First I'd like you to excuse me if this is a very basic question but I'm just a beginner.
I'd like to reduce the brightness of a led that indicates me that a circuit which is activated by a TTP223 board is in stand-by mode. I do this indication using the TTP's output as sink.
But in this topic I've found that post:
"IOH is spec'ed as VDD=3V and VOH=2.4V means you can pull -4 ma out of the pin, maximum. This would be current pulled through the device, out the pin, into your load. This would be what you'd get if you attached a resistor from the output pin to gnd.
The IOL spec is higher at 8 ma which means it can absorb more than it can push. Think of a resistor with one end to the output pin, the other end attached to VDD.
Not that you'd do this, but you could calculate the maximum values for your resistors this way:
Sourcing: IOH=-4ma, VOH=2.4. R=E/I so 2.4/.004 = 600 ohms
Sinking: IOL=8ma when VDD=3V, VOL=0.6V. VDD-VOL=2.4V/.008 = 300 ohms"
In my case the calculation is:
VDD=5v, VOL=0.6 ; VDD-VOL=4.4v ; 4.4/0.008 = 550 ohm
But I've got the ideal led's brightness with a 2k2 ohm resistor...
May I use this 2k2 resistance for that?
And just trying to understand this TTP's sink resource, as merely curiosity, as the recommend resistor for this led is 140 ohm (as calculated here), if I use less resistance than the 550 ohm, could that demage the chip?