9 Volt 2 LED Snakeye light

Hello

I do'nt know much about electronics,but, I'm trying to make this
9v alkaline snakeye with 2 infrared LEDS and I was wondering if someone
familiar with LED's can give me a few pointers. This is what I have to work with:

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Spec Sheet No. 1

2pcs LED's 5mm High Output infrared

Electrical Characteristics(25C°)

Radiant Power
output(100ma)]: 16Mw min
forward voltage: 1.2 v
forward current: 100 ma
viewing angle to 1/2 intensity: 45°


Absolute Maximum Ratings(25C)

forward voltage(20Ma): 1.6v
reverse voltage: 5v
forward current: 1.2A
reverse current: 10uA
wavelength: 940nm



Question No. 1:
The information above is telling me that it needs to pull 1.6 volts of tension
on a continuos current of 20 MA. So, for two LED's, how much resistance
should I provide? 47 ohms? Is there a formula to derive this from, given that
we have:

using the Absolute Max. Ratings
a) current needed is 20Ma per LED
b) voltage supplied in volts i.e. 9v
c) voltage required per LED 1.6v

It woulkd be nice, if possible, if someone that knows, could explain
most of the gibberish of both, the Electrical Characteristics as well as
the Absolute Maximum Ratings. I´m going with Absolute Maximum Ratings only,
but, it would be nice to know the rest as well.

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Spec Sheet No. 2

2pcs 47 ohm resistors

4 Band System

1 yellow band
1 violet band
1 blue band
1 gold band

Question No. 2:

I´m assuming the store clerk actually gave me a 47 ohm resistor. But,
for the skae of certainty, could someone check to see if it" is", in fact,
a 47 ohm resistor? How do I add up all those bands to mean 47 ohms?
Sorry for the greenthumb question.
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Spec Sheet No. 3

1pc 9 Volt Alkaline Battery

*Not purchased yet so specs on it are:

N/A

--------------------------------------------------------------------------------------------
Spec Sheet No. 4

1pc 9 Volt Battery snap connector

I.E the kind that snap to the top of a rectangular 9 v battery. with a Pos
and a Neg wre sticking out of it.

------------------------------------------------------------------------------------


Again, sorry for the greenthumb questions and thank you very much
for your input I just don't want to burn these LEDS since they were
hard to come by over here...and expensive too($ 3 U.S Dlls. each!).

Vormat

Well 47 ohm is yellow, violet, black, gold, are you sure of the colour on your resistor? - blue is unlikely!.

You calculate the resistor value with ohms law V = I x R, you can rearrange this as R = V / I.

You connect the two LED's in series, this gives a voltage drop of 3.2V (2 x 1,6V), and connect a resistor in series with them to the battery.

The battery is 9V, the LED's 3.2V, so subtracting them you get 5.8V, that's the voltage across the resistor - and the value needed in the formula.

You then need to decide on the current you want to feed the LED's with, the rating in the spec is the absolute maximum, so it's best to be lower than that. It all really depends what you are trying to do?.

Anyway, for this example we'll assume 100mA as listed in the spec, so this gives us 5.8V / 0.1A, which is 58 ohms - so 47 ohms would feed them too much current.

You are aware that IR LED's give off invisible light I take it?, I don't quite see what you are trying to do?.

Conversion from Millie Amper to Amper

Nigel

Thank you for your reply.

<You are aware that IR LED's give off invisible light I take it?, >

Now I am.

Thank you for pointing it out...Errrr confused Infrared with Ultraviolet, which, is what I wanted. oooof! See in the dark. Yep. that´s what Ill use those infrarerrrrds for!

So that I get my conversions correct.....
Okay, so i want to convert milli amps to amps:

0.001 A = 1 Ma (one thousands of an Ampere)
0.0001 A = 10 Ma (ten thousands of an Ampere)
0.00001 A = 100 Ma (one hundred thousands of an Ampere)

Well, now I'm really confused since you say 100 Ma is 0.1 A. But, either way, is the above conversion table correct?


