# Behavior of V and I in a circuit with one R and two LED, which one will glow?

#### xan137

Joined Feb 5, 2021
5
Hello,
I have a simple circuit with single source providing 9V, R=3,9KΩ, two LEDs in parallel (red 1,8V and blue 2,7V).
What will happen if we turn on the voltage source?

Are my thoughts correct?
I is the same on each branch, I resistor is I LED1 + I LED2

We can't tell I total, because we don't know the resistance of each LED, but
I = U / R → 9V / 3.9k Ω = 0,0023 A
We have 0,0011A (1,1mA) on each LED, correct?
I think I am wrong, because forward voltage of LEDs are different (1,8 and 2,7V), that means they have different resistance, that means we have more current on one branch.

Voltage is the same on each branch, do we have three voltage drops (R and two LED) or only two (R and parallel LEDs)?
Which voltage do we have on each LED? To calculate that we need to get voltage drop of LED, subtract 9V and we get voltage drop for both LEDs
Voltage drop R
V=I*R → 2,3mA * 3.9k Ω = ~ 8,97V
Oh... that doesn't work because we need R total... wait... if I multiple by 5000 (R total would be greater then value of one resistor) I get 11V.

Another way is to assume the voltage drop of the resistor, between the resistor we set a voltage and calculate the current.
 Voltage Current 1 0,00025 A or 0,25mA 2 0,0005 3 0,0007 4 0,0010 5 0,0012 6 0,0015 7 0,0017 8 0,0020 9 0,0023

The current would be divided for both branches, if we have e.g. 23mA on resistor, 11mA on each LED.
But how can that help me, If I don't know the voltage drop of resistor.

I am lost here. I know that the red LED will glow but I don't know why and I don't know how to get the answer through calculation.

Would be amazing, if someone could help me out.

#### ericgibbs

Joined Jan 29, 2010
16,002

#### Papabravo

Joined Feb 24, 2006
18,978
You will have the same voltage across both LEDs and different currents. To see what the currents might be you need to look at the LED datasheet which will tell you the approximate current for a given forward voltage. there are a couple of possibilities:
1. The voltage across the LEDs is less than the threshold of either one. Currents will be small and neither one is likely to "glow"
2. The voltage across the LEDs is between the thresholds of the two LEDs. One will glow and the other may glow dimly or not at all.
3. The voltage across the LEDs is above both thresholds and both will glow
The maximum current you can get through a 3.9K resistor is 2.31 mA. That current is too small to make most LEDs glow at all, let alone brightly. The LED currents in each branch will be less than this figure so you may not have enough to work with. You could of course try the experiment with 390 Ω for more visible results.

You can use the table above and the diode equation to construct a model suitable for calculation.
https://en.wikipedia.org/wiki/Shockley_diode_equation
Ask questions if you have trouble coming up with values to use in the equation. The following may also be of interest.
https://en.wikipedia.org/wiki/Diode_modelling#Shockley_diode_model

N.B. The equation is named after the guy who came up with it. Do not confuse this with Schottky which is another type of diode altogether.

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#### Delta Prime

Joined Nov 15, 2019
1,311
Hello there The answer may lie here - Shockley Diode Law.
Or here! parallel connection of a red, orange and green LED. If the value of R1 chosen were to result in a common voltage of 2.0 V applied across each of the three LEDs then we could calculate the expected current through each
The green LED has the highest VF of the three and at 2 V it will pass about 12 mA. It will be reasonably bright at this current.
The orange LED has a lower VF and it will pass about 27 mA. It will be very bright at this current and would be close to maximum continuous rating for a typical 3 or 5 mm LED. See LED current rating for a typical specification.
The red LED has the lowest VF and it will pass about 44 mA. This is above the 30 mA rating in the datasheet in the article above. The LED will be good and bright – for a time!

#### MrChips

Joined Oct 2, 2009
26,790
The LED with the lower Vf wins.
For 1st order approximation, ignore the other LED.
Calculate the current assuming you have the red LED alone.
The current through the blue LED will be much less.

