assuming the battery resistance is 0 ohms.

Assuming the battery resistance is 0 ohms.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 28, 2016, 06:53 am**

Post by:

Question 2But to be clear the current will be balanced between each branch(LED) based on voltage drop?

Total Current is 96ma

In practice you will only know what the current in each leg is by measuring it.

No two LEDs have the same forward voltage drop.

That is LED 6 won't get 90mA due to lacking an individual current controlling resistor while the other 4 split the rest of the current if the forward voltage drops are near the same?

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 28, 2016, 07:02 am**

Post by:

You never put LEDs in parallel.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Jan 28, 2016, 07:16 am**

Post by:

1a You forgot to include the power lost in the LEDs.

2. This circuit would not work well.

LEDs voltage and efficiency drops with temperature, like any diode.

As they begin to heat up, one will begin to draw away more and

more of the current, most likely burning up.

If the LEDs shared the current evenly, you would be right but they wouldn't.

3.

drawing 3

The total voltage around any loop is 0. 1.8V for LED. 15ma through a 150 resistor

if 2.25V.

2.25 + 1.8 = 4.05V.

5V - 4.05 = .95V to be drops on the unknown resistor.

The total current is 5 * .015 = .075ma

.95/.075 = 12.6... ohms.

4.

I don't quite follow your question. The LED is a relatively constant voltage device.

It would stay around 1.8V but its current would change if you changed the individual

resistor. If using circuit 3, changing any one resistor would have some effect on

all the others.

Still one can figure things out based on the sum of the voltage around any loop

is 0 and the sum of the current going in and out of any point is also 0.

That with ohms law will give you the results.

If you are worried about the fine details of the slight change in voltage of the

LED, the problem could be more difficult to solve for a circuit of type 3 as

it would need to be solved for several simultaneous equations with the non-linear

LEDs. Such equations are best solved with successive approximations.

Of course the fine result would only be good for the particular LEDs you had.

When looking at total power, remember, anything with both voltage and current

will use power.

Dwight

2. This circuit would not work well.

LEDs voltage and efficiency drops with temperature, like any diode.

As they begin to heat up, one will begin to draw away more and

more of the current, most likely burning up.

If the LEDs shared the current evenly, you would be right but they wouldn't.

3.

drawing 3

The total voltage around any loop is 0. 1.8V for LED. 15ma through a 150 resistor

if 2.25V.

2.25 + 1.8 = 4.05V.

5V - 4.05 = .95V to be drops on the unknown resistor.

The total current is 5 * .015 = .075ma

.95/.075 = 12.6... ohms.

4.

I don't quite follow your question. The LED is a relatively constant voltage device.

It would stay around 1.8V but its current would change if you changed the individual

resistor. If using circuit 3, changing any one resistor would have some effect on

all the others.

Still one can figure things out based on the sum of the voltage around any loop

is 0 and the sum of the current going in and out of any point is also 0.

That with ohms law will give you the results.

If you are worried about the fine details of the slight change in voltage of the

LED, the problem could be more difficult to solve for a circuit of type 3 as

it would need to be solved for several simultaneous equations with the non-linear

LEDs. Such equations are best solved with successive approximations.

Of course the fine result would only be good for the particular LEDs you had.

When looking at total power, remember, anything with both voltage and current

will use power.

Dwight

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 28, 2016, 09:15 am**

Post by:

1a You forgot to include the power lost in the LEDs.How do I include that?

2. This circuit would not work well.I agree and your explanation makes sense to me.

The total voltage around any loop is 0. 1.8V for LED. 15ma through a 150 resistorIt took me awhile, but I finally got what you are saying here.

if 2.25V.

2.25 + 1.8 = 4.05V.

5V - 4.05 = .95V to be drops on the unknown resistor.

The total current is 5 * .015 = .075ma

.95/.075 = 12.6... ohms.

And while your method is different than how I was trying to do it, had I used 75mA for total current instead of 74mA I would have arrived at the same answer. Even though my method was more luck based I think now.

Your explanation is excellent.

I don't quite follow your question.Agree it is not very clear, more so since it was based on my incorrect assumption on how I first answered question 3.

In my example schematic all of the branches were the same, so my question was basically what if one of the branches was different(using a different LED with different drop voltage and replacing the resistor to keep that particular loop current near the same as the other 4 so total circuit current would remain the same). Could a single resistor be added in series with the 5 parallel branches to lower the current of each branch to 15mA if one branch had a different voltage drop?

So four of the loops would have been just like you explained for question 3.

1.8V for the LED

15mA across a 150 ohm resistor for 2.25V

For a branch(loop) total of 4.05V

So 5V - 4.05V = .95V to drop across the unknown resistor

R=.95/.075

R=12.66

And that would work for all 4 red loops.

