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I would like to connect 32 3W LEDs together. All the LEDs are the same.

Each LED (32 of them):
750mA
2.2V
3W

Power Source:
Constant Current
3A
100W

I thought making 4 chains, 8 LEDs connected in serial in each chain and connect the 4 chains in parallel so 750mA falls on each chain. I may have to put a resistor on each chain according to the Voltage range of the power supply. I know I can connect all the LEDs in parallel but that is quite allot of soldering and more resistors in the project :)

I know that this may be risky if one of the LEDs fail. I thought of placing a 0.8A-0.9A fuse on each chain or even a current limiting resistor. I have no problem that all the LEDs won't work if one LED is burnt.

Will this work?

Gilad.

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  • \$\begingroup\$ Thank you all. I got very good answers and was convinced to move to a constant voltage supply and to use an LM317 to limit the current. \$\endgroup\$ – Gilad Sep 19 '11 at 11:27
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Summary

  • As proposed, your LEDs MAY die a bright and early life.

  • It's easy to help them to live long and prosper.


It is easy to drive LEDs properly.
Failing to do so can lead to very short lifetimes and uneven illumination.

LEDs should always be operated with either a constant current supply or a supply which is close enough to constant current for the purposes. Operating them from constant voltage directly invite substantial learning experiences which you might rather not have.

A small decrease in LED_Vf at a given voltage can lead to a very substantial change in LED current. If you parallel 4 LED strings and do not make any attempt to equalise currents then you can be almost certain that string currents will not be equal.

One easy way to set maximum current in a string is to use an LM317 and one resistor. This requires about 3.5 V of voltage drop to function so may or may not suit in your case. The following circuit is from fig 19 in the LM317 datasheet.

enter image description here

As shown, R1 sets maximum current - here for 750 mA R = 1.25/0.750 = 1.7 ohms. You could use the standard E12 value of 1.8 ohms or a lower value eg 1.5 ohms to cause limiting once imbalance got too high (I = V/R = 1.25/1.5 = 830 mA).

Without some form of per string limiting:

  • Current imbalance of up to about 2:1 per string MAY occur.

  • If any one LED blows total average string current rises to 3A/3 = 1A. If you again got 2:1 imbalance then currents may be eg 750 mA, 750 mA, 1.5A in the 3 strings. If your LEDs are rated at a true 3W and the Vf you have stated occurs in practice then Imax = P/V = 3/2.2 = 1.36A SO the LEDs would not be vastly over-rated at 1.5A. Still not a good idea though.

If you have lots of "headroom" you can use more voltage and a series R to drop voltage and stabilise current. An LM317 will give a much more stable result. The LEd shown here can be one LED or a string. Drop aross the LM317 circuit is 1.25V across the resistor plus the dropout voltage of about 2 Volts (graph on page 6 of datasheet).

enter image description here


LM317 or other current source power dissipation

Each LED string is specd as drawing 750 mA.
The LM317 has a minimum dropout of 2V and dropout will be as high as required to maintain constant current.

At 2V that's 1.5 Watt (2V x 0.75A) and at say 5V it's about 4W. A 10 degrees C/Watt heatsink will handle that OK. If you need higher than that there is something wrong with the design. The higher dissipation MAY be required if one string of 4 goes O/C but that's a fault condition. The heatsink may be thermally shared with all 4 x LM317 as they will only go into high voltage drop when something goes wrong and drop will vary depending on LED Vfs.

The LM317 has thermal shutdown and will gracefully turn off as/if required due to overload.

Note that the ability of the LM317 to "float" is part of what allows it to be used as a constant current source. It obtains it's internal power supply from the drop across the regulator. Other regulators may have a low minimum dropout BUT rely on a higher value from Vin to ground terminal to power them.

As soon as the string goes into constant current mode the regulator (regardless of dropout voltage) will be required to drop any "excess" voltage so even a low dropout regulator will be very little better off when it is called on "in anger".

I have used LM317's as constant current drivers for LED strings on numerous occasions (usually for LED testing) and find them very useful for this purpose, subject to proper thermal design. For currents above 1A (1.5A in some versions) an LM350 may be used (up to 3A afair).

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    \$\begingroup\$ At 3W per LED, I think the OP will have a hard time heat sinking the LM317. Better to get a lower dropout current regulator. \$\endgroup\$ – Kevin Vermeer Sep 19 '11 at 10:12
  • \$\begingroup\$ @Kevin Vermeer - please see addition to LM317 section. \$\endgroup\$ – Russell McMahon Sep 19 '11 at 10:29
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You do have to place the series resistor in each branch, even if the LEDs are all the same. The internal resistance of a LED is so small that a minute difference in voltage will cause most of the current to go through one branch (or hardly any, depending on whether the voltage of the one LED is lower or higher).
A constant current source is not a good idea for the reasons you mention: if one branch fails the others will see 33% more current. Fuses may help, but there's some tolerance on their nominal breaking current, so be sure that they don't break at 750ma and will break at 1A. Once the first fuse has gone the others will follow fast. I would recommend a constant voltage source, though.

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Ummm... your suggestion will work just fine and it may be ok if this is a hobby project. Your circuit will be less robust. When the first LED eventually fails open, current increases in the other strings and all will fail before long.

Depending on how the LEDs are binned this may or may not be an issue. If the LEDs are binned by Vf, you are unlikely to experience any problems for a long time.

If this is a commercial project, please disregard this advice, and do it properly.

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better off with a fixed voltage power supply current limited to 3 A. A minimum of Vo of 17.6 VDC will do the trick.

No resistors necessary for this voltage. Above a Vo of 17.6 VDC the equation for the series resistor is R = (Vo - 17.6)/0.75.

This will be fine.

Cars that have LED's in the lights also use a combination of series and parallel to get the reliability and brightness there.

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