#### crutschow

Joined Mar 14, 2008
30,438
I think I am wrong, because forward voltage of LEDs are different (1,8 and 2,7V), that means they have different resistance,
The don't necessarily have a different delta resistance (change in current with a change in voltage) but they do have different threshold voltages (voltage drop for a given small current).

#### Delta Prime

Joined Nov 15, 2019
1,311
The LED with the lower Vf wins.
For 1st order approximation, ignore the other LED.
Calculate the current assuming you have the red LED alone.
The current through the blue LED will be much less.
Hey ! I just got issued double secret probation for giving the answer to the student, and then it was deleted. Who is monitoring the moderators?

#### MrChips

Joined Oct 2, 2009
26,790
Hey ! I just got issued double secret probation for giving the answer to the student, and then it was deleted. Who is monitoring the moderators?
If I gave too much then I apologize.
Moderators are bound by the same Homework rules.

• Delta Prime and ericgibbs

#### xan137

Joined Feb 5, 2021
5
hi xan,
Look at this LED forward voltage chart.
...
E
Thank you, if I interpret you intention behind he chart correctly, I don't want to build the circuit. I want to understand the behavior of current and voltage in a circuit.
I want to know what is possible to calculate from the provided circuit, knowing the voltage source, resistor and if only the "VF Typical". In my case blue LED has 2,7 VF.

The maximum current you can get through a 3.9K resistor is 2.31 mA. That current is too small to make most LEDs glow at all, let alone brightly. The LED currents in each branch will be less than this figure so you may not have enough to work with. You could of course try the experiment with 390 Ω for more visible results.
That means the current provided for each LED is 1.15mA? I got the circuit from a video, he whole channel is for students from a university.
he connects the 9V battery and the 3.9k resistor and the LED glows.

The answer may lie here - Shockley Diode Law.
...
The diode - effectively - acts as a voltage controlled resistor.
Sorry but it feels to complicated for me. I tried to watch a video but I already get more and more questions.
Sure part of learning is to fight through theoretical text, no problem for me, but I think my beginner questions need a simple answer. E.g. what can I and what I can't calculate from the given circuit?
...
parallel connection of a red, orange and green LED. If the value of R1 chosen were to result in a common voltage of 2.0 V applied across each of the three LEDs then we could calculate the expected current through each
Sorry but I can't imagine how the voltage and current moves in your circuit. Does you circuit also has 9V, or 6V (3*2)?

The LED with the lower Vf wins.
For 1st order approximation, ignore the other LED.
Calculate the current assuming you have the red LED alone.
The current through the blue LED will be much less.
Yes I know that the red LED will glow.
I can't calculate the current, I need the resistance of all elements in the circuit.
My thoughts:
I = U / R → 9V / 3.9k Ω = 0,0023 A.
Both LED have some resistance, lets say 1k Ω
I = U / R → 9V / 4.9k Ω = 0,0018 A.
No matter what, the current will be smaller than 2.3mA.
The with smallest Vf will glow. I don't understand why, because it need less current. Is that something I should understand, because it is logical or I only can work that out having LED's datasheet?

It feels like my issue is very simple and you guys are just way to experienced and smart to explain the operation in the circuit on my level.
I appreciate your effort and I know the struggle, because I setup and fix computers issue for "non-computer" people.

#### Delta Prime

Joined Nov 15, 2019
1,311
I understand being bound by rules of this topic is there any special case or discretion to be used by moderators given the student is making a valiant effort?

#### ericgibbs

Joined Jan 29, 2010
16,002
Behavior of V and I in a circuit with one R and two LED, which one will glow?
hi xan,
I was attempting to suggest to you why only the RED LED, glowed and the BLUE LED didn't.

At their normal operating current, a BLUE has a forward voltage of ~3v and the RED ~1.8v.

So the RED would clamp the junction voltage to approx 1.8V , so the BLUE would not be passing sufficient current to make it glow.

The chart shows this 'effect' when you check the forward voltage for the different colours.

E

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• Delta Prime

#### Papabravo

Joined Feb 24, 2006
18,978
...
That means the current provided for each LED is 1.15mA? I got the circuit from a video, he whole channel is for students from a university.
he connects the 9V battery and the 3.9k resistor and the LED glows.
...
No this is wrong. The current will be different if the LEDs have different forward voltage thresholds as I outlined in my previous response.