But with loop 5 being blue the value would be different

3.0V for the LED

15mA across a 94 ohm resistor for 1.41V

Makes the loop total 4.41V

So 5V - 4.41V = .59V to drop across the unknown resistor

R= .59/.075

R=7.87 ohm

So one resistor will not equally solve all 5 loops unless all 5 loops are the same.

Which answers question 4(I think anyway).

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**aarg** on **Jan 28, 2016, 10:07 am**

Post by:

You never put LEDs in parallel.Sorry, that's not completely true. LEDs from the same fabrication process and batch can be, and are, safely paralleled in some LED modules. But unless you have matched LEDs, it's not a good idea.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Paul__B** on **Jan 28, 2016, 12:20 pm**

Post by:

Sorry, that's not completely true. LEDs from the same fabrication process and batch can be, and are, safely paralleled in some LED modules.Very commonly. Universal in those arrays rated at 10 W and above. Thermal runaway is limited by mounting on a common heatsink.

And in all the multi-LED torches you see. Because the LED also has an internal resistance.

Mind you, you notice the differences as the batteries go flat.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**MarkT** on **Jan 28, 2016, 01:05 pm**

Post by:

For all of these you actually have a non-linear set of equations to solve as the LED V/I curve is logarithmic.

However if you know the LED current is going to be close to a particular value (20mA) then you can approximate

by saying the voltage across the LED is fixed, and the rest is Ohm's and Kirchoff's laws (from which the

series and parallel resistance formulae are derived).

In reality Q2 will not work well as explained above, and noticably unequal current sharing is likely.

The constant voltage approximation for the LEDs is good enough - and at the end of the day a 10%

difference in current from what is intended cannot be detected by the human eye, and is less than

the variability in optical performance of LEDs anyway I suspect. You usually should try various

resistor values by hand to get the brightness level you actually want with the particular LEDs you

are using.

However if you know the LED current is going to be close to a particular value (20mA) then you can approximate

by saying the voltage across the LED is fixed, and the rest is Ohm's and Kirchoff's laws (from which the

series and parallel resistance formulae are derived).

In reality Q2 will not work well as explained above, and noticably unequal current sharing is likely.

The constant voltage approximation for the LEDs is good enough - and at the end of the day a 10%

difference in current from what is intended cannot be detected by the human eye, and is less than

the variability in optical performance of LEDs anyway I suspect. You usually should try various

resistor values by hand to get the brightness level you actually want with the particular LEDs you

are using.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**68tjs** on **Jan 28, 2016, 02:01 pm**

Post by:

Quote

Sorry, that's not completely true. LEDs from the same fabrication process and batch can be, and are, safely paralleled in some LED modules. But unless you have matched LEDs, it's not a good idea.Sorry, that's not completely true.

This is true if Leds are close on the same Wafer.

There are differences between distant areas on the same wafer.

Only one rule = never put LEDs in parallel.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Jan 29, 2016, 10:34 pm**

Post by:

How do I include that?Power on each LED is simple. If the current through it is 20ma

and the voltage is 1.8V:

1.8 * .02 = .036 watts.

if you have five bulbs:

5 * .036 = .18 watts

Just add this to the watts in the resistors.

Power is power. It doesn't make any difference if it is a resistor

or LED diode.

Dwight

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Jan 29, 2016, 10:47 pm**

Post by:

Sorry, that's not completely true.In any case, never put LEDs in parallel if out of a hand full of LEDs.

This is true if Leds are close on the same Wafer.

There are differences between distant areas on the same wafer.

Only one rule = never put LEDs in parallel.

Success is a false sense of future success.

The difference is often in a few 10s of mV. Adding a small resistor in

series with each is most likely enough.

Dwight

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**DrAzzy** on **Jan 29, 2016, 11:55 pm**

Post by:

Your rule is inconsistent with observed reality.Only one rule = never put LEDs in parallel.

LEDs of the same type (mfg/model/lot) can be paralled just fine, and this is done widely in production by organizations that are way too sloppy to be matching LEDs from the same part of the wafer or whathaveyou. Go take apart any commercial LED array (flashlight, floodlight, streetlight if you can get your hands on one, aftermarket LED replacements for incandescents) - LEDs get put in parallel all the goddamned time, and when the LEDs are all of the same type, it works well enough that it's fine. It works particularly well with the 1W and 3W dies, where the dynamic resistance in the operating regime is relatively high.

It can, of course, only be done with very similar leds, and you should be aware that LEDs are not all the same, so that you can react appropriately if you discover that you have a batch of LEDs are significantly different from eachother and it impacts functioning.

