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Is there a way to tell max volts an LED will take from a CC power supply?
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Beamin:
I was going through somethings and I found something I built but can't remember anything about it. Its a transistor, two resistors, a photocell, and a 1.5"X3" white LED backlight from ada fruit, can't really make out what values are on the parts.

So I want to figure out how bright this will go with out blowing it up, it works pretty well to light up rooms at night. I have no idea what kind of led this is or anyway to look up the data sheet. BUT I do have a power supply that I can set by 1ma and 0.01v. Setting the ma at some arbitrary number then seeing the power supply drop the voltage is there a way to get a close estimate of the LED max current brightness? At 3v the circuit pulls about 1ma to see it start to glow then at 5v with your hand blocking the sensor it pulls 3 ma. Sounds really low, or a backlights supposed to be dim low power leds? Looks like a single LED. I would like to crank up the voltage but not sure. Does the photocell care how much voltage it sees?
mariush:
Limit current to something reasonable like 10..20mA (pretty much any led will tolerate 10mA and 10mA is big enough for power supplies to measure)
Set voltage to some low value like 1v
connect led in circuit.
Slowly increase voltage while monitoring current
When you get close to the forward voltage, led will start to conduct and current will rise.

Leds have a very narrow region where they behave like partially open, like there's something chocking current flow
For example (using bogus numbers) a white 100mA 3.2v led may be fully closed up to 2.8v and then between 2.8v and 2.9v it may be partially on, letting 1..10mA flow through...
and once you go over 2.9v it may be fully open letting hundreds of mA go through if you don't limit current.
But, you can't rely on these values as they're dependent on led temperature. it may be 2.8v..2.9v at 25..30c ambient but if you just turned off the led after an hour of running and led is 50-60c warm, that narrow region most likely drifted to something like 2.75v..2.85v

If you limit psu to 10..20mA and ramp voltage slowly you'll see the current jump from nothing to 10..20mA. It's also big enough value to see individual diodes if led is made with multiple diodes in series... ex a 6v led may be made with 2 leds in series inside package.

If psu won't let you limit current, you can make your own with a linear regulator

ex see the plain LM317 regulator: http://www.ti.com/lit/ds/symlink/lm317.pdf

See page 12, section 8.3.3 ... how to limit current Iout = 1.2 / R1 ... so for example if you want to limit current to 20mA (0.02A) then  0.02 = 1.2/R1 so R1=1.2/0.02 = 60 ohm

This is not a standard E12 or E24 value but 56 ohm or 62 ohm is and they're close enough, instead of 20mA you have maybe 18..22mA which is still good enough. If worried, you can recalculate for 15mA and then pick closest resistor standard value.

So you can feed 5v..12v to lm317 and you get on the output some voltage with a current limit.
Now you can use a second LM317 in standard adjustable configuration (see page 10, 8.2) ... with potentiometer you can adust output voltage from minimum 1.25v to input voltage minus ~ 1..1.5v
And note that the regulator itself will consume around 5mA so if you limit current to 20mA only with first regulator and 2nd regulator consumes ~5mA+ then led may only get 10..15mA
bob91343:
That probably isn't the best way to test.  My method is to apply around 10V to the LED through a resistor of around 470 Ohms.  If you vary the voltage you will also vary the current.  Set the current at 10 mA and measure the voltage across the LED.

Connecting an LED directly to a power supply is too risky.  You can blow it up.  Best to have that series resistor.
Beamin:
The post I just typed got deleted grrr..  :palm: img too big it says. Anyways.

Im trying the above method now but how did I wire this? I drew diagram just now from what I can see, I'm assuming the photodiode is to the base(center pin of trans, but I can't read the number on it nor see resistor values) and either C or E, that's why its unlabeled. Shouldn't the photodiode have +Vcc hooked to it then to the base? Looks like I wired it between base and either e or C. Seems backwards?  The power supply does 1ma and 0.01v all the way down to zero, so there is also the circuit as protection, I turn the power supply on then block the sensor and watch the ma go up.

EDIT: Forgot: Why doesthe LED flash on when I first turn it on the power even though the sensor is in light? (Turns on with dark)
schmitt trigger:
What you drew as a "photo diode" is in reality a CdS photoresistive cell.
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