Powering LED's

Hi, Your LEDs are speced 3.0 to 3.8 Fwd. Wiring them in parallel means they would all have to drop the same voltage. There is no guarantee that this will be the case, in fact they most likely wouldn't be. So, no, do not wire them in parallel, and for the same reason as stated you would still be better off with a resistor otherwise you are relying solely on internal resistances. Since you do not have a lot of 'extra' voltage a small value (39 ohms) could be used on each LED then you could wire the combo's in parallel.

To know more accurately than you have calculated the expected useful time of the lighting system, you would need to know two things.

1. The current at which the LEDs produce a light that is not acceptable to you.
2. The discharge curve of your power cells at the current that you are using. BTW: You seem to imply that you believe that the 3.0 to 3.8 volts are operating ranges. This is not the case. This is the range of the expected voltage drop across the LED when it is fully conducting. Good Luck, Tom

>

You may get away with it, but fair chance you cook the LEDs. For one thing, LEDs get more conductive as they get hotter, and you may get "thermal runaway".

I would put a resistor in series with each LED and put the LED-resistor combos in parallel with each other. As for resistor value - I would guess

22 ohms. You can try different ones - measure the voltage across the resistor and divide that by resistance to determine the LED current. That will probably indicate current better than a milliammeter, which on a suitable current range will probably add enough resistance to change the current significantly.

- Don Klipstein ( snipped-for-privacy@misty.com)

Thanks guys, so now I have a few more questions:

To clarify, is this what Don is reccomending:

[+]LED,resistor[-] [+]LED,resistor[-] [+]LED,resistor[-] ...

and I'm still a bit confused about what resistor(s) to choose. I was reading this page on how to power LEDs:

and it leads me to believe that resistors are specified for both resistance (ohms) and wattage (watts). Since now I will be using resistors, I was thinking of going with 4 AA NiMH batteries to increase runtime. So this leads me to the question of which resistors to choose. Is the 39ohms, or the 22ohms resistor better suited to this task? and what wattage? and just so I can better understand this, is it better to pick a resistor rated to handle more ohms and more wattage? or less?

I noticed a section on that website I linked to above about combining both the series and parallel methods,

and was thinking of maybe going that route. Is that a good idea, and if so, how would that affect my resitor selections.

If I should go with the hybrid serial-parallel route, won't I need to have a battery that supplies more volage? 4 NiMH AA's would be 4.8V... not even enough for 2 LED's. (I plan to use all 20 LED's) I was thinking of perhaps buying a pair of rechargable 9V batteries, but 18V is only enough for about 5 LED's. I do not want to use too many batteries as it would weight-down my RC car... and since I have a large supply of rechargable AA's I would prefer to use them. Would I need to get another device to increase the voltage... what is it called, a transformer?

Sorry, I'll bet I'm really confusing stuff up right about now. So back to basics... Is there a nice webpage or book that describes the basic electronics vocabulary, and basic devices (diodes, resistors, transformers, capacitors, transistors, etc...) I think this is my main problem, because the webpage I linked to above did a great job of defining how the equations worked, it's just that I don't quite know what voltage, wattage, current, power, etc. are, and I'm going off of guesses and past expierences.

Thanks again,

--Farrell F.

Hi, Try 39 Ohms and if the LED is not bright enough try a lower value.

1/4 watt resistors are readily available and are more than sufficient for your case.

Let's educate you a little so that you may better understand the suggestions that are given to you.

The LEDs are diodes not light bulbs. The junction of the diode will emit light when current passes through it. The voltage across the junction will rise as current increases up to a point; at that point the voltage will remain fairly constant. This is the forward voltage the spec talks about. If you supply more voltage than that the junction will overheat and die. The resister in series will allow the extra voltage to "have somewhere to be". So the resister that you need is determined by the extra voltage that you have and the current that you need, R=V/I. You mentioned a 3.8 volts supply and LEDs that could max at 3.0 volts.Your resister would be .8V/.020A= 40 Ohms.

39 Ohms is the nearest standard value. The power dissipated would be P=VI. .8V*.020 A. = .016 Watts. For physical convenience I would use a 1/4 watt resister. If you use more voltage in your supply you would need to replace the .8v with the excess that you have. Forget about a transformer right now. The theory doesn't apply to what you are doing. Good Luck, Tom >

Thanks Tom!

So since you reccomend that I should not go with a transformer, I guess my assumption of how to wire my LED's in my previous post is correct? e.g.

[+]LED,resistor[-] [+]LED,resistor[-] [+]LED,resistor[-] ...

No? and does it matter which "side" of the LED the resistor will be attached to?

And now I think I will go with a 4 pack of 1.2V NiMH AA's as my power source, so will you please check my math and tell me if I am messing things up:

• 3.0V is the min my LED's are spec'd for, so would it be bad if I aim to provide them with 3.6V, just a little below their 3.8V max?

