This is a LED driver IC. It says it can deliver 20V output, but what does that mean? That the output pin will have 20V in respect to GND? I am asking, because the max supply voltage is 7V, so does it have an amplifier inside?
Link to datasheet: http://www.farnell.com/datasheets/1837098.pdf?_ga=2.60289641.55768663.1518169214-1262566742.1517153785
Answer
It's not delivering 20V, this is where you are getting confused. It's a constant current sink device which means that current will flow into the device and you can fix how much current will flow into the device (between 5mA to 100mA per output in the case of this part). I do see how this is a little difficult at first. It can almost seem like it's an input rather than an output at first glance.
Now, the voltage at each output pin can be up to 20V. So, after the volt drop across each LED in the chain from the supply voltage to the pin of the output, you can have up to an absolute maximum of 20V on the pin before you would damage the device.
In the diagram it shows a typical application. If you drop 2V across an LED, and you had a VL of 22V, you'd have 20V on the output pin.
In practice, you would chose a much smaller VL if you only had one LED per output. However, it does give you the opportunity to have lots of LEDs in series on some strings and fewer on another.
Also, please bear in mind that the larger the voltage at the output pin, the bigger the power dissipated in the package; you have to be really aware of how much you are dissipating per output and what the absolute overall maximum power dissipation is for the part. Just because you can sink up to 100mA and have a voltage of up to 20V, does not mean you can necessarily have both at the same time!
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