Let me see if I have this figured out correctly.

I have tied two AA batteries in a series and they read 3.03V. When powering an LED I'm scavenging from an old Marantz amp, I measure at the legs of the LED and get 2.93V. So, if I understand correctly, I have just determined that this LED is a .1V LED by measuring the "drop" across the LED.

Please, one of you more confident in your knowledge folks let me know if I am correct or if I need to be taking anything else into consideration, etc.

Man I love this stuff

Your understanding is a little backwards.

Basically, Voltage = Amperage x resistance. The Amperage will remain constant throughout the circut, however, the voltage will drop with a greater resistance, and the LED light serves as a resistor, with a resistance of .034 ohms.

That's basic high-school physics right there.

http://www.tpub.com/neets/

Lots of good reading for your entertainment.

Quote:



Thanks, I'm always looking for more places to edumacate myself. This will keep me busy the rest of my shift!

...oh, and thak you too Ebo. That highschool level lesson will be useful as well.

Not quite, my friend

Your voltmeter ALWAYS measures the voltage difference between two points; also known as drop. So the voltage drop between the batteries' terminals, with nothing connected, is 3.03V. Similarly, the voltage drop across the LED is 2.9V...

Now, here's the interesting bit. Since the LED is nothing but a fancy diode, the fact that current is passing through it means it's shorting the battery terminals! (that is, it's allowing an unlimited amount of current through, but reducing the voltage by X, the diode drop.) Thing is, like any diode, it can only withstand so much current passing through it... I've actually seen LEDs destroy themselves because of this, when connected to a power supply that will supply the current. To prevent this from happening, then, you put a resistor in series with the LED in order to limit the amount of current that can flow - the more current, the brighter the LED, but also the hotter the junction gets, and with enough current you literally fry the LED. LED's are usually rated for 20mA or so - that means you want to prevent more than about 20mA of current flowing through them. So, either the LED you're scavenging has a resistor built in to it (possible), or your batteries can't supply enough current to cause damage. (doubtful)

Quote:



Now that is interesting. If I understoof that correctly, a LED without built-in resistance should read nearly exactly the same value as when measured straight from battery source?

One thing I did notice about this pcb (seperate board for a series of LEDS) is that there were only a few resistors on it, and not all LEDS were traced through them. Perhaps some of the LEDS were and some weren't built with resistance. There are two distinct different builds visually from this board.

Quote:



Another very good source of information is This TI document ....err book.

Note: Filesize warning on this puppy.

Quote:



Between extreme b/w and tabbed browsing care of Mozilla I'm good.. but thanks for the warning.

..ok, I know the warning is for all out there, but I sure do love tabbed browsing.

Quote:



I'm guessing the LED is insanely bright at this point. I did this with a 9v battery and a low current LED...within a couple seconds, the LED was toast and my fingers were nearly burnt.

Let's see if I can help you out; we just went over this in physics a few days ago. Batteries, by nature, have a certain amount of internal resistance. It's usually small, a fraction of an ohm, but it drops the supplied voltage depending on the current draw. The equation:

V = Emf - I*r,

where V is the actual supplied voltage, Emf is the rated battery voltage, I is the drawn current, and r is the internal resistance.

Let's say the internal resistance was a tenth of an ohm. 2.93=3.03-I*0.1; I=1A, which should should have fried your LED.

So, what does this mean? The voltage drop across the battery (like you measured) is a function of how much current you're drawing and the internal resistance of the battery. And treat an LED like a short circuit when working with the schematic.

Another useful equation, this one used to determine the resistor value to place in series with the LED:

R = (V - V[f])/I,

where R is the value of the resistor, V is the source voltage, V[f] is the LED's forward voltage, and I is the desired current through the LED. Note that most LEDs can light visibly on around one milliamp, and brightly on 5mA. Keep battery life in mind; a portable amp usually draws 8-20mA, and you don't want to add significantly to that.

If you really want to find the forward voltage, use small batteries and a railsplitter to create lower voltage, until the LED doesn't light anymore. Don't forget the resistor

edit:clarity

Quote:



Well, not quite... since you have to understand that an LED when wired up correctly without a resistor is seen as a dead short, more or less! (an odd kind of shrot, of course, because it reduces the voltage by X amount, and vanishes if the voltage is below that amount...) In effect, the batteries see a short until the voltage drops below 1.7V, at which point they see no load...

Generally the LEDs with built-in resistors are 'modules'; i.e. an LED in some kind of permanent holder.

Quote:



Actually, no. It was as bright as you'd want, and it didn't ever get hot. I only ran it for about 30 seconds or so. If it matters, keep in mind that this LED is not from this decade!

