[sdiy] thanks for replies - Blue LED

[Michael Bacich]

On Oct 22, 2006, at 9:49 AM, John Luciani wrote:

> A problem with using the output of a hex inverter to source and  
> sink current
> is the output impedance is too high. It may be hard to find an  
> inverter with
> a low enough impedance that also meets your low cost requirement.

One way to raise the available current drive on a typical CMOS hex  
inverter package is to drive several of the individual inverters in  
parallel and connect their outputs together via resistors.  This will  
divide the load up between the separate inverters and hopefully, keep  
the load within the device's spec.  If you would normally drive your  
LED via one 1K resistor, and you want to use three of the inverters  
in parallel, then use one 3k resistor at the output of each inverter  
and tie the other side of those three resistors together and connect  
that junction to your LED.  If you use all six of the inverters, then  
use one 6K resistor at the output of each inverter, and tie all six  
6K's together to drive the LED.

I don't know how this figures into your cost requirements.