8x8 LED matrix using the smallest number of pins of my 8051

ahmad2005 · Oct 12, 2012

I'd like to create a 8x8 LED matrix using the smallest number of pins of my 8051. It'd be great to use two shift registers, one for rows and one for columns but it doesn't seem possible because according to my understanding shift registers can only source current, they cannot sink current.

How is this possible? I'm open to alternative solutions as long as the pin count stays as low as possible

Note: One of my friend request me to post this message here.

FoxyRick · Oct 12, 2012

ahmad2005 said:
according to my understanding shift registers can only source current, they cannot sink current.

Who told you that? Take a look at the 74HC595 datasheet:

www.nxp.com/documents/data_sheet/74HC_HCT595.pdf

You'll see on page 6 that the maximum output current is +/-35mA - so, it can source or sink 35mA.

The problem will be when you have more than one LED on and the 35mA is exceeded on the shift register that is connected to a full row of LEDs. You'll need to buffer it with transistors.

bigdogguru · Oct 12, 2012

It is certainly possible to utilize a shift register like the 74HC595 in the design of an LED matrix:

LED matrix using shift registers

**broken link removed**

Introduction to 74HC595 shift register – Controlling 16 LEDs

Another option is the use of I2C/SPI port expanders, like the Microchip series of devices:

Microchip Port Expanders

BigDog

8x8 LED matrix using the smallest number of pins of my 8051

ahmad2005

Member level 1

FoxyRick

Advanced Member level 4

bigdogguru

Administrator

Similar threads

8x8 LED matrix using the smallest number of pins of my 8051

ahmad2005

Member level 1

FoxyRick

Advanced Member level 4

bigdogguru

Administrator

Similar threads

Privacy & Transparency

Privacy & Transparency