EDIT: It might(??) help to consider a forward biased diode as a non-linear resistor whose resistance decreases with increasing forward voltage. It can be modeled in simulation with a behavioral current source (sink)

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• Delta Prime

#### MrChips

Joined Oct 2, 2009
26,790
Look at the graph in post #5.
The red LED turns on first because it has the lowest Vf.
It then acts like a voltage regulator.
Use this as a 1st iteration while ignoring the other LEDs.

Current I = (Vs - Vf) / Rs

This is your starting point.

#### Audioguru again

Joined Oct 21, 2019
4,930
Why are you thinking that an LED has a resistance? It has a threshold voltage instead.
If a blue LED does not get its about 3V threshold voltage that is clamped to 1.8V by the red LED then the blue LED does not conduct.
Instead of 16 posts in this forum, didn't you try it??

#### MrAl

Joined Jun 17, 2014
9,343
Hello,
I have a simple circuit with single source providing 9V, R=3,9KΩ, two LEDs in parallel (red 1,8V and blue 2,7V).
What will happen if we turn on the voltage source?

Are my thoughts correct?
I is the same on each branch, I resistor is I LED1 + I LED2

We can't tell I total, because we don't know the resistance of each LED, but
I = U / R → 9V / 3.9k Ω = 0,0023 A
We have 0,0011A (1,1mA) on each LED, correct?
I think I am wrong, because forward voltage of LEDs are different (1,8 and 2,7V), that means they have different resistance, that means we have more current on one branch.

Voltage is the same on each branch, do we have three voltage drops (R and two LED) or only two (R and parallel LEDs)?
Which voltage do we have on each LED? To calculate that we need to get voltage drop of LED, subtract 9V and we get voltage drop for both LEDs
Voltage drop R
V=I*R → 2,3mA * 3.9k Ω = ~ 8,97V
Oh... that doesn't work because we need R total... wait... if I multiple by 5000 (R total would be greater then value of one resistor) I get 11V.

Another way is to assume the voltage drop of the resistor, between the resistor we set a voltage and calculate the current.
 Voltage Current 1 0,00025 A or 0,25mA 2 0,0005 3 0,0007 4 0,0010 5 0,0012 6 0,0015 7 0,0017 8 0,0020 9 0,0023

The current would be divided for both branches, if we have e.g. 23mA on resistor, 11mA on each LED.
But how can that help me, If I don't know the voltage drop of resistor.

I am lost here. I know that the red LED will glow but I don't know why and I don't know how to get the answer through calculation.

Would be amazing, if someone could help me out.
Probably the thing you really want to know is which LED lights up, or do both light up with possibly one being dimmer than the other. The theoretical answer to that question depends on a characteristic called the "first light" current and that is the current at which the LED exhibits the very first light which may be very dim but still lit up a little. This spec may or may not be listed on the data sheet however and it could vary quite a bit between different LEDs.

As to the current, the theoretical answer to that lies in the equations for the two LEDs.
This gets a little more complicated and the results will vary widely because LEDs and diodes have curves that vary by a lot depending not only on current but also temperature and process variations.

Just as a quick example, here is one way to get a feel for what is happening. Note that if you want to understand this you should look up the spec sheet for the two LEDs and go from there. This is a more or less random example with two somewhat randomly chosen LEDs that may or may not exist in real life although they are still representative of how two different LEDs with different characteristic voltages (1.8 and 3.0 volts) might behave in your particular circuit.

