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I want to make a grid of 12 LEDS which is powered by a USB cable... I know that the voltage is too high for one LED..the cpu stops the voltage after 2 sec.

So what is the setup i need to use to do so..

I thought connecting 4 rows (parallel) of 3 serial LEDS to the usb cable? Is it ok?

The LEDs are of default type. USB gives 5v so 3 serialed leds will get 1.66v each (which is close to 1.5v when testing leds) 4 parallel rows won't change the voltage but what about the current? does it matter?

If possible a solution without resistors will be great..

thanks, Eli.

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    \$\begingroup\$ Can you tell which LEDs you're using? A link to a datasheet would be great. \$\endgroup\$
    – stevenvh
    Jul 17 '11 at 7:18
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There's no such thing as a "default type LED". LEDs exist in all shapes and (literally) colors, and it's mainly color which determines the voltage.
If you have a 1.4V LED you can't just put 1.66V on it. Either your LED will die in smoke, or your power supply will cut off (which seems to happen here if I understand correctly).
You need resistors. They will limit the current through your LEDs, you can't do without.

Calculation of resistor values:
Suppose 1.4V LEDs requiring 20mA. Then 3 LEDs in series will have 4.2V over them, leaving 0.8V for the resistor. \$ \frac{0.8V}{20mA}=40\Omega\$ . Pick the closest E12 value: 39\$\Omega\$.
Power \$P = V \times I = 0.8V \times 20mA = 16mW \mbox{ } \$ , so any resistor type will do (even an 01005, if you would feel tempted...).

If your LEDs' voltage is close to 1.6V the voltage over your resistor is too small for a proper current setting; a small variation in LED voltage will result in a large change in current, and hence brightness. In that case only put 2 LEDs in series.

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    \$\begingroup\$ Note that with 2 LEDs per string there will be 6 strings. If he has 20mA LEDs and wants to light them fully, then that would require 120mA total. That's more than the 100mA you get by default from USB power without negotiating for more. Desktop computers usually just connect USB to 5V with a polyfuse, but laptops can be very power concious and may only provide what the USB spec requires. \$\endgroup\$ Jul 17 '11 at 16:43
  • \$\begingroup\$ @Olin - Good point, thanks for that. (I thought about it, but didn't mention, because I thought the limit was higher.) \$\endgroup\$
    – stevenvh
    Jul 17 '11 at 16:47
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I don't understand your aversion to resistors. They are simple, cheap and very reliable.

If you're worried about efficiency, and circuit complexity is not an issue, you might want to look at Linear Technologies' LED drivers. (As an example; several other companies make LED drivers as well.)

The thing is, light output from an LED is related to the current through the LED, so the best solution is a current regulator (which most LED drivers are). Forward voltages have way too much tolerance for a voltage regulator to be very useful.

Remember, they're still diodes, and once they hit their knee voltage, small changes in voltage will mean large changes in current, and vice versa. If the forward voltage is 0.1-0.2 V off spec, it will mean a huge change in current (and brightness and power dissipation) to get to the regulated voltage.

The reason we use resistors so much is that they are way less complex than a current regulator, and very compatible with a regulated voltage rail. The idea is that you choose the current you want in the LED, and together with Ohm's law and the forward voltage, you figure out your resistor value. So long as the voltage across the resistor is much larger than the tolerance in the LED's forward voltage (1-2 volts is usually plenty), you'll get pretty repeatable results for brightness and current consumption. Don't forget to make sure the resistor can dissipate enough power (usually a problem for high source voltages or small surface mount resistors).

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    \$\begingroup\$ His aversion to resistors may have to do with currently not having them in his box... \$\endgroup\$
    – stevenvh
    Jul 17 '11 at 14:52
  • \$\begingroup\$ We can speculate all day, though. I could speculate that he'd prefer buying some resistors to burning out his LEDs... \$\endgroup\$ Jul 17 '11 at 18:38

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