<Well 47 ohm is yellow, violet, black, gold, are you sure of the colour on your resistor?- blue is unlikely!. >

Hmmmm. Bad flourescent lighting over here. Power co. , you know. Can´t seem to carry a stable load sometimes....either that or a cruddy littel resistor card that not only does'nt explain how to calculate Ohms from the given bands...but, also included the black band as Blue. Go figure.

By the way:

Yellow Band = ?
Violet Band = ?
Black or Blue Band = ?
Gold Band = ?
------------------
Total 47 ohms (Now, that...I know!)

Sorry for the bother and Thank You Again.

Vormat

0.001 A = 1 Ma (one thousands of an Ampere)
0.0001 A = 10 Ma (ten thousands of an Ampere)
0.00001 A = 100 Ma (one hundred thousands of an Ampere)

Click to expand...

Your conversions are backwords it goes more like this.

0.001 A = 1 Ma (one thousands of an Ampere)
0.010 A = 10 Ma (ten thousands of an Ampere)
0.100 A = 100 Ma (one hundred thousands of an Ampere)

Millie Ampere to Ampere Conversion

magickaldan


Thank you for your input.

Okay, this little conversion table is correct, I think:

0.00001(one hundred thousands of an Ampere) is not equal to 100 Ma . Correct ?


0.001 A = (one thousands of an Ampere or one Ma)
0.0001 A = (ten thousands of an Ampere or one tenth of an Ma)...not ten Ma
0.00001 A = (one hundred thousands of an Ampere or one hundreds of an Ma)...not 100 Ma

Only need to work with 0.001 Ampere which is 1 Millie Ampere

The simple math is what´s wrong with my first table...


Convert to Millie Ampers:
100Ma = 1Ma x 100

Convert to Ampers:
0.1A = 0.001A x 100

Okay, now I got:

2Pcs LEDS (White Light ) with absolutely no specs on them...Could´nt find any white lite LEDS over here with specs, unfortunately. These look really cheap. Of the Grab Bag variety.

2Pcs 56 ohm resistors

1 Lime Green Band = ?
1 Light Blue Band = ?
1 Black Band = ?
1 Gold Band = ?
--------------------
Total 56 ohms...so said the store clerk.

1Pc 9 Volt Battery Clip

1Pc 9 volt battery. Not alkaline, though.

thanks again

Vormat

Ok Got it to work

Okay, thanks to everybody.


Shorted one LED trying to light it up, but, I got it to work O.K.

On a 9 volt non Alkaline battery, it will light up some, not too bright though, with as little as 5 Millie Amperes of current.

Here's how:

On the anode leg(+) of the LED, the long leg, strap on a 1000 ohm (IK) resistor then strap that to the pos + or red wire lead coming out of the 9v Battery clip.

With the cathode leg(-) of the LED, the short leg, strap it to the neg or black wire lead coming out of the 9v Battery Clip.

Snap the Batery clip to the 9v Battery itself.

That's all there is to it.


Here are the calculations(LED'S had no specs):

* Where the assumed voltage Drop across one White LED = 3.6v
*Where the given Battery voltage is 9 v.
*Remaining voltage is 5.4 v

I = V/R
I = 9v - 3.6v/ 1000 ohms
I = 0.0054 Amperes

Convert Amperes to Millie Ampers:

Ma = Amperes/ 0.001
Ma = 0.0054 Amperes/ 0.001
Ma = 5.4


Well, I'm not too satisfied with the light output. Can't really do anything with it in the dark. Exept look for a key hole on a dark night. I'm going to try it with more current. These Grab bag LEDS should work OK and not short out with 20-25 Ma of continuos current. Should look a lot brighter too. But, at 5 Ma and a 9 v battery, I'm curious as too how long that battery will last.

So, is there any formula that I can use to try to "guesstimate" how long one LED will light up on 5 Ma of current with a 9v non alkaline Battery?

thanks again


Vormat

Re: Ok Got it to work

Vormat said:

Well, I'm not too satisfied with the light output. Can't really do anything with it in the dark. Exept look for a key hole on a dark night. I'm going to try it with more current. These Grab bag LEDS should work OK and not short out with 20-25 Ma of continuos current. Should look a lot brighter too. But, at 5 Ma and a 9 v battery, I'm curious as too how long that battery will last.

Click to expand...

The only way to know battery life is to measure it, calculations are often massively wrong - not even worth bothering with.

