A cautionary tale with cheap ebay Voltage Regulators.

You're driving a LED based on voltage, then you're making it worse by using a regulator circuit that is specified at +/- 2.5% regulation. Grumpy Mike would kick your butt for doing it that way. There's nothing about a LED V/I curve that suggests anything but eventual failure.

It's also not necessarily a knock-off/counterfit LM2596 issue. Nat Semi specs their LM2596 at +/- 4% regulation:

Other features include a guaranteed ± 4% tolerance on out-
put voltage under specified input voltage and output load
conditions...

cjdelphi:
I've blown several high power (>1amp cree) LED's with these, when they get hot (or becomes shorted) the regulated voltage increases, so after running it for an hour, the temp slowly creeped up, with heat came a slow rising voltage until death....

Sounds to me like you're using the wrong sort of board and overloading it. Anything will fail if it gets hot enough.

LEDs are supposed to be driven by current, not voltage. They make those boards with constant-current output (adjustable):

The boards with three adjusters on them also have a shutdown voltage, for charging batteries. ie. The board supplies a constant current then shuts down if the output voltage reaches a certain voltage level.

If you want to go the whole hog (and you should if you're using expensive Cree LEDs you can even get them with heatsinks, just in case.

cjdelphi:
what i do and do not do with my LED is my own buisness...

Sure...

cjdelphi:
secondly... You're wrong what killed my LED's is not current at all but heat.... these power supplies got hot to drive the led which increases the light, truth is even at 3 amps which i can assure you these can not deliver at 4.2v ... even at 12v and even if it could... it's well within max current specifications... but no all i required was a lot bigger heat sink... heat and heat alone caused the death as the regulator crept up in voltage so did the current death from heat NOT anything else

If the thing was getting too hot then you were overloading it.

What counts is the power dissipation (wattage), not the amps. The board may be rated for 2A max but 2A at 5V isn't the same as 2A at 20V. The voltage conversion is only 90% efficient at best, the rest turns into heat.

If you look at the board you posted there no way that board can handle a 20 volt 2 amp drop that's 40 watts
the chip should shut down, lower its voltage not increase it.

I've blown several high power (>1amp cree) LED's with these, when they get hot (or becomes shorted) the regulated voltage increases

If the regulated voltage increases when those devices are shorted, those are miracle devices.

What you are likely observing, without realizing it, is why a serial resistor is sometimes needed with leds: to prevent thermal runaway.

20 volt 2 amp drop that's 40 watts

This is a switching mode regulator.

There's onboard thermal regulation which prevents anything excess current being delivered to the LED.

That doesn't mean the less-than-excessive voltage from the regulator wouldn't cause excessive current / power dissipation on the led.

No, the board does not regulate current output at all: they call it a voltage regulator for a reason.

case closed.

I am actually still curious as to how its output voltage increased when you shorted it.

It is so out of this world.

want a video?

I can show a stable voltage on video to the device you're powering, but decrease the resistance the voltage increases.

I can show a stable voltage on video to the device you're powering, but decrease the resistance the voltage increases.

That's probably quite different than saying that the voltage increases when you short them.

dhenry:

20 volt 2 amp drop that's 40 watts

This is a switching mode regulator.

There's onboard thermal regulation which prevents anything excess current being delivered to the LED.

That doesn't mean the less-than-excessive voltage from the regulator wouldn't cause excessive current / power dissipation on the led.

No, the board does not regulate current output at all: they call it a voltage regulator for a reason.

Read the data sheet it said if the output is shorted the chip decreases the voltage

you saying ohm 's law don't apply dherry if the voltage goes down the current can't go up.

20 volt 1 ohm load 20amps 5 volt 1 ohm load 5 amps

Now whats this say

(3) The oscillator frequency reduces to approximately 18 kHz in the event of an output short or an overload which causes the regulated
output voltage to drop approximately 40% from the nominal output voltage. This self protection feature lowers the average power
dissipation of the IC by lowering the minimum duty cycle from 5% down to approximately 2%.

dhenry:

I can show a stable voltage on video to the device you're powering, but decrease the resistance the voltage increases.

That's probably quite different than saying that the voltage increases when you short them.

the voltage sure does rise when you short them, i'm uploading a video of me just brushing 2 crockadile clips together, i then make it switch between voltage in and voltage out (it alternates). then as i short the clips
you'll see the voltage jump, the voltage reader is correct, confirmed it by the multimeter, but when it gets hot powering a device with a large current the voltage does also increase, but it has to be a large current, eg an arduino alone wont be effected too little current.... but if you start connecting bigger items, motors and things? I dunno, i only discovered it by accident when shorting it once time and blowing a small 5mm LED never thought much of it until this board which displays the voltage on automatically arrived..

So i was playing with it and noticed that the 5? pin regulator on this was behaving the same as the even cheaper boards with an even less wattage rated IC switcher... but still this board can supply more watts big whoop, not interested, what I am interested in is why the voltage increases when shorted briefly or after long periods of heat the voltage goes up, device goes um poof.

the voltage sure does rise when you short them

Sounds like you have defined "short" uniquely.

"A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an excessive electric current/overcurrent"

  • wiki

touching the clips results in max amp discharge causing the wires to give off heat. is this not a "short" ? I'm serious.....

You probably want to focus on the part where it talks about very low / no resistance.

You cannot have a voltage drop over a wire with no resistance.

huh? (video... look at the video)

A cautionary tale with cheap ebay Voltage Regulators

The Title... or more correectly how to abuse a component without really trying.
The regulator is rated at 2A output current... for short periods of time because there is no effective heat sink.
Read ALL the data sheet INCLUDING the thermal considerations. The voltage drift is most likely due to components on the board getting hot. The OP once mentioned that the module was too hot to touch and it is my opinion that IF as I believe this is the case, then the module needs to have a heat sink soldered to the bottom of the switcher. A 12 mm X 25 mm strip of copper about .6 mm thick with a hole 1/4 from one end a 90 deg at the half point, (middle) soldered to the switcher and mounted to a good heat radiator.
IMO the component is being abused and it is very simple to use it properly.
I appear here to be a hard nose but that is my training, I had to learn the hard way to be Certain that anything I did was 100% Perfect. I am non trying to be the expert, more to share my thoughts only, If I appear otherwise it is unintentional and I apologize.
{Edit, RKJ}

Bob

dhenry:
You probably want to focus on the part where it talks about very low / no resistance.

You cannot have a voltage drop over a wire with no resistance.

A wire with no resistance? Can you spare me a little of yours as my supply of super-conductors has run low?

Lefty

Well, it is a switching regulator, so... if you apply a large load to the output, then remove that load, it is likely to overshoot. With an undersized output filtering stage, it's entirely possible that its regulation suffers when faces with current transients. Now, if the voltage increases steadily with a consistent heavy load on the output (say, a 5 ohm resistor) then it's probably a component behaving out of tolerance. Maybe a saturated inductor? Or comparator oscillation? I don't know.

It would be interesting to put a scope on the feedback voltage divider to see what the comparator input looks like under overload conditions.

Whether this is "abnormal abuse" or something that should be handled properly by the IC's protection circuitry is debatable. I would like to know that if something in the load circuit failed, the PSU would do everything it its power to behave according to the three-tiered model: Protect the user, protect the load, and protect itself -- in that order. As a realist, I don't necessarily expect that in a regulator bought for spare change from a no-name overseas merchant via eBay.

It would also be interesting to wire up the same circuit with known-real chip and parts and see if it behaves similarly...