Why no posts here?

[Obelix]

What, nobody building keyboards?

N-e-Way, so if the "power on" pin (KWCV) is internally pulled up then why would it need to be terminated by the specified 470k/100k voltage divider? Why should it not be left N.C. when unused?

[Guido Körber]

Check the data sheet section 4.3 it defines what the pin is for.

During power up it is used to determine if the chip is connected to USB or some other interface.

[Obelix]

The data sheet specifies it as being internally pulled up, implying it is active low, if left unconnected shouldn't the KW reliably default to one state?

What else would be the purpose of an internal pull up?

[Admin]

I just noticed that we have a numbering error in the data sheet paragraphs...

Here is a quote right from the data sheet that contains all the answers:

PowerOn
Input for power key, also used to determine if device should suspend on power up.
Connect a switch closing to ground here to generate the power key scancode. An internal pull up resistor is activated upon device reset.
At power up KeyWarrior checks this pin before activating the internal pull up. If this pin is tied to +5V with 100K? or less KeyWarrior enters a suspend mode which it leaves only when detecting USB bus activity.
When the pin is pulled to Gnd KeyWarrior goes active immediately upon reset. Do not tie this pin directly to Gnd, always use a pull down resistor of 100K? or more, even if no power key will be used.

[Obelix]

So instead of the external divider it would be appropriate, when unused, to use only the 470k ( >100k) resistor from this pin to ground?

[Guido Körber]

Yes. The two resistors in the standard circuit are actually not intended as a voltage divider, the 100k pull up gets power only if the USB is connected and due to the nearly 1:5 ratio the chip then sees a high on this pin.

[Obelix]

Thank you for clearing that up and for your time.