1
\$\begingroup\$

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

enter image description here

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.

enter image description here

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

enter image description here

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

enter image description here

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:

enter image description here

\$\endgroup\$
17
  • 1
    \$\begingroup\$ Why did you buy them if there is no datasheet? \$\endgroup\$
    – Andy aka
    Sep 16, 2014 at 7:29
  • 3
    \$\begingroup\$ Only boring people only ever buy components with datasheets from trusted manufacturers and sourced via trusted distributors. Customers usually prefer their designers to be boring :-). When it's for yourself the risk equation may vary. And the results probably will as well, alas. \$\endgroup\$
    – Russell McMahon
    Sep 16, 2014 at 13:13
  • 1
    \$\begingroup\$ @RussellMcMahon I have edited my question with test measurements. \$\endgroup\$
    – tgun926
    Sep 16, 2014 at 23:58
  • 1
    \$\begingroup\$ Do not use the ammeter - it's resistance and therefore affect is finite and unknown. Use a SMALL resiatce series resistor - say 1 Ohm in place of the ammeter and measure voltage across it. Better is a 0.1 Ohm if you can resolve voltage OK ( 10 mA = 1 mV). You may destroy a low wattage low Ohm resistor rapidly if you short it across the supply. Heating is lowish (2A x 0.1R = 0.4W) so a 0.5W+ may survive. At 1 Ohm Power at 2A = 4W. | Please repeat tests with series R and provide ALL voltages for each test. So VLED = , Vseries_R= ?, V1 Ohm = ? Vpsu loaded = ? .... \$\endgroup\$
    – Russell McMahon
    Sep 18, 2014 at 4:48
  • 1
    \$\begingroup\$ .... Note that while technically you don't need the 1 Ohm or 0.1 Ohm current sense R when using the eg 4.7 Ohm series R (as you can calculate I from the drop across the 4.7 Ohm) if you O always include the eg 0.1 Ohm you get a point of reference. If eg V across 4.7R is 0.47V and V across 0.1R is 9.2 mV then there is a difference of about 8% at I_4.7R = 0.47/4.7 = 100 mA but I_0.1R = 9.2mV/0.1 = 92 mA. \$\endgroup\$
    – Russell McMahon
    Sep 18, 2014 at 4:51

3 Answers 3

2
\$\begingroup\$

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

\$\endgroup\$
4
  • \$\begingroup\$ I'm still confused. If there isn't a series resistor, why didn't the LED blow out when I connected it direct to the supply? \$\endgroup\$
    – tgun926
    Sep 16, 2014 at 8:13
  • \$\begingroup\$ @tgun926 depending on the forward voltage at a certain current of the LED, and the voltage you supplied, it may have not reached the point of "holy crap lets allow unlimited current into us!" \$\endgroup\$
    – KyranF
    Sep 16, 2014 at 9:33
  • \$\begingroup\$ @tgun926 I see that you gave it 5V and it's forward voltage is ~3.3V, yes indeed it should have short circuited. Either your supply wasnt strong enough, or your multimeter "ameter" has a lot of resistance. \$\endgroup\$
    – KyranF
    Sep 16, 2014 at 9:35
  • \$\begingroup\$ @tgun926 "Blowout" is relative. The power supply provides current limiting. Short term they may survive. With heatsink they may last quite a while. That's one more thing you haven't mentioned. without one they may last for minutes. Or not. So far we do not know the current and voltage together when operated direct off supply because you have not told us. Add a small resistor and find out. ACTUAL LED current and voltage with series R's is just a matter of your measuring and telling us, When you do we can help more. \$\endgroup\$
    – Russell McMahon
    Sep 16, 2014 at 13:18
2
\$\begingroup\$

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.

\$\endgroup\$
-1
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ Aaron, please observe that the Question form original post (OP) was made many years ago (2014). There are already answers and comments. So, it would be expected Your Answer to add value or bring a new insight to the question and answer (Q&A). I did not see them in your answer. Eventually due to language barrier, the phrasing is confusing. But I identified some technical problems in your explanation, with wrong concepts. If you read Russel’s answer and other similar Q&A, you would realize such inadequate concepts/descriptions as “irrespective of voltage, the LED will light up”. \$\endgroup\$
    – EJE
    Dec 7, 2022 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.