The light output is going to be low, these are LED's!, not torch bulbs. You can obviously increase the light output by increasing the current, but at the expense of decreasing battery life - but it's never going to be great!.

The biggest problem with battery life is your high battery voltage, you're wasting most of the battery power in the resistor (about two thirds), so it's an extremely poor design.

9 volt LED

Nigel


<The biggest problem with battery life is your high battery voltage, you're wasting most of the battery power in the resistor (about two thirds), so it's an extremely poor design.>

Yes. Running one LED on 5 Ma of current from a 9 volt battery is a ppor design.

I tried it with aprox. 21 Ma and 3 pcs. 82 ohm resistors hooked up in series. The light output is somewhat better and can actually be used for reading something "very up close" on a dark night. Of course, it is an exttremely poor set up with no reflector and some of the light is wasted because it comes straight out the back of the LED.

But, it in "no way" compares to a Mag incandescent , krypton clone torch I compared it to. At only $5 U.S, the Mag clone is "very bright" and casts "one heck" of a beam on just 2 AA batteries !

Maybe one of those 12000 mcd "Millie Candela" pure white LEDS will do better.....if I can actually find one around these parts, which, I doubt.


Just two simple greenthumb questions, if possible, is current also divided between two Leds like voltage is, in a series circuit ? For example, if I specify 20 milliAmps for two LEDS, then 10 MA will go to each?


thanks for your input.

Vormat

Re: 9 volt LED

Vormat said:

But, it in "no way" compares to a Mag incandescent , krypton clone torch I compared it to. At only $5 U.S, the Mag clone is "very bright" and casts "one heck" of a beam on just 2 AA batteries !

Click to expand...

LED's don't compare with bulbs, a bulb is a GREAT deal brighter than an LED - but an LED has a massively longer life span.

Just two simple greenthumb questions, if possible, is current also divided between two Leds like voltage is, in a series circuit ? For example, if I specify 20 milliAmps for two LEDS, then 10 MA will go to each?

Click to expand...

No, in series each LED gets the same current - but you get double the voltage drop. This is what you need in your design, more voltage across LED's and less across the resistor.

9 volt LED

Nigel


Thank you again for your input.

<LED's don't compare with bulbs, a bulb is a GREAT deal brighter than an LED - but an LED has a massively longer life span. >

True. Very true.


<No, in series each LED gets the same current - but you get double the voltage drop. This is what you need in your design, more voltage across LED's and less across the resistor.>

O.K, Thanks.
In parallel, if 20 Ma are specified for two LEDS then 20 MA will go to each as well? Although I read somewhere that in parallel, because the LEDS can come from different batches, they will have different light intensities. Not too sure if this is true, though.


Vormat

Re: 9 volt LED

Vormat said:

O.K, Thanks.
In parallel, if 20 Ma are specified for two LEDS then 20 MA will go to each as well? Although I read somewhere that in parallel, because the LEDS can come from different batches, they will have different light intensities. Not too sure if this is true, though.
Vormat

Click to expand...

In parallel, 2 led's will have a voltage drop equal to 1 led, but the current will be the same as for 2.
In series, 2 led's will have a voltage drop equal to 2 led's, but the current will be the same as for 1.

About the intensity, I never saw 2 led's of the same type that had a visually diffirent intensity. So there pretty close, to close to notice any diffirence.

Re: 9 volt LED

Vormat said:

In parallel, if 20 Ma are specified for two LEDS then 20 MA will go to each as well? Although I read somewhere that in parallel, because the LEDS can come from different batches, they will have different light intensities. Not too sure if this is true, though.

Click to expand...

No, in parallel each LED would take part of the current, and the total would be 20mA.

It's extremely bad practice to connect LED's in parallel though, one is very likely to take most of the current, to the extent that only one might light up. You should provide seperate feed resistors for each LED if you want to do this - so have two resistors, each double the previous value, one feeding each LED.

9 volt LED

Exo

Thank you for your reply. Although, I'm not too sure I usnderstood what you posted, I'll put it in my own words in order to nail down this parallel and series stuff.


