# Need circuit board help

Discussion in 'The Projects Forum' started by b3ans, Apr 3, 2008.

1. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
Hi I am trying to learn about circuit boards and I am going to try to make one. But I have a question before I try. I was going to have 6 leds but have every 2 hooked up to a 33 ohm 1/2 Wat resistor. Using 5v coming from a usb drive will this work. If you need i can post a link to what the circuit board looks like.

2. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807

What is the color, and maximum and typical current @ voltage ratings for your LEDs?

Have you measured your LEDs for Vf? (forward voltage) They can vary quite a bit, even in the same lot.

You can verify the Vf of your LEDs by testing them with a power supply - the power supply could be as simple as several batteries (I suggest at least 6v)
Take the typical Vf specification, and subtract that from your supply's voltage, and then divide the result by the current rating to determine the current limiting resistor:
Rlimit = (Voltage Supply - Vf(LED)) / Current Rating

Example: if your LEDs are rated 20mA @ 2.1v (typical), and you have a fresh 9v battery handy:
Rlimit = (9V - 2.1V) / 20mA
Rlimit = (9-2.1) / 0.02
Rlimit = 6.9 / 0.02
Rlimit = 345 Ohms
Use the next larger standard value, or 360 Ohms.

A table of standard resistor values is here:
http://www.logwell.com/tech/components/resistor_values.html
Use the E24 series, which is what's typically available in stores (5% tolerance)

Connect and measure each LED in series with the resistor, measure & record the Vf.
You should pair the lowest reading LED with the highest reading one, the next lowest with the next highest, etc. This will help considerably to ensure even brightness from all of your LEDs.
Then you can calculate the series Rlimit value you need much more accurately.
For example, let's say the highest Vf was 2.21, lowest was 1.98. Your intended supply voltage is 5.
Rlimit = (Voltage Supply - Vf(LED1+LED2...)) / Current Rating
Rlimit = (5 - Vf(1.98+2.21)) / 20mA
Rlimit = (5 - 4.19) / 0.02
Rlimit = 0.81 / 0.02
Rlimit = 40.5
The next larger standard value is 43 Ohms.
To calculate what the actual current will be:
I(LED) = V/R
I(LED) = 0.81 / 43
I(LED) ~= 18.8mA
Note that a 5% tolerance 43 Ohm resistor might measure as low as 40.85 or as high as 45.15.
You'll wind up with the LEDs being as bright and evenly lit as possible with a long service life.

3. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
My leds are 2.2 volts so I think that will work. Thank you for your help.

4. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
I looked at radio shack and they only have a 47 ohm 1/2W resistor with 5% tolerence. would that work too or will it make it worse.

5. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
Ok, you said that your LEDs are 2.2 volts - but you didn't specify their current rating.

Is that 2.2 volts measured, or a specified maximum, or a specified typical?

When you are working with tolerances that close (ie: your remaining voltage being under a volt) you really should measure your LED's to determine what their actual Vf is. The reason is that any variances in the power supply voltage will show up as much more significant current fluctuations through the limiting resistor than if there were more than a volt "left over". Therefore, a small error in limiting resistor selection becomes much more significant.

If your LED rating is "maximum" 20mA @ 2.2v, your actual Vf may be significantly lower.

In that case, I strongly recommend getting each LED's Vf using the procedure I documented above.

Let's see what kind of current rating you might need on your resistors. We'll assume for the moment that your LEDs ARE 2.2v at 20mA, since I have no further information.

Rlimit (5v - (2.2+2.2) / 20mA
Rlimit = 30
E(remaining) = 5V-(2.2+2.2)
E(remaining) = 0.6V
P(Rlimit) = E(remaining) * I
P(Rlimit) = 0.6V * 20mA
P(Rlimit) = 0.012 Watts, or 12mW
In this case, you could use 1/10W resistors with no problem.

6. ### b3ans Thread Starter Member

Apr 3, 2008
17
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T-1-3/4 size (5mm)
Green lens color
Viewing angle of 30°
10 mA current max
Typical voltage 2.2V, with a maximum of 2.4V Typical MCD is 20.
This 5mm Green LED has a typical wavelength of 565mm.

This is all the information i have so does that mean that i have 6 mW instead of 12. And if so what kind of resistor would that be.Sorry for the stupid questions and thanks for your help.

7. ### Audioguru Expert

Dec 20, 2007
11,093
1,290
A max current rating of only 10mA is very low for a 5mm LED. Most are 30mA to 40mA.
A light output of only 20mcd is also very low. Most today are in the thousands.

8. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
I agree with Audioguru's assertion that 10mA is very low. However, they may be "new old stock", with a LOT of emphasis on the "old" portion.
So, let's plug the 10mA into the equation.
Rlimit (5v - (2.2+2.2) / 10mA
Rlimit = 60
Closest standard 5% value is 62 Ohms. They may vary between 58.9 and 65.1 Ohms.

P(Rlimit) = E(remaining) * I
P(Rlimit) = 0.6V * 10mA
P(Rlimit) = 0.006 Watts, or 6mW

They won't be very bright, but you'll certainly be able to see them glowing - and they should last a long time.

9. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
Thank you for your help i will see if i can try it out today. Next time i will try to get better leds though.

Apr 3, 2008
17
0
11. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
So, let me ask YOU a few questions...  1) How much current will you have going through two LEDs if you use a 68 Ohm resistor instead of a 60 or 62 Ohm resistor?

2) The amount of light you get from an LED is roughly porportional to the amount of current flowing through it. How much will the light output be reduced by using the greater value of resistance?

3) Can you think of a way to use a combination of 2 resistors they DO stock in order to get 60 Ohms?

12. ### Audioguru Expert

Dec 20, 2007
11,093
1,290
If your supply is 5.0V and your LEDs are 2.2V then a 68 ohm resistor in series with each one will produce a current of (5.0V - 2.2V)/68 ohms= 41mA and the LEDs will blow up.

If the supply is 5.0V and two 2.2V Leds are in series and in series with a 68 ohm resistor then the current will be (5.0V - 4.4V)/68 ohms= 8.8mA which will make the LEDs look dim. But your LEDs are old and low current so that is how they will be.

The 68 ohm resistor will dissipate 0.115W or 0.005W. A small 1/4W resistor would work as well as a larger 1/2W resistor.

If your LEDs are actually 2.5V or higher then they will be exteremely dim or will not light is two are in series with a 5.0V supply.

I used 1/2W resistors 40 years ago with vacuum tubes. I have used 1/4W resistors with solid state ever since. RadioShack is 40 years behind the times. That is another reason why they are no longer in Canada.

Do you see how simple is the arithmatic?

13. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
i am using a parralell circuit and i used an led calculator lol and it says i can use a 1/4W 5% 33ohm resistor but instead can i use a 1/2W is that ok if not i will try to figure out the math lol. i did one of those equations and i think 18.1 mA will be going through the leds so it should be fine i hope.

14. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
The LED calculator was assuming your LEDs were capable of 20mA current.
Two 2.2v LEDs in series on a 5v supply leaves 0.6v for your Rlimit to drop.
0.6/0.02 = 30 Ohms, so the program told you to use 33 Ohms, which is the closest standard size that is larger.

You can use a 68 Ohm 1/2 Watt resistor in series with two LEDs that are also in series.

Radio shack does carry a limited supply of 1/8W resistors now.
You can use one 100 Ohm and one 150 Ohm resistor in parallel to get 60 Ohms.
There are basically two ways to calculate resistance in parallel.
If you have only two resistors:
Rtot = (R1 * R2) / (R1 + R2)
Rtot = (100 * 150) / (100 + 150)
Rtot = 15000 / 250
Rtot = 60 Ohms
If you have two or more resistors:
Rtot = 1 / (1 / R1 + 1 / R2 +.... 1 / Rn)
Rtot = 1 / (1 / 100 + 1 / 150)
Rtot = 1 / (0.01 + 0.00666...)
Rtot = 1 / 0.01666...
Rtot = 60 Ohms

Here is the page for their 150 Ohm 1/8 W resistors:
Here is the page for their 100 Ohm 1/8 W resistors:

But before you actually build the board and solder things in, don't you want to see what your LED's measure first?

15. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
I forgot i put 20 MA in the calculator. now lol it says to use 68 ohm 5% 1/4W resistor so can i use that but a 1/2 watt instead thanks for the help.

by the way this is what my skamatic looks like lol i wanted to keep it simple lol 16. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
Yikes! Your LEDs are in parallel, not series!

If you used the 68 Ohm resistors, you would VERY quickly burn out the LEDs.
Rlimit=(5-2.2)/10mA
Rlimit=280 Ohms!
But, you are trying to connect LEDs in parallel - this is not a good idea, because the actual Vf of LEDs can vary considerably.
If you used 280 Ohms, and your LEDs just HAPPENED to be perfectly matched, then each LED would get just 5mA.
So, Rlimit actually needs to be 140 Ohms.
But instead, let's fix your board so that the LEDs are in series.

17. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
Are you shure that will not work with a 68 ohm resistor. if not i will try to design a series circuit. lol i hate led calculators

18. ### SgtWookie Expert

Jul 17, 2007
22,201
1,807
I am ABSOLUTELY certain that if you soldered in your LEDs and 68 Ohm resistors that your LEDs would try to split about 41mA per parallel pair - and it would not be an even split, because you haven't attempted to match your LEDs - or at least not reported that you have.

So, one LED in a pair might get 25mA, the other 15mA. Since you say they are rated for
10mA maximum, that will mean they will have far too much current going through them. One will quickly burn out, and as soon as it does, the OTHER LED in parallel with the burned-out LED will get the full dose of 41mA. It won't last long at all like that.

19. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
Ok i will start designing that right away. Thank you I would be very upset if that happend lol.

20. ### b3ans Thread Starter Member

Apr 3, 2008
17
0
Ok i thought i was keeping it simple but now it is reall simple i was going to use the two 1/8W resistors like you showed : )

here is a picture: 