Light Output from LED

If they were light-emitting resistors, they would be abbreviated to LERs. But they're not. They are light-emitting diodes, which is why they're called LEDs. So I don't know what you're doing or what you're asking without seeing a circuit diagram with component part numbers attached. Do you have the component manufacturer's datasheets for them or are you trying to fish for them on here?

Actually a light emitting resistor is abbreviated as 'incandescent lamp'.

I'm not quite sure what you are asking when you say 'for the same amount of current'.

If one of the LEDs has a built-in resistor (some do) and it is connected in series with its LED chip, then yes, when that LED is connected in series with other LEDs, the same current flowing through that resistor and that LED will be flowing through the other LEDs.

In a series circuit the current is the same,so what i am asking is, if one of the leds is resititive, then the ouput light intensity would be different from another cirucit that doesnt have a resisitive led.

Sorry: we don't know what you mean by a resistive LED.

Do you mean one with a resistor already connected in series to give the correct current when the thing is connected to a given voltage?

LEDs are essentially current driven devices. Look at the data sheet for one: it will have a forward voltage and a normal operation forward current.

Let's say the forward voltage is 1½ volts and the normal operating forward current is 20 mA.

And lets suppose you have a 5V supply. If you wanted to light one LED you would:-

take 1½V from 5V leaving 3½V;
Calculate the value of the resistor you require R=V/I = 3½/0.020 = 175 Ω;
The next common value is 180 Ω, so you would end up by driving the LED with 19.4 mA

If you wanted to light two LED you would:-

take 2x1½V from 5V leaving 2V;
Calculate the value of the resistor you require R=V/I = 2/0.020 = 100Ω;
100Ω is a common value so your LEDs would each be driven by 20 mA.

If you wanted to light three LED you would:-

take 3x1½V from 5V leaving ½V;
Calculate the value of the resistor you require R=V/I = ½/0.020 = 25 Ω;
The next common value is 27 Ω, so you would end up by driving the LEDs with 18.5 mA

Here's a circuit that would drive a super bright LED with a forward voltage drop of 3½ V with a current of 14 mA (OK just ignore the fact that 600 Ω is not a common value).

There are lots of on line calculators for LED series resistors.

http://www.google.co.uk/#hl=en&output=search&sclient=psy-ab&q=led+resistor+calculator&psj=1&oq=LED+resistor+calculator&aq=0&aqi=g9g-v1&aql=&gs_sm=1&gs_upl=3625l12782l0l16579l23l23l0l5l5l0l250l2595l0.17.1l18l0&gs_l=hp.1.0.0l9j0i15.3625l12782l0l16579l23l23l0l5l5l0l250l2595l0j17j1l18l0&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&fp=9df159aac7d46fa5&biw=1344&bih=826

If your driver board is effectively a current source then we need a little more information.

I think I understand your question. Let's first consider the case of using the mythical ideal current source (I) connected to three series connected forward biased identical LED. This produces L lumen of visible light with each LED producing L/3 lumen. The voltage drop across each LED is Vd. The voltage drop across the current source will be 3*Vd. If an LED is replaced by a resistor (R) the voltage drop across the current source now becomes 2*Vd+IR. Each LED still produces L/3 lumens. If the ideal current source is replaced by a real current source then there will be a maximum voltage this real source can produce and still be providing the same current. So one must now know the limitations of the current source and the resistor value to now know if each LED still produce L/3 lumens.

Just try it.

I'm not the expert in the sector but I know some strange behaviors of these little things.

To start: as explained by some of the colleages: a diode is not a resistor.

The light output of an LED is in direct relation with the current through the junction.

but the voltage drop over the LED is not.

That is why the 3 LED in series systems driven by 12V DC on a busbar have a resistor in serie.

Driving an LED needs indeed close control of the current.

But you can trick the thing by going in the pulse: cranck out more light with the same average power.

Humans don't see the flikering of the light if the frequence is sufficiently high (>25 hz is sufficient), be aware that birds do see 100hz tL tubes go on off.

LED's are killed by the thermal effect (joule effect) of the current/voltage combination. but this joule effect has to be integrated: the average power over time will kill. The instant power can be high, if time is short.

To be short: current drive circuits are the way to go, constant current is ok but pulsed current is delivering a higher efficiency.

What i meant by resistive led is that the solder paste attaching the led to the ceramic board is resisitive, so i call it a resistive led. So if one of the LED's is resistive, wouldnt the light output from the three leds be compromised, even though the driver circuit provides a constant current?

Thanks

No.

You would just have the additional voltage drop across the unwanted series resistor.

If the "bad" joint is more than a fraction of an Ohm then you need to re-solder the joint.

Ok, I'll repeat what you wrote in a different way to clarify what you really are asking. You have three LED but one of these LED has a poor solder connection so that the resistance of this joint is no longer an insignificant voltage drop compared to the voltage drop of each LED. All three LED remain in the circuit and the current through this now four series components (3 LED +1 R) is controlled by a real current source. Clearly if this added resistive voltage drop makes the current source to reach its maximum voltage then less light will be produced because less current will actually be going through the three diodes. The current source is no longer acting as a current source.

If instead one has not exceeded the voltage limit of the current source and the same current as before flows through all three LED then the two LED without the bad solder connection will produce the same amount of light. The question now becomes very interesting and not as obvious as one might think. The same current will be flowing through this third LED but the temperature will not be the same as the other two LED. The poor solder connection will be adding a thermal load to this junction along with the self heating by the junction voltage drop. Will the internal quantum efficiency that produces light be the same for this hot LED and thus the identical light level be produced? My gut says that it will be the same efficiency or that the difference will be difficult to detect but I must research this scenario and quantum equations in more detail.

One thing I am certain of, the added thermal load will significantly reduce the lifetime of this hot LED.

The light output from an LED is a function of the current through it, so if you have a constant current driver, the light output from each LED should be constant.

You can't talk about the resistance of an LED because, unlike with a resistor, the voltage across an LED is not proportional to the current. At low voltage there is very little current. As the voltage is increased the current increases at first slowly, then very rapidly. The curve is the classic exponential curve. This "knee point" of voltage is the forward voltage of the diode, and it varies according to the color of the LED. This has to do with the energy of the light photons, and it varies from about 1.7 volts for infrared LEDs to 3.8 for ulta-blue.

Assuming that the constant current remains the same, there would be no difference. The voltage across the LEDs would be higher for the resistive one (only on the board), but measuring across the pads of the LEDs should give the same or similar values.(of course this assumes that the LEDs' pads wrap around from bottom to top.)