Nigel

Thanks again for your input. So, just to get this straight:

In sereis, each user in a circuit "gets" what it is asking for or is correctly specified for it, provided it's availabe. This holds true for both tension(voltage) as well as current(amperage). So, consumption of voltage and curent is multiplied by the number of users

Thus, 2 leds hooked up in series in a circuit, will require twice the consumption of one led for both voltage and amperage.

In parallel, each user in a circuit "shares" with other users in the circuit ,what is correctly specified for the entire circuit, provided it's availabe. This holds true for both tension(voltage) as well as current(amperage). So, consumption of voltage(tension) and current(amperage) is "divided" by the number of users.

Thus, 2 leds hooked up in parallel in a circuit, will require half the consumption of one led for both voltage and amperage. Although it is bad practice in low voltage circuits to put anything(as is) in parallel that does'nt consume but only a very small fraction of the specified or available supply of voltage and amperage.

Aghh! Hope I made sense there.

Thanks again!

Vormat

Re: 9 volt LED

Vormat said:

In sereis, each user in a circuit "gets" what it is asking for or is correctly specified for it, provided it's availabe. This holds true for both tension(voltage) as well as current(amperage). So, consumption of voltage and curent is multiplied by the number of users

Thus, 2 leds hooked up in series in a circuit, will require twice the consumption of one led for both voltage and amperage.

Click to expand...

No, In a series circuit the current or 'amperage' is limited by the part with the highest resistance. So for 2 led's in series you'll require the voltage for both leds (Vf-led1 + Vf-led2). but the current should only be that of 1 led - You have to add the resistor to limit the current to the led, also in series.

Vormat said:

In parallel, each user in a circuit "shares" with other users in the circuit ,what is correctly specified for the entire circuit, provided it's availabe. This holds true for both tension(voltage) as well as current(amperage). So, consumption of voltage(tension) and current(amperage) is "divided" by the number of users.

Thus, 2 leds hooked up in parallel in a circuit, will require half the consumption of one led for both voltage and amperage. Although it is bad practice in low voltage circuits to put anything(as is) in parallel that does'nt consume but only a very small fraction of the specified or available supply of voltage and amperage.

Click to expand...

Again no, Connecting 2 leds in parallel to a power supply would be exacty the same then connecting 1 led to it. The 2 chains are connected to the power supply independent of each other. Every led should have it's own resistor.

What Nigel intended by saying "it is bad practice to connect leds in parallel" is that connecting 2 leds in parallel, with only one shared resistor is bad.

9 volt led

Exo and Nigel

Thanks for your input. But, I understand what Nigel is saying:

Nigel said:

<No, in parallel each LED would take part of the current, and the total would be 20mA.>

Thus, if 20 milliamps are specified for 2 leds in parallel then each takes 10 Ma. Thus, specified current(amperage) is divided between the two users, which, in this case are two leds.
(Not taking into account the extra precautions needed, like different ohmage resistors on the "in" side of each led)

But, you say:

<Again no, Connecting 2 leds in parallel to a power supply would be exacty the same then connecting 1 led to it. >

Are you saying the same thing as Nigel, but, in different words?

As for how I tried to explain it, bad choice of words on my part.......So,

In either case, I thank both of you for your input, I'm gonna look up an electronics tutorial on the WEB. That should get me started.

Vormat

9 volt led

Never mind....sorted it out:

SERIES CIRCUIT
(Rule A) Components in a series circuit share the same curren(Amperage)t I Total = I 1 = I 2 = . . . I n
(Rule B) Total resistance(Ohmage) in a series circuit is equal to the sum of the individual resistances : R Total = R 1 + R 2 + . . R n
(Rule C )Total tension (Voltage) in a series circuit is equal to the sum of the individual voltage drops : E Total = E 1 + E 2 + . . E n


PARALLEL CIRCUIT
(Rule A) Components in a parallel circuit share the same tension ( Voltage): E Total = E 1 = E 2 = . . E n
(Rule B) Total resistance(Ohmage) in a parallel circuit is less than any of the individual resistances: RTotal = 1 / (1/R1 + 1/R2 + . .1/Rn)
(Rule C ) Total current (Amperage) in a parallel circuit is equal to the sum of the individual branch currents: I Total = I 1 + I 2 + . .I n.