We start with two 20ma nominal current diode equations one for each diode, and you can tell which one is which by looking at the exponent and i used these just for that reason so they are easy to distinguish from each other:

The two representative LED (diode) implicit current equations are:
Id1=2.4685007189237009e-6*(e^(3*(Vd-2*Id1))-1)
Id2=8.4635102888342643e-4*(e^(1.8*(Vd-Id2))-1)

where Id1 and Id2 are the two diode currents and Vd is the voltage common to both. We have three unknowns, Id1, Id2, and Vd, and Vd is also an unknown because we just dont know it yet as you noticed.
We have three unknowns but only two equations, so we need one more equation. From the circuit itself we can get this 3rd equation because we know that the sum of the LED currents must equal the current through the resistor, and so Vd is the source voltage (9v) minus the voltage drop in the resistor caused by the two currents. This leads us to the 3rd simple equation:
Id1+Id2=(E-Vd)/R

where E is the source voltage and R is the series resistor value in Ohms.
Now we just substitute E=9 and R=3900, but i used R=390 Ohms to see some more typical results (you will have to do 3900 if you really want too, and i didnt want to solve this completely for you yet).

Doing that and taking all three equations we end up with:
[Id1=2.4685007189237009e-6*(e^(3*(Vd-2*Id1))-1),Id2=8.4635102888342643e-4*(e^(1.8*(Vd-Id2))-1),(9-Vd)/390=Id2+Id1]

and because we have all those exponentials we have to solve this numerically. Using some math software or doing it manually we get:
Id1=4.6128070020549152e-4
Id2=0.018138291432195
Vd=1.746166868363842

(and remember this is using a 390 Ohm resistor not 3.9k).

So we see that the 3v LED draws a very small current about 0.5ma while the 1.8v LED draws nearly the full current of 20ma (18ma). Obviously the 1.8v LED will glow quite bright while the 3v one will only light up a tiny amount (for two LEDs with similar efficacy).

Also remember that LED characteristics vary a lot so this may not match your LED results exactly but still gives you an idea how different the two currents can be. In this example we see a difference of 39 to 1 in the two currents.

As a final note, the diode equations i used here result in somewhat smooth diode IV curves so a run of the mill numerical solver can probably always handle this. If the diode curves are sharper however a numerical solver may start to fail due to the rate of convergence being too slow. In that case you can solve for Vd in terms of the two currents, then substitute that in the other two, then solve for one of the currents in terms of the other current, then finally we end up with one equation in one variable which is one of the currents and then we can solve for that current numerically, then insert that into the other equation to get the other current, then insert the two currents into the 3rd equation to get the voltage common to both. That solves for all three variables.
It is very interesting to do this you should try at least once if you are into the math.

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#### BobaMosfet

Joined Jul 1, 2009
2,007
Hello,
I have a simple circuit with single source providing 9V, R=3,9KΩ, two LEDs in parallel (red 1,8V and blue 2,7V).
What will happen if we turn on the voltage source?

Are my thoughts correct?
I is the same on each branch, I resistor is I LED1 + I LED2

We can't tell I total, because we don't know the resistance of each LED, but
I = U / R → 9V / 3.9k Ω = 0,0023 A
We have 0,0011A (1,1mA) on each LED, correct?
I think I am wrong, because forward voltage of LEDs are different (1,8 and 2,7V), that means they have different resistance, that means we have more current on one branch.

Voltage is the same on each branch, do we have three voltage drops (R and two LED) or only two (R and parallel LEDs)?
Which voltage do we have on each LED? To calculate that we need to get voltage drop of LED, subtract 9V and we get voltage drop for both LEDs
Voltage drop R
V=I*R → 2,3mA * 3.9k Ω = ~ 8,97V
Oh... that doesn't work because we need R total... wait... if I multiple by 5000 (R total would be greater then value of one resistor) I get 11V.

Another way is to assume the voltage drop of the resistor, between the resistor we set a voltage and calculate the current.
 Voltage Current 1 0,00025 A or 0,25mA 2 0,0005 3 0,0007 4 0,0010 5 0,0012 6 0,0015 7 0,0017 8 0,0020 9 0,0023

The current would be divided for both branches, if we have e.g. 23mA on resistor, 11mA on each LED.
But how can that help me, If I don't know the voltage drop of resistor.

I am lost here. I know that the red LED will glow but I don't know why and I don't know how to get the answer through calculation.

