...such LEDs are driven by a current of about 5...10mA, in order to minimize the whole current consumption of board and to not overload the drivers.

Assume you want to drive 6 LEDs by a 74HC04, containing 6 inverters. Each LED is individually connected to Vcc via a current limting resistor, with one end of resistor connected to Vcc, the other end of resistor connected to the anode of LED and the cathode of LED to the according inverter output. Again, each LED has its own current limiting resistor and its own inverter output, making the LED shine, whenever the according inverter output goes low.

When now all 6 inverters go low at their outputs, 6 LED currents have to be handled by the 74HC04. Datasheet says, that 50mA max is allowed to flow through the common ground pin to ground. So, each LED must only draw 50mA / 6 = 8.3mA. Choosing 5mA LED current is recommended then.

For this LED we want briefly calculate the current limiting resistor:

http://www.kingbright.com/manager/u...143816.12)

In such an application for a rough estimation of current limiting resistor you usually take into consideration the typical forward voltage drop. The maximum forward voltage drop, on the other hand, would result in less current and is uncritical. Assuming a typical forward voltage drop of 2.1V for our green LED at 20mA, the current limiting resistor should be about (5V - 2.1V) / 5mA = 580 Ohm. So, a 560Ohm resistor is a good choice.

Even when assuming that the forward voltage drop at 5mA is actually somewhat smaller than 2.1V, concretely about 1.95V, 560 Ohm yields a current of about (5V - 1.95V) / 560 Ohm = 5.4mA, which is well within safe limits.

Antoher way to determine this current limiting resistor is to assume a worst case scenario where the forward voltage drop is unnaturally small and that nearly all the Vcc is dropping across the current limiting resistor. So, let's assume that the forward voltage drop is only 1V, which no green LED will ever reach. This gives a current of (5V - 1V) / 560 Ohm = 7.1mA, which is still below the 8.3mA from above.

Due to the fact, that the forward voltage drop will differ from LEd to LED, even if they are from the same type, you can end with a display, where the LEDs are shining unevenly bright. This can be prevented by a preselection of LEDs: Just take a handful and check one after the other by simply comparing their brightness. To do this, take two such inverters from above, have one LED ("the original") permamently connected to the one driver and clamp to other driver all the LEDs you want to test. Take those, which are as bright as "the original" one.

Even if all the LEDs from one colour are shining evenly bright, there can still be a difference from colour to colour. Then you could slightly modify the current limiting resistor for one colour. But do only increase the value, otherwise the drive current might become too big.

Example: Your red LEDs are shining much brighter than your green LEDs. From begin on both have 560Ohm resistors. Then give 620Ohm or 680Ohm for the red LEDs a try.

Kai