Jump to content

# 10 LEDs and a resistor

## Recommended Posts

Ok, I need a little help here.

I have 103v leds together with a 300 ohm resistor, if my math was correct, I would get 3v, but sadly, I don't it takes 7 300 ohm resistors clustered togeather to get 3v w/10 led load.

1 300 ohm resistor does not provide the nessisary power to light all 10 leds. Using...

(9v - 3v) / 20ma = 300ohm

My best guess is that the above equation was not meant for 10 leds. So what resistor will provide 3-3.5v to 10 leds from a 9v power source, with-out the 7 resistors and generating lots of heat.

Any ideas?

--Daniel L

##### Share on other sites

Ok, I think I understand your concern. First off, I am assuming the seven resistors are connected in parallel, meaning that they are connected side by side and not end to end. That actually lowers the resistance because it gives more ways for the current to flow. You need less total resistance because you are hooking up the LEDs in parallel, so the current of each LED adds up. You have 10 LEDs, so the total current going through the resistor is 200 mA. So, you technically need a 30 ohm resistor, which you got close to from the multiple resistors in parallel.

However, putting LEDs in parallel is often times a not so good idea. LEDs are pretty uniform now, so odds are you won't have a problem, but if the LEDs are slightly different, it could burn them out.

In any case though, a couple hundred mAs is not going to generate a lot of heat whether you use one 30 ohm resistor, or seven 300 ohm resistors.

If I get some time tonight, I'll make a drawing of a better way to wire them up. It's too complicated to type out.

##### Share on other sites

I'm assuming that the 3 volts is the Forward Voltage (Vf) of your LEDs. If you connected all the LEDs in series, you'd have a total Vf of 30 volts, so your 9 volt supplywouldn't provide enough power.

Best bet is to use a 300 ohm resistor for each LED and connect each resistor/LED pair to the + and - of your 9 volt supply.

Hope this makes sense.

Good luck!

Kym

##### Share on other sites

Take a look at this...

joel

the guru suggests Supply is 9v

Voltage drop is 1.7 per led

Current is 10ma

makes R1 56 ohm

##### Share on other sites

Ok, guys, here's the scoop, since all the resistors are of the same VALUE, then your parallel resistance formula simplifies to

R-equivalent= (Resistance in ohms)/( # resistors).

The above only works if they are the same value.

So, your equivalent resistance is thus 300ohms/7 resistors= 42 ohms.

SO... why does a 42 ohm resistor work for you when a 300 ohm resistor does not??? It's simple!

Using Ohm's law, V=i*R, your current with the 7 resistors tied in parallel, would thus be, solving for i:

i= Voltage across resistors/ total resistance

9v - 3vvolts=i * 42.8 and solving for i, your current is:

6volts across resistor/.0428 kohm= 140 mA of current flowing.

I did .0428 above, because you keep the units as volts, ma, k-ohms.

This 140mAmakes perfect sense, because LEDs require generally, about 20 mA before they start flowing.

Current in parallel circuits adds up linearly, so divide that 140mA by 7 LEDs, you get 20 mA flowing in each LED. This is why your LEDs did not power up until you had 140 mA total, because each of your 7 chains required 20 mA, so 7 of them in parallel need 20 mA * 7 =140 mA supply from the source.

The way the store bought LED strings work, is they split it into 2 zones, so 60 LEDs would have 2 chains of 30 LEDs each wired together. This allows you 90 volts of droppage across the 30 3-volt LEDs, which still leaves you room with a 120V AC supply.

You need to check your data sheet for the LEDs and determine when forward voltage Vf is satisfied, what the current requirements are, then choose the appropriate resistor value, that when plugged into Ohm's law equation, gives you the proper current that the data sheet requires.

HERE'S WHY 300 OHMS DID NOT WORK FOR YOU

With the 300 ohm resistor, the current flow through the resistor is

6v/.3k = 20 MA, which was not enough current to allow a forward voltage to jumpstart 7 LEDs THAT NEED 20 Ma EACH. You got 7 babies and 7 bottles, but only one bottle has milk in it! LOL!

But if you disconnect 6 of the LEDS, I bet your circuit would light the 7the LED!

Lastly, one other point we need to touch on, is you should buy high quality LEDs from the same lot, because they need to be matched so that you are not forcing 2.7 volt VF onto an LED that needs 2.4volts, next to an LED that needs 3 volts. Ideally, perfectly matched LEDs from the same lot should all have exactly 3.0 Volts Vf. Good manufacturers with 6-sigma manufacturing techniques will make these LEDs with tight tolerances so that they all act the same.

##### Share on other sites

jeffostroff wrote:

Ok, guys, here's the scoop, .....snip

Jeff,

Thank you for the very good information. I just printed a copy of this post to keep on the wall next to my workbench.

Jeff

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account. Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
• Create New...