Would be amazing, if someone could help me out.
This stuff is usually poorly explained because few people understand the mechanics well enough to actually understand the math they are using. Math is useless if you don't understand how it gets the answer. Energy is very straightforward. I will explain exactly how your circuit will behave (below). Get this, it will help you in electronics-

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

The simple answer to your schematic and problem is this- This is a current problem. Ignore the resistor for the moment. Understand that electricity travels all paths, but the majority of energy will travel the path of least resistance- in this case, that is your 1.8V RED LED. Because it represents the path of least resistance, virtually all the current the resistor allows will go through the RED LED and the BLUE LED will starve. it will get virtually nothing.

In fact, because this holds true, it doesn't matter what the resistor value is, the RED LED will be driven until it fries before the BLUE LED is given sufficient current to light up. 3.9K-Ohms drops the current to almost nothing. Even if you used a 225-Ohm resistor, allowing up to 40mA through, it would use that to fry the RED LED because it is a lower-resistance. Only when it fries, and becomes a higher level resistance than the BLUE LED will the BLUE one light.

Voltage and Current do not (without being made to do so) work independently of one another.

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#### MrAl

Joined Jun 17, 2014
9,343
Another idea is just to compare the two LED graphical voltage curves. Slide one curve over the other until the voltages are the same and add the currents and multiply by the series resistor resistance and subtract that from the power source (9v) and if the two voltages match then we have the solution to both the voltage and the two currents.

Another math way is to start with the two LED function curves for voltage and simply equate them (because the voltage for both must be the same in the circuit) and a second equation that solves for the LED voltage due to the voltage drop in the series resistor (again as above) and those two equations can be somewhat easily solved because with that information we can solve for one LED current and then back substitute to get the other values the second LEd current and the two LED voltages (which are both the same).

#### xan137

Joined Feb 5, 2021
5
Thank you very much for all you answers, it has been a long time since I found a community where I got so much help. Give me some time and I will beginners as well.

Thanks @MrAI but your math overwhelmed me, because this circuit was shown on this channel so early (before he explained how to calculate resistance), I thought the math would be on similar level. I am also not sure how to work with such number: 4.6128070020549152e-4 it must be 0,00046 if I am not wrong.
I also understood everything your wrote and replaced correctly all the variables with x,y,z and entered them in few online equations solver, with no luck.

@BobaMosfet thank you for the book recommendation, I will check it out.
I read around 100 pages of All-in-One Electronics Guide by Cammen Chan, it felt like a reference book not a guide.
Currently I am reading Electronics For Dummies by Cathleen Shamieh, page 170/600 and it is easy and quite nice.

All the answers merged to the following thoughts:
Only one LED is emitting visible light, because it drains all of the 23mA ( I = U / R → 9V / 3.9k Ω = 0,0023 A) or
18mA ( @MrChips I = (Vs - Vf) / Rs; (9V-1.8V)/3.9k Ω) not sure what we get here, is it the correct current, because in the calculation above I didn't added the resistance of the LED?
Higher Vf, more resistance and smaller current. Red LED has smaller resistance and all the current flows though red.

What about voltage drop? Because the LED needs the 1.8V, we have a Vdrop of 7.2V at the resistor? But we have a branch, we should have same voltage on another branch.
2x1.8V=3.6V-9V we get 5.4 voltage drop at resistor.
If we use Ohm's law
3.9k*0.002A=7.8V
3.9k**0.0023A=8.97V
3.9k*0.0018A=7.02V (I = (Vs - Vf) / Rs)....... 9-7.02-1.8=0.18V left for blue
this value looks a lot better, because I think there is no current flowing to another LED, because we don't have 0.7V to overcome barrier potential of the second LED. But I am wrong, MrAI calculated a current of 0.5mA but he used a much smaller resistor... so his version of the circuit the blue led could has more than 0.7V. How do I even calculate the voltage on this branch? U=R*I, 390 Ω*0.0005A= 0.195V. Why is there a current flowing when we have not enough voltage to overcome the depletion layer?

I am sorry but I don't know what happens to the voltage in this circuit. Because red is glowing, it has a Vdrop of at least 1.8V, which voltages have the other LED?
The asked question was why red glows, I learned because it has smaller resistance, based on the smallest LED Vf value. But now I am not sure what to think about the voltage behavior in this circuit.

Thanks again guys. I am also available for a private lesson on any VOIP software just in case... I live in Germany UTC+1