I am working up a circuit to drive an array of LEDs using shift registers and a darlington driver. The entire circuit is powered by a 7805DT 5V 1.5Amp voltage regulator for clean power to my micro controller. The uC drives the input to the shift register and the MIC2981.

The LED anodes ( 8 columns ) are driven by a MIC2981 which claims to source 350 mA, and the cathodes are driven by the shift register HEF4794B which can sink 40mA per pin. This works well except when driving the segments of a 7 segment display 67-1486-ND. This display claims 25mA max and has a 4V forward voltage, so something like a 56Ohm resistor per segment should be good.

However, when I hook it up, with the MIC2981 driving the common anode, and one pin per segment of the shift register on the cathodes, with a 56 Ohm resistors, the LED is barely visible. Removing the current limiting resistors entirely still does not make the segments bright enough.

However, when I hooked up the Vin of the MIC2981 to my "dirty power" ( 6V source to the voltage regulator ), then the LED lit well with 56 Ohm resistors. Seems like the 1.5A regulator should handle everything and the 5V supply should work, I am obviously missing something here... What should I check, what am I doing wrong, any ideas?

 

It sounds like you may be running out of voltage headroom. With a 4V forward LED running from only a 5V rail, you have only 1V left over to share between the cathode driver, anode driver, and ballast resistor.

The voltage loss of the MIC2981 looks pretty steep - 2V? http://www.micrel.com/_PDF/mic2981.pdf

 

Yep, you're going to get a minimum of a 1.3v drop across the MIC2981 even with very little current, as the MIC2981 has a Darlington follower in the output stage (see the schematic on the bottom right of the 1st page of the datasheet Adjuster linked to.)

5v - (1.3v+4v) = -0.3v. Oops!

 

Your "dirty" power is only 6V but a 7805 regulator needs at least 7V. The output of your 7805 is probably only 4V.