# LED Electrical Flow

Discussion in 'General Electronics Chat' started by Exodus00FF, Apr 5, 2009.

1. ### Exodus00FF Thread Starter New Member

Mar 19, 2009
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I have some questions about Overdriving, I am still not feeling to confident about building circuits and I feeling that everything I'm doing is just lucky guessing. I've attached a gif of a possible LED lighting schematic.

I used an LED calculator to help generate this. I entered the following specifications about the LEDs.

Source Voltage: 34v
Forward Voltage: 3.4 Typical
Forward Current: 100ma
Total Number of LEDS: 50

When these led calculators build these diagrams aren't they technically overdriving them? I really do not understand how you overdrive an LED.

From how I see it, the first LEDs have 34 Volts running through them. The Next set have a 3.4 voltage drop? But since the first set in the series are 34v passing through aren't they over driven?

If they are not overdriven, what would I need to change in my diagram to make my LEDs overdriven? I really don't get it, and I would like to at least know how to avoid over driving LEDS. My assumption about electrical components to this point has been that you can run as many amps through something and they only draw as many amps as are needed, but overdriving seems to basically through this Idea that I have out of the window.

In my diagram since I am using an LM317 (1 Amp Current Regulator) can avoid the use of resistors?

On the LM317 or LM317 what is the Adjustment line for?

On other ICs what is a vRef for?

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2. ### SgtWookie Expert

Jul 17, 2007
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It won't work.
Was it on a website, or a downloaded tool? In either case, what is the URL to the site?

OK, and then it built you a schematic with 5 strings of 10 LEDs with an LM317 that had no connection to the ADJ pin. Just an LM317 by itself has a drop of about 1.7v from IN to OUT.
So, 10 x 3.4v = 34v, +1.7v = 35.7v. The generated schematic requires 35.7v from a 34v source voltage, but the regulator won't regulate because there are no provisions for regulation.

Well, if the duty cycle (ON vs OFF time) is low and at fairly high frequency (say, 10kHz) you can often use higher current than is specified for the typical average specified. The details should be in the datasheet for the particular LED in question.

At their rated forward current, the LEDs will have a nominal 3.4v drop across them. Since you have 10 LEDs in series per string, each drops 3.4v across itself; as a result the entire 34v supply is divided up between the LEDs in the string. However, you either need to use a current limiting resistor, or an active current regulator to ensure that the LEDs don't get too much current.

An LED's Vf will decrease as the LED warms up. The hotter it gets, the lower Vf gets, which causes more current to flow, which gets the LED hotter, etc. This is called "thermal runaway", and can quickly lead to your LED getting burned up. Having fixed resistors or active current regulators in series with the LED string prevents this from happening.

Tungsten filament light bulbs have a positive temperature coefficient; the hotter the filament gets, the more resistance it has. Because of this, they tolerate fluctuations quite well.

Resistors have a rather neutral temperature coefficient; they will have about the same resistance from cold temps right up until they get overheated.

Semiconductors like LEDs, diodes, and transistors generally have a negative temperature coefficient; the hotter they get, the lower Vf for a given current they have. This can quickly lead to thermal runaway. MOSFETs are an exception; they have a positive temperature coefficent.

They're actually 1.5A voltage regulators, which can be wired as current regulators. When used as a current regulator, they will drop a minimum of 3v across themselves.

The source voltage is connected to IN, the load is connected to the ADJ terminal, and a resistor (R1) is connected from the OUT terminal to the ADJ terminal.
R1 = Vref / DesiredCurrent
where:
Vref = the individual regulators' voltage between OUT and ADJ; nominally 1.25, but may be anywhere from 1.2v to 1.3v.
DesiredCurrent is >= 10mA and <=1.5A.

The LM317 attempts to keep the voltage difference between OUT and ADJ to a nominal 1.25v by sourcing current from the OUT terminal.

That is the regulator's reference voltage.

http://www.national.com
and find the page for the LM117/LM317 regulator. Download the Datasheet and Application Notes.

3. ### Exodus00FF Thread Starter New Member

Mar 19, 2009
9
0

I think I neglected to mention that I got the original Schematic from an online LED Calculator, BUT I removed to resistor part on the end which was set for 1/4w 1ohm resistors. Then added the LM317 at the begining, because I thought I would be able to limit the current that way.

ok so basically the ADJ pin samples the out line, and what ever resistance you apply to the adjustment pin FROM the VO is what VO will be?

So if you have .8 Amps running through the Adj pin (which you do by applying resistance to VO and supply it back to the Adj pin,) then VO will be .8 amps? Is that correct?

