10 LEDS for 9v

[C21]

The supply volt i am currently using is 9v battery and i want to hook up 10 5mm LEDS (2 Blue, 2 White, 2 Yellow, and 4 Red) to the 9v. The resistor is already install on each LEDS (330 Ohm on blue and white, 390 Ohm on yellow and red). Only problem is the LEDS has higher voltage than 9v correct? I think so because i have already tried and the battery got HOT. 9v battery is the only option to i want to use. So is there any way i can make this work? Will i need something additional than resistor? I have seen 12 5mm LEDS (2 white, 2 blue, 4 yellow and 4 red) connect direct to 7.2v and work like charm!

Any help is much appreciate.
Thanks!

 

[C21]

Help please, I am in bit rush to get this finish.

 

[Gunner12]

Well, the normal voltage for a white and blue LED is 3.7v. Green is a bit lower and Red is around 1.7v.

What are you trying to make?

:welcome:

 

[frenzee]

9V across a 330-Ohm resistor will produce only .25 W of heat, and far, far less than that in the battery itself. If the battery got hot, something, somewhere must be shorted out. How did you connect things? Pics, diagrams would be helpful.

 

[Mr Happy]

The typical drive current for a small LED is 20 mA. You have 10 of them, so the 9 V battery will be supplying about 200 mA. I think that's rather a lot for a 9 V battery to supply, so it won't last very long. I'd suggest putting an additional 500R to 1k resistor in series with each LED to limit the current a bit more.

Just a moment ago I put red, green and yellow LEDs in series with a 1k resistor on a 9 V battery and they all lit up quite well. So you could try putting some of your LEDs in series too. If you only want to see them lit up you don't need much current to drive a 5 mm LED.

 

[C21]

I have the specification paper that comes with LED.

White = 3.0-3.2 Forward Voltage
Blue = 3.2-3.4 Forward Voltage
Red = 1.9-2.1 Forward Voltage
Yellow = 1.9-2.1 Forward Voltage

I am building this for radio control. Here the diagram i have design.

 

[chimo]

Is that a normal 9V (MN1604, 6LR61) type battery?

If so you are asking it to supply an aweful lot of power. You have 10 LEDs drawing about 18mA each for a total of 180mA (assuming no battery sag - not realistic).

9V x 0.18A = 1.62 Watts!

If the answer to my first question is yes - that's why the battery would be hot.

 

[C21]

I am using energizer 9v battery.

I am assume this setup with the resistor i have choose is not enough. So i will need a higher resistor as Mr Happy has mention but in parallel.

This thing is getting difficulty as far as i go.

 

[VidPro]

your answer is right there
"2 white, 2 blue, 4 yellow and 4 red"

use a resister calculator like this one (of many) http://home.cogeco.ca/~rpaisley4/LEDcalc.html

Supply Voltage | Voltage Drop Across LED | Desired LED Current

put in about 9.2 volts into the supply voltage
for 2 Whites put in about 6.2 (2x3.1v led) in the voltage drop
in desired current put in about 20ma
then
tear off the resisters you have and reconfigure it for 2 leds in series.
- --{~~}---@-----@----- +

do the same thing for the others, using
3x red with a 6.6v drop across the 3 red leds
and so on. 3x yellow, 2x blue

configure each series set, then test them seperate,
then combine them in parellel connecting what you tested all at once.

(-) |--{~~}-------------| (+)
(-) |--{~~}-----------| (+)
(-) |--{~~}--:naughty:--:naughty:--:naughty:-----| (+)
(-) |--{~~}---:scowl:-----:scowl:-----| (+)


drive current at PEAK of the 9V should be a max of 20ma, if you expect a wee 9v to survive longer and not droop down with such a quantity of leds, you will go for a 10ma MAX, then the alkaline can handle it as it droops.
use the calculator with a battery graph or knowing about how the battery will reduce in voltage under the load. that way you can calculate about how it will operate at 9v AND at 8.4 when its getting weak and going down.

always test with a bit more resistance at first, or use a variable resistor , which you can get locally at any ol radio shack.

with the leds in series you wont waste so much power on something that doesnt put out light, with resisters you will control the current a bit, and with the right quanity of series, and the right resistance, this can sorta work.

 

[C21]

Resistor on neg??

I don't really like the series way because that take up too much of wires, I want to use less wire as much as possible because i will be making more.

I have scanned off the LED's specification.

Also i have make other diagram, What do you think?

 

[Mr Happy]

Resistor on neg??

It doesn't matter where the resistor goes, as long as it is in the circuit.

I don't really like the series way because that take up too much of wires, I want to use less wire as much as possible because i will be making more.

This is not logical. Series wiring always uses less total wire than parallel wiring.

I have scanned off the LED's specification.

<snip>

Also i have make other diagram,

<snip>

I'm curious -- why did you put the white and blue LEDs in series when they have the highest Vf, and the yellow by itself when it has a very low Vf?

What do you think?

I can't escape thinking you are a bit out of your depth here. I think you should start with something a bit simpler and learn a little more about the basics of electrical circuits and LEDs before you get so ambitious.

 

[TorchBoy]

Re: 10 LEDs for 9V

(-) |--{~~}-------------| (+)
(-) |--{~~}-----------| (+)
(-) |--{~~}--:naughty:--:naughty:--:naughty:-----| (+)
(-) |--{~~}---:scowl:-----:scowl:-----| (+)

Awesome use of smilies! :twothumbs

I don't really like the series way because that take up too much of wires, I want to use less wire as much as possible because i will be making more.

That doesn't make much sense. You know, at present you're struggling (to understand how) to make one.

I can't escape thinking you are a bit out of your depth here. I think you should start with something a bit simpler and learn a little more about the basics of electrical circuits and LEDs before you get so ambitious.

+1. You don't seem to understand the difference between connecting the LEDs in parallel and series, or what you are asking the battery to do if you put them all in parallel, as chimo explained in post 7.

Is there some physical arrangement of the LEDs that you're trying to get?

 

[C21]

Do anybody have any idea what the ohms and watt of this resistor? I am thinking maybe 470 ohms 1/4 watt??

Also does the thick of wire matters for LEDS?

Color band is yellow, purple, black, black, brown. or backward.

 

[2xTrinity]

It looks like you have all your LEDs in parallel. That means, your are wasting about 2/3rds of the power, since most of the LEDs have a Vf of about 3V, and you are dropping the difference between that and your power supply (9V - 3V = 6v) in your resistors.

If you want all of these to be lit simultaneously, I'd suggest grouping them up and wiring them in series. For example, grouping a two reds and a white together would give you a voltage of around 7V, then you only need to drop 2V with resistors, so you'll save on power draw.

If these are LEDs going to be switched on and off separately, to use as say indicators, they will need to be in parallel. In that case, you're better off using 3AAAs in series to power these, rather than a 9V battery. In that case, you won't have to waste as much with resistance, and you'll hvae a lot more capacity.

 

[TorchBoy]

Do anybody have any idea what the ohms and watt of this resistor? I am thinking maybe 470 ohms 1/4 watt??

Also does the thick of wire matters for LEDS?

Color band is yellow, purple, black, black, brown. or backward.

Looks right. Resistor Colour Code. The last brown stripe means 1% tolerance. I normally double check with a multimeter in case I've read the colour wrong with whatever light I'm working under.

If you're talking about current ~20mA and less than a metre then just about any wire will do. Even light duty hookup wire (10 strands x 0.12mm, or 27 gauge) has only 0.17 ohm per metre. Here's a table.