# 36 LED Circuit

Discussion in 'Homework Help' started by User1.0, Dec 9, 2011.

1. ### User1.0 Thread Starter New Member

Aug 10, 2011
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My assignment is to create the schematics for a doll house with four rooms each with a switch and 9 lights.

Voltage Source: 12 VDC
Voltage per LED 3.1V
Current per LED 20 mA

To ensure I understand everything, my math is as follows:
9 LEDs per room at 3.1 V per LED Allows for 3 in series.
12-3(3.1) = 12-9.3=2.7 V
2.7 V /.02 A= 135 OHM

Does that all look right? Are my calculations correct?

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
You should not put LEDs in parallel by themselves. They have too much individual variation amongst units. You need a resistor per LED string for good design practice.

LEDs, 555s, Flashers, and Light Chasers

3. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
300
That is not right, because for each room you have three strings in parallel, each of three LEDs in series. What is the total current for each room?

On top of that, as somebody else has just pointed out, each string should have its own resistor.

Finally, the resistors probably don't need to be anything fancier than an E24 value. What's 135 ohms, the exact value as calculated?

4. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
So, you are suggesting that I simply put a resistor on with each set of LEDs. You also stated that E24 will suffice. Are you suggesting to go to a 130 Ohm? If I were to make the above changes, would that suffice for proper running this circuit?

5. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Uhh, no. LEDs are current controlled devices. If you were splitting the current 3 ways you would need a resistor 1/3 the value to give the LEDs the right current.

I am assuming 20ma per LED. If each LED drops 3.5VDC then you have a total Vf drop 10.5VDC. This leaves 1.5VDC, which is the voltage you would use to control the current.

1.5VDC/0.02A=75Ω

The only reservations I have is the 1.5VDC is not very guaranteed due to the fact LEDs are not that predictable, which means there could be quite a bit of variation in the current.

I get the feeling from this design you are fairly new to LEDs. This is why I wrote the tutorial I linked to earlier. I recommend reading the first 2 chapters.

Your design would have had 11ma, split 3 ways it would have been around around 4ma each LED, which is way under powering the LEDs. You may have meant to do this, it is hard to say.

Resistors are quick and simple, and work well. In the real world you will need to verify the currents to make sure they are really in the ballpark.

6. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
My bad. I hadn't seen the link. I will look into that, and review a bit more about the basics.

My using a 135Ω was the same reason you used 75Ω. I calculate each at 3.1 Volts instead (as per the assignment) so I think I had that right.

I had no intentions on amperage. As I new to circuitry altogether, what may have been viewed as intentional was more ignorance.

I'll study some more information and get back to it.

Thank you for your assistance.
-Shane

7. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
OK, I've done some research, relearned what I've so easily forgotten, read your suggested readings, and have come back with a better schematic, I hope!

As I said, I've forgotten the basics on electricity. I forgot what exactly makes amps drop or not drop. I think I've got it now!

My new math led me to place 4.3Ω on each line of LEDs (Thanks Bill).
I first took my 20mA and multiplied it by 3 for the three branches of LEDs:
.02A*3=.06A
I then multiplied that by the four separate rooms:
.06A*4=.24A
I then divided my 12V by .24A for 50Ω total. divided back by 4 and 3 to 4.1666Ω.
Closest E24: 4.3Ω

Is this right now? I am a bit more optimistic, as the simulation in MultiSim actually worked! This is for a school project, and will be turned in as just the image above. No actual breadboard will be made, though I want it to be good enough to be made into one anyway.

8. ### Adjuster Well-Known Member

Dec 26, 2010
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300
You seem to have misunderstood something. 4.3Ω is too low and will probably lead to the LEDs burning out quickly.

Each series string of LEDs requires 20mA. The input voltage is 12V.

Each string drops perhaps 3*3.1V=9.3V (your figures: rather low for typical white LEDs) or 3*3.5V =10.5V (Bill Marsden's figures).

The difference between the input voltage and the LED string voltage divided by the required current gives the resistance. I might be inclined to start by using your lower voltage: with too much resistance the LEDs may be dim, but nothing will burn up.