I think this was much more of a problem in the past - LED process control has improved greatly.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 30, 2016, 05:34 am**

Post by:

I still am having difficulty understanding why my answer for total resistance in circuit 1 in the OP is incorrect when supposedly my answer for total current was correct.

(http://s5.postimg.org/60f156ruf/res2.png)

Information for the above circuits:

Question 1: For circuit 1 would the R_{tot} always be equal to 220 ohm if the normal operating parameters of the resistor are followed?

Or to ask somewhat differently, will changing voltage only change the current of circuit 1 while the resistance remains constant?

For the current for circuit 1 it is simply an application of Ohm's Law.

I=V/R

I=5/220

I=22.7mA

Question 2: The forward voltage of the LED has to be considered when determining the current of circuit 2. Yes/No?

I=(V_{in} - V_{f}) / R

I=3/220

I=13.6mA

For circuit #3 determining the current should work the same as in circuit 2.

I=(V_{in} - V_{f}) / R

I=2/150

I=13.3mA

Question 3:What is the R_{tot} of circuit 3?

Total voltage / total current gives resistance ;)So lets have more circuits.

Assuming the battery resistance is 0 ohms.

(http://s5.postimg.org/60f156ruf/res2.png)

Information for the above circuits:

- Supply voltage is 5V.
- Forward voltage of red LED is 2V.
- Forward voltage of blue LED is 3V.

Question 1: For circuit 1 would the R

Or to ask somewhat differently, will changing voltage only change the current of circuit 1 while the resistance remains constant?

For the current for circuit 1 it is simply an application of Ohm's Law.

I=V/R

I=5/220

I=22.7mA

Question 2: The forward voltage of the LED has to be considered when determining the current of circuit 2. Yes/No?

I=(V

I=3/220

I=13.6mA

For circuit #3 determining the current should work the same as in circuit 2.

I=(V

I=2/150

I=13.3mA

Question 3:What is the R

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 30, 2016, 05:54 am**

Post by:

Quote

I=(Vin - Vf) / RIf the supply voltage is 5 volts and the calculated supply current draw is 13.3 ma:

I=2/150

I=13.3mA

Question 3:What is the Rtot of circuit 3?

R = V/A = 5/.0133 = 376 ohms = Rtot

150 ohms for the resistor and 376 - 150 = 226 for the LED

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 30, 2016, 06:39 am**

Post by:

If a LED has that much internal resistance why would it need a current limiting resistor in series?

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 30, 2016, 07:19 am**

Post by:

A LED is a non linear solid state device.

Connecting a voltage to the LED greater than its Vf will damage the structure and it will be destroyed.

Connecting a voltage less than Vf results in no current flow (just leakage current) through the LED.

Therefore you must limit the current through the LED when you have a supply voltage greater than Vf.

You limit the current with a series resistor.

Connecting a voltage to the LED greater than its Vf will damage the structure and it will be destroyed.

Connecting a voltage less than Vf results in no current flow (just leakage current) through the LED.

Therefore you must limit the current through the LED when you have a supply voltage greater than Vf.

You limit the current with a series resistor.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Jan 30, 2016, 10:59 am**

Post by:

I still don't quite see how a LED can have that much resistance. At 5V the LED alone would be limiting the circuit current to 22.1mA

A small internal resistance for the LED makes more sense to me.

Interesting conversation about it here. (http://electronics.stackexchange.com/questions/76367/accounting-for-led-resistance)

A small internal resistance for the LED makes more sense to me.

Interesting conversation about it here. (http://electronics.stackexchange.com/questions/76367/accounting-for-led-resistance)

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Jan 30, 2016, 09:30 pm**

Post by:

@Frostline

The point is a LED is complicated to model.

I am giving you an equivalent resistance.

Wait till you get into inductive and capacitive reactance and resonance.

When dealing with non linear or reactive components, you can only calculate equivalent resistance.

Try this, put your device (LED in this case) into a closed black box that you cannot see into.

You are allowed the measure the applied voltage and the current draw.

How do you figure out the resistance of what's in your box?

Question:

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

(http://forum.arduino.cc/index.php?action=dlattach;topic=375121.0;attach=152896)

Click this LINK (http://imgur.com/yYZytTn) to see what is in the black box.

.

The point is a LED is complicated to model.

I am giving you an equivalent resistance.

Wait till you get into inductive and capacitive reactance and resonance.

When dealing with non linear or reactive components, you can only calculate equivalent resistance.

Try this, put your device (LED in this case) into a closed black box that you cannot see into.

You are allowed the measure the applied voltage and the current draw.

How do you figure out the resistance of what's in your box?

Question:

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

(http://forum.arduino.cc/index.php?action=dlattach;topic=375121.0;attach=152896)

Click this LINK (http://imgur.com/yYZytTn) to see what is in the black box.

