Do these LEDs have a constant current driver built in?

Question

I bought these LEDs from eBay and connected them all (5) in parallel with a 4.7 ohm series resistor each.

This should have given me a total current draw of 1.7A, but when I measured it I was only getting 0.4A.

I then connected an LED directly (via the multimeter) to the 5V power supply, expecting the LED to blow. Instead, the LED drew 0.7A, as specified in the description. There is no datasheet, and I'm unfamiliar with LEDs having a current regulator built in.

So why didn't my LEDs blow?

In case the link dies in the future, here are the specs:

• Forward Voltage:3.2-3.4V
• Forward Current:700MA

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Edit: I did the following test measurements. There are some discrepancies I believe which are associated with the quality of the multimeter I am using.

Note: The voltage measurements were done with the 'ammeter' removed.

• \$I_{circuit} = 230mA\$
• \$V_{supply} = 4.95V\$
• \$V_{led} = 3.3V\$
• \$V_r = 1.23V\$
• Ground pad: Warm

Note: The voltage measurements were done with the 'ammeter' removed.

• \$I_{circuit} = 140mA\$
• \$V_{supply} = 5.05V\$
• \$V_{led} = 3.17V\$
• \$V_r = 1.63V\$
• Ground pad: Warm

Note: The voltage measurements were done with the 'ammeter' removed.

• \$I_{circuit} = 650mA\$
• \$V_{supply} = 4.63V\$
• \$V_{led} = 4.63V\$
• Ground pad: Very hot.

Further note, that the LED when directly connected to the power supply, without the ammeter still does not die (held for 5 seconds.)

The LED is soldered onto a PCB like so:

Answer 1

You have probably been sold LEDs with false specs.

They look like LEDs that are usually 1 W rated.

A number of other ebay sellers are selling LEDs of similar appearance and description but with 1 W or 2W ratings.

Your seller describes them as 3W but also says 3.2 - 3.4V, 700 m. As 3.4 x 0.7 = 2.4W and as ebay sellers are usually unlikely to understate their specs and as 700 mA is exactly double the 350 mA that others claim, th chances that the seller is being creative is finite.

See here ebay - 1 Watt

1 Watt

Numerous more.

Your low current results are somewhat puzzling but you have more data than you have provided.

400 mA for 5 = 80 mA each . 80 mA, 4.7R = 0.376 V ~= 0.4V.
So 4.6V across LED at 80 mA. As you measure 4.2V with one LED with direct psu connection so ILED >> 40 mA, you are doing other than you claim or the LED is exhibiting negative resistance (about 0 chance).
BUT if your 4.7R were really 47R ten Vr = 3.76V and VLED =~~ 1.24V so not likely.

If you connect all 5 again with series Rs and measure Vsupply, Vled and Vr you will have a better idea of what is happening.

A series R would help explain what you see BUT is very unlikely.

Connecting one LED directly across the charger produces maybe 6 x rated I if Imax = 350 mA or ~~= 3 x Iax if 700 mA. This is liable to be enough to rapidly degrade the LED but may or may or may not cause a sudden hard fault. If you MUST do that add a small series R so and measure RLED and Rpsu you can calculate current

Answer 2

It's not that unusual. If you connect the LED to a 12V supply, you'll get smoke. It is not uncommon, however, to see keychain LED torches that comprise a switch, a battery and an LED - no resistor. So why does the LED not blow?

The answer lies in the bulk resistance of the semiconductor that the LED is made of. Diodes also have a forward voltage drop (usually defined in the datasheet), which means the actual voltage you see across the limiting resistor is less than the applied voltage. In the case of silicon diodes, this voltage is usually between 0.6V and 1V.

All semiconductors have a feature known as "bulk resistance", which is the minimum resistance you will see when the semiconductor is fully conducting. There is also a dynamic resistance component that varies inversely to the applied current. This all makes diodes, and LEDs in particular, a bit more complex than you would first imagine. Diodes are not either "off" or "on", their conduction varies as the applied voltage - and current - changes. Their conduction curve is very non-linear, which is why they are useful as rectifiers. As the forward voltage rises from zero, there is almost no change in the forward current - the resistance is very high - until you get near the forward voltage of the diode (the "knee" voltage). The resistance then drops pretty smartly and the current increases until, just above the knee voltage, the resistance drops to its minimum bulk resistance value.

LEDs take this to a whole new level. The use of exotic materials such as Silicon Carbide, Gallium Arsenide and Gallium Aluminium Indium Phosphate means that the forward voltage drop across the LED can be as high as 4V. The current curve around the knee voltage is also much less sharp and the bulk resistance is quite a lot higher than with silicon rectifier diodes.

So from your description, I suspect that your LEDs are still well inside the knee part of the current curve at 5V. You will probably find that some LEDs will draw more current than others and that some may even evaporate when connected directly across the 5V power supply. As I mentioned at the start, if you connect one across a 12V supply you will most likely get a bright flash and some smoke.

Answer 3

LEDs work depending on the type of power source. There are current limiting power supplies (i.e current is constant) and there are voltage limiting power supplies (i.e voltage is held constant.) Most LED circuits are current controlled. Irrespective of the voltage, the LED will light up because the voltage supply, automatically adjusts to the LED connected in series while the current is held constant. If the voltage is held constant, then the LED can blow up (i.e in a voltage limiting power supply.)

To answer your question: Yes, LEDs are controlled by current.