# Powering LED's

Discussion in 'Electronic Basics' started by [email protected], May 15, 2005.

1. ### Guest

I'm working on making a very bright set of "headlights" for my
remote-controlled car. I decided to go with white LED's because my RC
car is power'd by a nitro engine that produces some good vibration
which would probably make incandesent lights not last very long.

I bought 20 white LED's which are spec'd for a forward voltage of 3.0V
min, and 3.8V max. I read that I should wire them in series, and not
wire them in parallel, but I'm planning to power them with 3 NiMH AA's
(3x1.2v = 3.8V). Since that would be the same as the max forward
current, can I get away with parallel wiring, and not using resistors?

I am new to electronics, and from what I have read online, it seems
that resistors are used when your power source produces more voltage
than your electronics can handle. So if I am correct, 3 NiMH AA cells
should produce a tolerable voltage at all times (perhaps too little
when they get drained). So am I correct in assuming that I can wire all
the LED's in parallel, without resistors, to the 3 NiMH AA's?

For more information about the LED's I bought, the eBay page for them
has tables of info on them:

http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=66952&item=7514693502&rd=1

Lastly, I'm curious about how long my 3 NiMH AA's will provide good
power to the LED's. It looks like my LED's will be drawing 20mA each,
and I plan to use all 20 LED's, so that will be 400mA total. My NiHM
AA's are 2400mAh, so my first assumption is that they would last about
6 hours. But then I realized that would be about 6 hours to fully drain
my batteries. So this leads me to the question of when my batteries
will start to get to the minimum forward volage for my LED's: 3.0V. I
have no idea how to calculate that... any tips?

Thank you,
--Farrell F.

2. ### Tom BiasiGuest

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

3. ### Don KlipsteinGuest

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 ()

4. ### Guest

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

http://wolfstone.halloweenhost.com/TechBase/litlpo_PoweringLEDs.html

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,
http://wolfstone.halloweenhost.com/...html#WiringLargeNumbersOfLEDsInSeriesParallel
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.

5. ### Tom BiasiGuest

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

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

6. ### Guest

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.

7. ### Tom BiasiGuest

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

8. ### Rich GriseGuest

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

9. ### Guest

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.

10. ### Tom BiasiGuest

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

11. ### Peter BennettGuest

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)

12. ### Rich GriseGuest

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

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