Ok, Now I'm stumped....

[charlie_r]

Can anyone explain why I haven't blown my 1W LEDs on my testing platform?

12V in, to LM317 set as constant current @ 100mA.

3 series connected LEDs, 2.1Vf each, for total LED Vf of 6.3V

3V loss in the 317 in CC mode leaving 3V unaccounted for.

Why are my LEDs still alive? What I've been taught is if you overvolt OR overcurrent an LED you will kill it.

Not to mention that according to all the rules I've been taught, voltage drop through a circuit MUST equal total supply. So where is the missing 3V?

 

[Grossel]

Probably a bad measurement somewhere.

Why not upload a schematic. There is probably just something obvious you haven't discovered.

 

[charlie_r]

Here's the schematic:


View attachment 61331

Supposedly, according to **broken link removed**, the regulator drops 3V.

From what pins would I make the measurement? Since I'm taking the current from the adjust pin, would I go from in to adj? Or in to out?

 

[charlie_r]

Here is what I measured at the LM317:

In to Out = 3.72

In to Adj = 4.96

Out to Adj = 1.23

Out to Adj is close to the expected value of 1.25

Even with the 4.96, there is still 0.78V missing. Where did it go?

BTW, kinda hard to see to measure the V-drop on those LEDs. Even at 100mA they are blindingly bright.

 

[Grossel]

Where is the "Adj" point you're talking about?

Without seing a scetch with probes, I still beleive that your measurment isn't properly done.

What kind of voltmeter are you using? An analog one will cause aditional voltage drops, so that all measured voltages added up will be slightly less than expected.

 

[Roff]

Here is what I measured at the LM317:

In to Out = 3.72

In to Adj = 4.96

Out to Adj = 1.23

Out to Adj is close to the expected value of 1.25

Even with the 4.96, there is still 0.78V missing. Where did it go?

BTW, kinda hard to see to measure the V-drop on those LEDs. Even at 100mA they are blindingly bright.

You don't have to look at the LEDs. Look at the meter.

 

[crutschow]

Here is what I measured at the LM317:

In to Out = 3.72

In to Adj = 4.96

Out to Adj = 1.23

Out to Adj is close to the expected value of 1.25

Even with the 4.96, there is still 0.78V missing. Where did it go?

BTW, kinda hard to see to measure the V-drop on those LEDs. Even at 100mA they are blindingly bright.

So measure from the right side of the R1 output resistor to ground. That will give you the total drop across the LEDs.

 

[charlie_r]

Meter is a GE 2524 DMM. Yes a cheap wally world unit.

Have to look at the LEDs in order to get to the pads....then look at the meter....

Any way I look at it, it appears to me that as long as I am not pushing too much current, AND have less voltage drop across the LEDs than the supply voltage, it should be ok. So does it really matter where I'm losing the volt or so? I don't know. At 100mA, they are bright enough for what I want.

Oh, a correction for that schematic: the 3.9Ω resistor is for 320mA, the circuit as built has a 12.4Ω there.

Got it, crutschow, you posted while I was replying....

 

[Mr RB]

If you have 1.25v across the Radj resistor, and it is indeed 3.9 ohms, then the LM317 is operating properly as a constant current source and there is exactly 1.25/3.9 = 320mA of constant current through the LEDs. The circuit is working correctly.

As for "losing voltage" you should do the simple test;
measure PSU voltage to the circuit (at the circuit input NOT at the PSU), measure combined LED voltage, measure voltage across the constant current source.
It will always solve correctly as;
Vpsu = Vcc + Vleds

 

[ronsimpson]

The constant current source has its input at the IN pin.
The constant current source has its output at the ADJ pin not the OUT pin. (output is the bottom of the resistor that is connected to the adj pin)

The constant current source consists of LM317 and resistor.

The out pin is 1.25 volts higher than the adj pin.

 

[Diver300]

Can anyone explain why I haven't blown my 1W LEDs on my testing platform?

12V in, to LM317 set as constant current @ 100mA.

3 series connected LEDs, 2.1Vf each, for total LED Vf of 6.3V

3V loss in the 317 in CC mode leaving 3V unaccounted for.

Why are my LEDs still alive? What I've been taught is if you overvolt OR overcurrent an LED you will kill it.

Not to mention that according to all the rules I've been taught, voltage drop through a circuit MUST equal total supply. So where is the missing 3V?

If an LED is working, it is impossible to get too much forward voltage across it if you keep the current to less than its rating.

You don't quote the full specifications of the LEDs, but they are probably something like this:-

https://www.avagotech.com/docs/AV02-1640EN

Now those quote a Vf of between 1.7 and 2.3 volts at 350 mA. The graph also shows that the typical voltage is around 2.2 V at 350 mA, dropping to 1.8 V at 100 mA.

However the voltage rating of an LED is not what you should supply it with. You should supply it with current, and the voltage is what the circuit has to put up with.

Most electrical devices are the other way round. They are rated at a voltage, and the current that they take is what the power supply has to allow for. For instance, you plug a heater or a mobile phone charger into the same socket in the wall. Both are rated at mains voltage, the heater takes lots more current.

LEDs are connected to circuits like yours that supply constant current. That is why the LEDs are in series, forcing them all to take the same current. The LM317 with the resistor will supply the same current whatever the load voltage is. If you short out all the LEDs, the current will still be the same.

The voltage of the LEDs is useful for calculating the supply voltage you need. The minimum voltage across the LM317 plus the resistor is about 3 V, leaving about 9 V, so you could probably run 4 LEDs in series. If you put 5 LEDs, there would be less than 2 V across the LM317 and resistor, so the current would be less.

The heat generated by the LM317 depends on the voltage drop, so if you did short out all the LEDs, the LM317 would get a lot hotter. The minimum LED voltage should be used to calculate the maximum voltage drop on the LM317, and therefore the amount of heat it will produce.

 

[charlie_r]

now THAT is the info I needed. Thank you.

And thank you all for replying.

 

[audioguru]

Your link is not correct. It says that the MINIMUM voltage from the input of the current regulator circuit to its output (the LED) is 3.0V but actually it is 3.25V and is higher with higher output current. The current regulator adjusts its output voltage so that the current is what is set. Its input can be as high as what makes it too hot.

Your LEDs are not 2.1V each. Some of them might be but others will be 1.9V or less and others will be 2.3V or more.