However you figure it, since you are using a separate resistor per chain, you must not do this dividing by 4 and 3. Have you not studied series and parallel circuits, or perhaps forgotten them? It would help you to get this stuff well understood, as it is basic to knowing how electric systems work.

Good luck.

Last edited: Dec 10, 2011
9. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
I'm not sure where my math is flawed. So I have the remainder of 2.7V. divided by the .02A: 135Ω.

Some math I use beings me to higher value resistors per series. Others bring me to lower values.

so... 135Ω/3 = 45Ω. If I were to put a 45Ω resistor on each series, that would do it correctly?

I found http://led.linear1.org/led.wiz from one of Wookie's posts. This states that I should use a 150Ω per series. I'm not doubting you're input necessarily. I'm just curious as to what I'm missing here.

I pulled out my breadboard and tried a few of my guesses. Again, I am not sure what is "right" and "wrong"

I've watched and read "beginner" circuit tutorials on series/parallel circuits, but have still come out not understanding as well as I should...

10. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
4.3Ω is 1/10th of the 45Ω you state that is needed, but you have 4R3Ω resistors in the schematic.

Here are my calculations:

$R=\frac{V_{drop}}{Current}\\
\\
R=\frac{12-9.3V}{20mA}\\
\\
R=135\Omega$

Per String of Three

11. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
You are figuring 3.1VDC drop per LED. If these are white LEDs that seems very low to me, but since the numbers were given to the OP they are what are used. Just be aware these numbers are abnormal for this kind of component. It also shows why I was recommending verifying the Vf so strongly.

Again, LEDs are current controlled devices, so it is the current flowing through the LED that is critical. The voltage drop is used to calculate the current.

TWG (thatoneguy) has shown the math, it is accurate given the starting conditions.

To repeat myself again, if you were to actually going to build this design I would strongly suggest you verify the LED Vf, since the numbers are pretty far off of the majority of white LEDs out there.

I am a strong advocate of resistors, due to the basic simplicity. An alternate way is to build simple constant current source, which will feed the designated current to the LEDs. If you want to explore that option I'll be glad to show some examples.

Last edited: Dec 11, 2011
12. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
I personally calc White/Blue LEDs at 3.4V and 18mA.

You can't tell that much difference in brightness between 18mA and 20mA, and if your LEDs or resistors are a bit out of tolerance, there won't be a major meltdown.

For What it's worth.

Using that rule of thumb, you'd want 100Ω resistors per string of 3, but higher resistances can only help. For example, if you plug in a non-regulated 12V 2A transformer type supply, the voltage can be as high as 16V until loaded down. These are other parameters we do not have knowledge of, but can only give you the formulas and clues of what to measure first and look out for before you hit the "ON" switch.

13. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
Yes, I was making corrections to my previous post without uploading another picture. Sorry for not specifying.

So, TWG says 135Ω per string of three (As I asked in my second post) but I understood your post, Bill, to say that I needed to divide by three.
That is why I placed the 45Ω resistors on each string.

Again, these unrealistic figures were given as part of the assignment. I will not be building a physical circuit. This is an impractical assignment in an impractical class basically led by a teacher's assistant...

14. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
For each 9 LED set.

How many LEDs are in parallel?

How much voltage is across those parallel sets?

What is the maximum current that can flow through any LED?

Remember, current in series is the same, current in parallel strings can be different (tolerances, etc.)

Which seems the correct answer to you knowing the above?

15. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
If you found one where I was actually recommending that site, it must have been a very old post of mine.

That calculator is flawed. I have specifically recommended against it's use for several years. The problem with it is than in many circumstances, it will leave practically zero volts to drop across the current limiting resistor, and use a 1 Ohm current limiting resistor. I tried to convince the site owner to fix the program, but they refused.