Thanks for the equations and what not as well. That will give me something interesting to work through tonight at work.. work, oops. I better run!

An LED doesn't have zero resistance, so you don't have to worry about it shorting out your batteries.

You can't measure the resistance of a diode directly since its non linear at the low voltages used by a multimeter. However, you can see that an LED is releasing light energy, and that energy must come from somewhere. (conservation of energy) So in order for the LED to release light , it must use up electrical power, where P= (I squared )* R. If the LED had zero resistance, then P will always equal zero, and no light could possibly be emitted...


A diode normally has infinite resistance at low voltages, then when the voltage rises above a cut-in voltage (forward voltage) specific to the diode, its resistance drops rapidly to a constant value (non zero).

LED's can only handle small currents, and will burn out if the current rises above ~5mA. I have no idea what the resistance of an LED is, but i'm guessing its in the range of 0.5 to 1.5 kilohm, which means 3 volts should be safe for most LED's, but go any higher and you might burn out the LED.

I stand corrected.

All the explanation does not suit my knowledge. Except stereth's.

You are not measuring the LED forward bias voltage but measuring the battery suply voltage.

The voltage drop is due to the internal resistance of the battery when high current passed through. Remember that voltmeter has high resistance so that when you measure at no load will give you higher voltage.

Also normal LED requires 3 volt to forward bias. It won't burn out after long usage, adding resistor in series with the LED is when you have higher than 3 volt supply, eg. 5V.

I square R is heat, not the energy to light up your LED.
Power equation is IxIx(R_led+R_battery) + V(voltage for light)I

You won't feel a hot LED (if correct voltage being applied) since it is more efficient than your house flourescent lamp.

If there is 0 resistance, the current will go to infinite. Since the battery is also called current limiting supply, you'll get warm battery after a few seconds. Don't try with 120/240V mains!

If someone post ridiculous answer without reading my post, i'll kill him!!! Literally...

LED forward voltages vary from 1 point something to 5V or even more. It depends for one on the color of the LED (which is determined by dopant material, so it's obvious why forward voltage will differ). So not all LEDs will behave the same when loaded with 3V. Some might not even light up. But most diodes do have less than 3V forward voltage, and at 3V they are going to be very bright and their life will be shortened.

And indeed what you were measuring is still the battery voltage minues the product of internal resistance and current. Real batteries are modelled as an ideal voltage source with zero resistance, in series with a (hopefully small) resistor which is its internal resistance. Voltage of ideal source equals that of non-loaded battery - as measured with high impedance multimeter. Equation for this circuit says that this voltage equals sum of internal resistance of the battery and the LED's resistance, times the current through the loop. If you measure current as well, you'll be able to find out internal resistance of battery as well as resistance of the diode AT THAT VOLTAGE. As thomas said, it is a non-linear device (you should be able to find I versus V plot for it in its datasheet) so resistance will vary depending on voltage applied. See here for examle:

A sample red LED datasheet

The steepness of the linear part of the curve (Fig. 2) determines the resistance of the diode in its normal operating region and further. You can estimate resistance as for example
(1.8V - 1.6V) / (30 - 7 mA) = 8.7 Ohm. You can see two things - first, if you keep increasing voltage, current will go up VERY fast. A diode might handle excessive current for a while depending on how bad it is but we all know what will happen in the end. Also note that as the current rises, so does the voltage drop on the internal resistance, therefore the actual voltage applied on the diode is less than one might think - 0.1V less in your case.

Without measuring current or knowing one of the parameters, you don't have enough information to determine resistances.

This also explains why a LED can be used as relatively stable voltage reference. If you put it in series with a resistor, and apply a voltage that is not stable, the current through the diode and resistor will vary, however, the voltage on the diode will vary much less than that. Steeper the curve, less the variation. In fact, let's calculate.

Take the operating point of LED to be say 1.7V. Current through diode is 20mA. Say the "unregulated" voltage is 10V. To get 20mA, we need 500Ohm resistance. Since diode resistance is 8.7Ohm, we need to add 491.3Ohm resistor in series with diode.

Say now that the voltage increases to 10.5V. Current through the LED/resistor will be 10.5V / 500Ohm (diode resistance is constant in this region), which yields 21mA. Extra 1mA over 8.7Ohm resistance yields only 8.7mV rise in voltage accross the diode. And so from 500mV change we got only 8.7mV change. In fact the rate of change of diode voltage will be equal to the proportion of diode resistance to the sum of diode and current setting resistor resistance. So 8.7 / 500 in this case. For any variation of input voltage, the diode voltage will change by that voltage multiplied with this ratio - 1.74%.