• 3.6V per LED, 4.8V power source leaves me with 1.2V to resist.

1.2V/.02A=60 Ohms

1.2V*.02A=.024 Watts

And since I am not familiar with standard resistor specifications, I'm guess I should still go with a 1/4th Watt resistor, for approx. 60 Ohms... and should I round up or down?

I looked online at Radioshack.com and found 100 Ohm 1/4 Watt resistors... do they sound about right?

Thanks,

--Farrell F.

Hi, Your math is correct. You can get a standard resistor in 10% of 56 Ohms and in 5% of 62 Ohms. 1/4 watt is fine. It doesn't matter which side the resister goes on but it does matter which side gets the plus and minus. Just some additional information: You indicate that you may still think the

3.0 to 3.8 is a min/max supply rating for your devices. It is a range of expected forward drop voltage. The forward drop can be anywhere in that range for any given device. Measure them in operation. You may need to calculate a different resistor for each LED if you want uniform brightness. Regards, Tom >

No, Yes. ;-) I've always put the resistor in the + lead, just 'cause it feels better to have the LED cathode closer to ground.

No, they're in series, and every point in a series circuit passes the same current.

NO!

LEDs are rated by _current_ - the voltage is just what happens to appear across a given LED when the rated current is passed through it. In this case, you'd calculate the resistor value at the low end of that range, for a current of, say 20 mA. So 4.8V - 3.0V is 1.8V, divided by .02A is 90 ohms. 91 is the closest standard value. If your LED happens to be one of the ones that drops 3.8V at the rated 20 mA current, then there will be 1.0V across your 90 ohm resistor, which will give you 11 mA, but (A) The only LEDs I've done the experiment were so close to as bright at 11 mA as they were at 20 mA that it was practically the same, and (B) you're on the safe side for any LED within that range.

As Tom Biasi has suggested, it would probably be a good idea to set up a circuit that provides a constant 20 mA, and measure the voltage drop across a batch of LEDs. It could be very educational.

This is all assuming that your LEDs are rated for 20 mA, of course. If yours have a different spec, just use that value in the formulas.

If you use a 12V supply, then for 20 mA with an LED that drops 3V, you'd use a (12 - 3)/.02 = 450 ohm resistor, which if your LED drops 3.8V, would still give you 18.2 mA, which is much closer to a constant current supply for your tests.

Hope This Helps! Rich

Thanks yet again :)

A few more questions, hope you don't mind...

By 5% and 10% I assume you mean the carbon film percentage? What does this mean, and how does it effect the results?

And what exactly is "drop" voltage? The amount used by the LED?

Thanks,

--Farrell F.

The 5% or 10% is the tolerance on the resistance - a 5% resistor may be up to 5% above or below its marked value (but I think most metal or crabon film resistors will be within 1% of the marked value, regardless of the indicated tolerance)

Yes.

--
Peter Bennett VE7CEI 
email: peterbb4 (at) interchange.ubc.ca        
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html 
Newsgroup new user info: http://vancouver-webpages.com/nnq

Hi, "Drop" is a term meaning the voltage across a device caused by the current going through it times the resistance. If you have 1 volt across the resister in this series circuit the LED will "see" 1 volt less than the supply, hence the term "drop". The percentage is how close to the specified value the resister really is. Ex.: A 100 ohm resister at 5% may really be 95 to 105 ohms. A 100 ohm resister at 10% may be between 90 and 110 ohms. (5% of 100 is 5 ohms 100 plus or minus 5 ohms.) Don't concern yourself with this for this application. Regards, Tom

Well, the voltage across the LED is the voltage "dropped" by the LED; the resistor will have its own voltage drop. It's not like volts get dropped off a cliff, or down a well, or something - it's exactly analogous to "pressure drop", which is used in pluming and HVAC all of the time. It's the differernce in pressure between one end of a component and the other. Volts don't get "used" other than to provide the pressure that pushes the current through the circuit.

Think of a tall water pipe, with ports with pressure gauges arranged along its height. At the top, there's zero water pressure, and at the bottom, there's some amount of pressure just caused by the weight of the column of water. This is how altimeters work - at altitude, there's less air on top pressing down on itself. (and on you, and your altimeter).

Now, if you have a 33ft. column of water, the water pressure at the bottom will be 15 PSI. The water pressure half-way up the column (16 1/2 feet) will be half that, or 7.5 PSI. So there's a pressure drop of

7.5 PSI from the top of the column to the middle, and another 7.5 PSI pressure drop from the middle to the bottom.

That's all voltage drop is - a difference in pressure, which is called "electromotive force" - and as you remmeber from high school, force is just a push - it's there whether the object is moving or not.

Hope This Helps! Rich