This part I actually do understand... I think. This is refering to Pulse Width Modulation, and the Duty Time is refering to the % of time the circuit is On for a Hertz (Cycle)? Please correct me if I am wrong.

Now since the circuit provided is parallel, that means all the individual Parallels have the same voltage, Correct?
I mean to say the first line and the last line both have 34 Volts available to them?

But each Led, uses up so much Voltage and Current. When ever electricity flows through one of the LEDs I will loose 3.4 volts and .1 Amps in that Series?

How can I know what is an acceptable amount of Current to allow to pass through the series?

I don't see anything on the specifications for the LEDs that I am referencing, that lists Maximum allowable Current (or something similar)

The Standard Conditions for this LED are written that 100mA is the typical Current used, but as I view the Maximum Ratings for this particular LED, it shows that the Max Forward Current is 100 mA. Does this mean that in a Series I must have this LED alone? I.E. Nothing after the first one in the parallel?
http://cgi.ebay.com/50p-5mm-0-5W-Mu...h=item370183950178&_trksid=p4634.c0.m14.l1262

It seems that one of my Goals is to use heat sinks and PWM to keep the LEDs as cool as possible. If I don't then they will heat up, and draw even more current? Is that correct?

I am confused on the second part. You mention that as the LED gets hotter it Lowers the Forward Voltage. Wouldn't lowering voltage cause less energy (Current) to pass through the LED? I'm probably imagining this incorrectly.

Am I correct in assuming the only way to overdrive an LED is to allow it to get hot? If it gets hot then it begins to draw too much energy, and then burns out quicker?

Your correct, I must have been looking at one of the Other related models, I know I was at one point looking at LM317L's which are 100mA regulators, and must've gotten confused somewhere down the line .

So from this to correct the circuits, I need to properly apply energy to the adj pin. Then and also apply resistance to drop the current from 1.5A to 1A? OR can I just apply resistance to the Adj Pin and have the 317 do the current regulation to 1A?

Is that similar to the LM317's adjustment pin, except where the adjustment pin regulates current it regulates voltage?

Hey Sarge, Thanks for your help I think you might actually be getting through my thick skull

4. ### Exodus00FF Thread Starter New Member

Mar 19, 2009
9
0
Can you please at least keep it relevant to my original topic.

5. ### SgtWookie Expert

Jul 17, 2007
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Yes, you missed that part.

No. I'm afraid that I don't have time to explain it this morning.

Please go to National Semiconductor's site, and download the datasheet for the LM117/LM317. It's a large datasheet, and it has lots of application schematics.

Yes.

Yes.

Power is dissipated. Voltage is dropped. Current flows.

You're not losing it. Voltage is being dropped across the LEDs. Current flows through them.

The specs on the website indicate 100mA @ 3.2v typical. However, I think that the specs are inflated, and they will have a short life if they are run that hard. I would stick to around 75mA.

Someone got carried away with 100mA when they were typing up the specifications. Note that those aren't 0.5W LEDs; they're 0.32W if you multiply 100mA x 3.2v.

As I said above, I think the specifications have been inflated.

I'd suggest running them at 70mA or 80mA instead of 100mA.

Using large pads and wide traces on the circuit board will help a great deal to dissipate heat.

If the voltage across an LED is held constant and the Vf of the LED decreases, the current through the LED increases. That is why current regulation is used.

To much current through an LED will either burn it out right away, or cause it to become dim sooner than it shoud.

No. Current through a series circuit is the same all the way through.
Voltage across parallel circuits is the same all the way across.

You really need to have a look at the datasheet, and do some experiments. I'm afraid that I just don't have time to explain it this morning.

6. ### Exodus00FF Thread Starter New Member

Mar 19, 2009
9
0
Thanks Sarge,

After reading your post, I realized that I don't know half as much as I though about electricity in general, and I'm trying to go back to basics right now.

With my understanding of electricityThanks Sarge,

After reading your post, I realized that I don't know half as much as I though about electricity in general, and I'm trying to go back to basics right now.

Particularly Voltage. The way I envision Voltage, it is like water pressure in a hose. If you drop the pressure less water (amps) comes out of the hose, but this seems to be the opposite case with electronics. As you said the voltage drops then the current rises. I need to wrap my head around that concept.

7. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Voltage = pressure (EMF, or electromotive force). It IS analogous to water pressure.
Current = flow (Amperes; electron flow). It IS analogous to water flow.

A bunch of LEDs in series is sort of like a fish ladder (aka fish steps, fish steps) on a dam. Instead of the large height difference of the dam, the fish ladder/steps/passes divides up the height into a bunch of small pools that the fish can jump high enough to get to the next higher pool.

