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

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ahmad2005

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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.
 
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FoxyRick

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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.
 

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