This is the way that I calculate LED strings:

1) Get the maximum number of LEDs in a string: If your supply voltage is stable (regulated), then subtract 1 volt from it, and divide the remainder by the typical Vf of your LEDs, and round the result down; eg:
Max_LEDs = INT( (Vsupply - 1v) / Typical_Vf)
Max_LEDs = (12v - 1v) / 3.1v
Max_LEDs = 11 / 3.1 = 3.548 = 3 LEDs

2) Compute the total voltage drop of the number of LEDs returned in step 1; in your case:
Vf_tot = 3 * 3.1 = 9.3v

3) Find the remaining voltage to drop across a current limiting resistor by subtracting Vf_tot from Vsupply:
Vrlimit = Vsupply - Vf_tot = 12v-9.3v = 2.7v

4) Determine the proper value for the current limiting resistor using Ohm's Law; R=E/I, so Rlimit = Vrlimit / Desired_Current = 2.7v/20mA = 135 Ohms. You already arrived at this number.

135 Ohms is the resistance required for EACH STRING OF 3 LEDs.

5) Find the closest standard resistance value:
Look at a decade table of standard resistance values:
http://www.logwell.com/tech/components/resistor_values.html
(bookmark that page!)
You will find that in the E24 values, the closest that is greater than or equal to 135 is 150 Ohms.
You should not go below the required resistance, as the life of the LEDs may be reduced due to too much current flowing.

The closest E24 value is 150 Ohms.
If you wish, you can use resistors in series or parallel to get closer to your target resistance. 15 + 120 Ohms in series = 135 Ohms, two 270 Ohm resistors in parallel is also 135 Ohms.

6) Determine the wattage rating required for Rlimit.
Power in Watts = EI. We know that V(Rlimit) = 2.7v. I(Rlimit) = 2.7v/Rlimit. If you use a 150 Ohm resistor, then 2.7v/150 = 18mA; 2.7v*18mA=48.6mW.
For reliability, you multiply that by 1.6 and wind up with 77.76mW.
You could use a 1/10th Watt or greater rated resistor.

16. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
So, now I shoulduse 135Ω on each string? Again, this is what I understood Bill to suggest not to do!

He stated that this would drop my voltage to far, and that I should instead divide by three for my final resistance.

I'm really not trying to be difficult here. I'm just hearing people say that my answers are incorrect, then give me a result I've already deduced.

17. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
300
NO, you were told you not to use 135Ω for three strings in parallel. This was for three reasons:-

1. Because 135Ω could not supply the total current required for three strings (three times 20mA).
2. Because it is better practice to use a separate resistor for each string.
3. Bill Marsden pointed out that typical white LEDs drop more voltage than your 3.1V value, so the resistor for a single string might need to be less than 135Ω to get the full 20mA. (But 135Ω per single string could only result in the current being a bit low, unlike the case where it supplies three strings.)
You should indeed use a resistor in series with each string, calculated to set each string current to 20mA.

Depending on the actual LED forward voltage, the best practical value for this might lie between somewhat more than your 135Ω value, and maybe a little less than Bill's 75Ω.

It is always safest to use a higher resistor and lower current though, so why not take SgtWookie's suggestion of 150Ω, and see what you get?

Last edited: Dec 11, 2011
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18. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
300
Think about each series string of three LEDs plus a resistor as a unit that needs to be correct in itself. The fact that there are other strings in parallel with it does not change things, apart from increasing the total current taken from the supply.

Only if a single resistor feeds several things in parallel do you need to add currents together. For instance, if you had stuck to your original not very good practice of using a single resistor per three series strings in parallel, the resistor would have been lower (12V-3*3.1V)/(3*20mA) = 45Ω

19. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
An addition to this thought, if this had been done the currents would not be equal through every LED string, so you can have what is a cascade failure. One string fails the other two gets 50% or more current. When another string fails the remaining string gets all the current.

20. ### User1.0 Thread Starter New Member

Aug 10, 2011
13
1
I completely understand the fact that I should have a resistor on each string of LEDs. That makes perfect sense for reasons of failure and fluctuation.

What I want to know is why is it that when I suggested in my second post to place a 135Ω resistor (I understand the reasons to round up to 150Ω) on each string that I was corrected saying that I should divide by 3
Bill:
Giving me a resulting 45Ω to be bumped up, but now I am being told to do what I said then???

I guess it is what it is. I will use a 150Ω resistor per string of lights. I just don't know why I was wrong when I suggested the same (aside from the rounding up)