# 5x5x5 Led Cude Powering Question

Hello im a programmer but new to electronics, so here is my question.

I have build a 5x5x5 led cube using 4x 74HC595 my problem is that when i turn on more than 10-20 leds at the same time the leds are not bright enough. Im powering the led cube form arduino uno. I have used 5 resistors total placed on the cathode of each layer. Each resistor is 5ohms.

If i have used 25 resistors one for each anode would the leds be brighter? In confused please help.

my problem is that when i turn on more than 10-20 leds at the same time the leds are not bright enough. Im powering the led cube form arduino uno.

Well, that's why.

Each resistor is 5ohms.

That's generally way too low.

If i have used 25 resistors one for each anode would the leds be brighter?

More resistors/more resistance == less current. Less current == dimmer.

Why 5ohms is too low, if i had used a more ohms resistor the light would be even more low, correct?

Why 5ohms is too low

The purpose of the resistor is to limit the amount of current an LED can draw. A 5 ohm resistor does a lousy job of limiting current.

The problem that you have is that there simply isn't enough current, limited by resistors or not, to power the LEDs. Kind of like fighting a forest fire with a squirt gun.

I built a 5x5x5 blue led cube. The problem you are having is that the brightness will vary based on the duty cycle of your scanning, and the number of led being commanded on at any one of the five levels. You really should have a dedicated resistor wire to all 125 leds, that would eliminated the variation of brightness due to the number of led selected on for each level.

The solution I used, which is the best was to use shift registers that had constant current regulators on each of the output pins that would automatically adjust the output current to whatever value you wanted via a single 'current programming' resistor wired to the shift register.

Here is a link to a cube project that I used for controlling the cube, but of course I used an arduino standalone chip to drive the thing rather then a PIC micro that this project uses. Anyway the constant current drivers work great and allow scanning without any resistors in the cathode or anode circuits of the LED. The arduino turns on each row in turn for 2 milliseconds via a transistor wired to +5vdc, while the 25 bit outputs a active low constant current output shift register turns on one vertical row of 5 levels of leds. Works great, very even brightness no matter what the pattern being displayed is.

http://picprojects.org.uk/projects/lc/index.htm

Lefty

Very cool project, and great advice from Lefty. Just wanted to add some LED basics. 1) The LED will be defined by its output brightness in millicandles. Its tough to quantify except to compare LEDS to each other , the more mcds, the brighter it can be. 2) The LED will have a rating called Vf. This is the forward voltage, how much volts it takes when turned on ( positive voltage applied to the anode + side ). If the LED has a Vf of 3 volts, & If the arduino drives it with 5 volts, there are 2 volts leftover that have to be used by a resistor ( if a device regulates the LED current, the resistor isnt needed.) it takes knowing #2 & #3 to pick a resistance. 3) The LED has a rating called If - this is the continuous forward current. If the LED is driven to ON, and stays on for several seconds or more, you want to keep the LED under this rating to not eventually burn it out. to pick this resistor if there are 2 volts leftover for the resistor to use, divide by the If ( in mA) to pick the resistor in kilo-ohms. For example, if If = 20 mA, 2 V / 20 mA = .1 kohms ( that is 100 ohms ) for prototyping arduino projects, I generally just use a 220 ohm resistor for indication. 4) If the LED is used in a for short times like in a scanned or pulsed circuit, there is a maximum value for current called Peak Forward Current ( for example 1/10th duty cycle, 0.1ms pulse width) where the resistor can be sized using the peak current value that is much higher, for example 70 mA ( 3 1/2 times the continuous rating ) and the resistor value can be much smaller to allow this peak current.