What water goes into the top of the fish ladder/steps/passes flows out the bottom. Each "step" of the fish ladder has the same amount of water (current) flowing through it as any other step does.

If the water pressures from the water surface to the level of the water in the next lower pool were added up, it would equal the drop in water pressure from the water surface in the dam down to the depth of what the surface was below the dam.

Perhaps I'm not doing such a good job in explaining things here, but a similar phenomenon occurs with the voltage drop across the LEDs and the electrical current through the LEDs.

8. ### Exodus00FF Thread Starter New Member

Mar 19, 2009
9
0

I reviewed the National LM317 Data sheet, and I got some ideas on what I need to do to improve my original schematic.

First I am planning on changing my LEDs

I've found some that list the Forward Current as 20ma and 3.2V Forward Voltage. Peak Forward Current is 100ma.

Forward Current: 20ma
Forward Voltage: 3.2 volts
Max Forward Current: 30ma

So I've made changes to my schematic, based on our previous conversations. According to the data sheet I should expect approximately 2 volts to be dropped by the LM317 Regulator. After each Led I will be dropping 3.2 Volts each.

All together I will need to consume 2v + (3.2v*5LEDs) = 18 Volts

To ensure that the LM317 works I consulted a calculator. http://www.reuk.co.uk/LM317-Current-Calculator.htm

I would have to have a 62ohm resistor to limit the current to 20mA.

Now as I've gathered so far in this circuit, Current should stay constant, and Voltage should be the only thing that drops. Since Current stays the same throughout the series 20mA should be the perfect setting for these new Leds that I'm using.

Does this sound right? I hope I'm on the right track now...

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9. ### SgtWookie Expert

Jul 17, 2007
22,183
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That's not quite correct.
In voltage regulation mode (Vin to Vout) the voltage drop is approximately 1.7v. However, when used in current regulation mode, you have the Vout-Vin drop plus the Vref drop, which can vary from 1.2v to 1.3v, for a total of at least 3v. Leave an extra volt for "headroom", so from a 34v supply, you will have 30v left for your LEDs.

So then, how many LEDs can you power from the remaining 30v?
LEDs = 30 / Vf(LED) = 30/3.2 = 9.375; round down to an integer value = 9.
9.3.2 = 28.8v.
Let's see how much power the regulator will be dissipating.
Regulator Power(Watts) = (Vsupply - TotalVf(LED)) x Current = (34 - 28.8) x 20mA = 5.2 x 0.02 = 0.104 Watts. This is easily what can be handled by an LM317L, which is a 100mA LM317 regulator in a TO-92 package. I wouldn't recommend this regulator for the Luxeon-type high-current LEDs.

The calculator you used isn't terrible, but it doesn't give the full picture, either.

When calculating the value of the resistor (R1) that goes from Vout to ADJ, you really need to know the Vref of the particular regulator that you are dealing with. Nominally, Vref is 1.25v; however it MAY vary from 1.2v to 1.3v. For best accuracy then, you must determine the Vref of your particular regulator while it is in use.

Determining the Vref of your particular regulator:
1) Connect a 62 Ohm resistor from Vout to the ADJ terminal.
2) Connect a power source of 4v to 6v between Vin and ADJ.
3) Measure the voltage between ADJ and Vout. This is Vref.

Determining R1:
R1 = Vref / DesiredCurrent
For an LM317L, 5mA <= DesiredCurrent <= 100mA
For an LM317T, 10mA <= DesiredCurrent <= 1.5A

Let's say you measured an LM317L's Vref to be 1.27V.
Since you want 20mA current, R1 = 1.27 / 20mA = 1.27 / 0.02 = 63.5 Ohms.

Let's talk about precision of the resistance for a moment. Say you got lazy, and decided to just use a 62 Ohm resistor with that regulator that had a Vref of 1.27v.
CurrentOut = Vref / R1, so 1.27 / 62 = 20.484mA. That's not too bad if the 62 Ohm resistor actually measures 62 Ohms; but remember that standard resistors have a tolerance; gold banded resistors' tolerance is +/-5%. That 5% resistor might measure from 58.9 Ohms to 65.1 Ohms. Run the above calculations with the high and low values to see what the current variation might be.

If you're still planning on using a 34v supply, that's a lot of power to dissipate in the regulator.
Power(Regulator) = (34v-18v) x 20mA = 16 x 0.02 = 320mW
Power(LEDs) = 18v x 20mA = 360mW
Efficiency = Power(LEDs) x100 / TotalPower = 360mW x100 / 680mW = 52.94%. You're wasting nearly half of the power in the regulator as heat.