.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Feb 01, 2016, 12:36 am**

Post by:

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

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Feb 01, 2016, 03:45 am**

Post by:

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 V_{f}? Or at least a mention of of it somewhere?

Like this datasheet (http://www.vishay.com/docs/83171/tlur640.pdf) 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.

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 V_{f} 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 (http://www.learningaboutelectronics.com/Articles/Battery-internal-resistance) 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?

If it (the LED) had resistance characteristics why is there no graph of resistance vs voltages beyond V

Like this datasheet (http://www.vishay.com/docs/83171/tlur640.pdf) 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.

Quote

1. How much resistance does a 1N4007 diode have when it is forward biased?1. Using this datasheet (http://www.diodes.com/_files/datasheets/ds28002.pdf) the diode has a forward voltage of 1V @ 1A.

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

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 V

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 (http://www.learningaboutelectronics.com/Articles/Battery-internal-resistance) 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?

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**larryd** on **Feb 01, 2016, 07:28 am**

Post by:

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 ;)

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 ;)

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Feb 01, 2016, 07:33 am**

Post by:

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

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Paul__B** on **Feb 01, 2016, 12:03 pm**

Post by:

I think what is making this so difficult for me to grasp is there is no mention of resistance really on LED datasheets.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.

If it (the LED) had resistance characteristics why is there no graph of resistance vs voltages beyond V_{f}?

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.

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Paulcet** on **Feb 01, 2016, 03:58 pm**

Post by:

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*

Title: **"Resistance is Futile!"**

Post by:**Paul__B** on **Feb 01, 2016, 10:32 pm**

Post by:

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.

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**polymorph** on **Feb 02, 2016, 09:10 pm**

Post by:

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

(http://www.electronics-tutorials.ws/diode/diode12.gif?81223b)

(http://www.electronics-tutorials.ws/diode/diode12.gif?81223b)

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Feb 03, 2016, 03:41 pm**

Post by:

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

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**polymorph** on **Feb 04, 2016, 12:55 am**

Post by:

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

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

The knee is certainly not at 20mA.

Another more accurate graph:

(http://girr.org/girr/tips/tips7/led_vf.gif)

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

The knee is certainly not at 20mA.

Another more accurate graph:

(http://girr.org/girr/tips/tips7/led_vf.gif)

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Feb 04, 2016, 04:12 am**

Post by:

with a proper graph, on can see what size resistor toWhat defines a proper graph?

use, with a non-linear device.

I ask because look at the current vs forward voltage graph for this red LED. (http://www.kingbrightusa.com/images/catalog/SPEC/WP813ID.pdf)

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 currentGoing 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

(http://s5.postimg.org/r1t0d3omv/load_Line.png)

you can use that currentSo since point C is about 15mA it would be 5/.015 or 333.3?

divided into the supply voltage to tell you the desired resistance.

This is called a load line( it still uses ohm's law ).

But what is that value telling me?

Say we have this circuit

(http://s5.postimg.org/3ouyul8jb/resist2.png)

Using the LED from the datasheet V

And I want to limit the current to 12.5mA

So formula is V-V

R=240

So I would need a 240Ω resistor in the circuit above.

According to LarryD my total circuit resistance is V

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

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

One can draw a new line for each resistor size and seeHow? Where did resistor value have influence where I drew the load line?

where it crosses the LED's line.

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Paul__B** on **Feb 04, 2016, 12:51 pm**

Post by:

Sorry, you drew your graph wrong.

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

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dlloyd** on **Feb 04, 2016, 02:00 pm**

Post by:

Quote

So since point C is about 15mA it would be 5/.015 or 333.3?(5-1.5)V/0.15mA = 233Ω

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Feb 04, 2016, 02:44 pm**

Post by:

Sorry, you drew your graph wrong.You are half right.

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

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

(http://s5.postimg.org/41i86eid3/resist3.png)

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

( 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).

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Feb 04, 2016, 02:51 pm**

Post by:

(5-1.5)V/0.15mA = 233ΩThis is because I failed to scale the X axis to 0 (the subtracting of the 1.5 from source voltage) ?

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**polymorph** on **Feb 05, 2016, 12:19 am**

Post by:

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

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Feb 10, 2016, 02:02 am**

Post by:

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

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

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**Frostline** on **Feb 10, 2016, 02:37 am**

Post by:

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

in current is.

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 (http://www.hebeiltd.com.cn/led.datasheet/1025MY8C.pdf) 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

But R

Title: **Re: Trying to understand current dispersal control across a parallel circuit **

Post by:**dwightthinker** on **Feb 12, 2016, 02:06 am**

Post by:

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

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