A small problem about driving RGB LED Matrix

[CanOzbek]

Hello,

I am trying to drive an 8x8 RGB Common Cathode LED matrix. I am planning to use four 74HC595 shift registers. The problem is that I want to get the maximum brightness out of the LEDs but I do not know which IC to use between output of shift registers and LEDs.

I am using 5050 smd RGB LEDs (https://dlnmh9ip6v2uc.cloudfront.net/datasheets/Components/LED/5060BRG4.pdf)

I know that shift registers provide 20mA and 20mA is sufficient for the LEDs, however as 8x8 LED matrix is driven by one column at a time, the brightness is diminished by 8, so I need more current. (That's why in the datasheet of the LED there is a parameter called : "Peak pulsing current (1/8 duty f=1kHz) " I think )

The schematics of the driver circuit and the 8x8 matrix are :
**broken link removed**
**broken link removed**

What should I use in the box with the question mark in the schematic of the driver circuit? I am planning to use UDN2981 for the anodes and ULN2803 for the cathodes of the LEDs. Would this configuration provide me maximum brightness? I would appreciate if you could help me.

Thanks.

 

[bluelasers]

The shift registers can only provide 6mA per channel if 74HC, and 8mA per channel if 74AHC. One channel can provide 35mA, but you should assume all channels active at once, which would mean that the chip would die if used beyond the value listed earlier. If the chip does not die instantly it will have a reduced life, as it is operating outside of specification. Max source current is 70mA. Some current is used for the internal logic and registers. To get the full current you will need source and sink transistors. You can use darlington arrays, but I wouldn't. While they are compact, they are expensive and have a high voltage drop.

I would use three shift registers, 24 P channel transistor, with 24 current limiting resistor for the columns. The P channels transistors only have to source 20mA so just about any should do. For the rows I would use one shift register with 8 N channel transistors. These transistors have to be strong. They could have to sink 20*8*3=480mA. So a ULN2803 may be better here. You can still drive it with a shift register.

This circuit would be easier if you had a common anode matrix. I hope this helped.

 

[skenn_ie]

Some years ago, I was the engineer and technician who soldered the stuff. I used TPIC6B595 shift-register-drivers as the sink-side driver, with simple transistor-from-demux as the source.