Reducing the voltage of a PWM signal
The easiest to think of would be the resistive divider you suggest.

To get a good dynamic output you need to not make the resistors too large, since the internal schematic shows a 50k resistor from the reference, so using 110k and 330k would mean you cannot pull the CTRL pin close enough to 0V to really consider it PWM-ing:







simulate this circuit &ndash; Schematic created using CircuitLab

I added a Zener diode there to hard-clip any erroneous signal glitches to something near 1.4V, so that you won't instantly melt your devices if something happens you didn't foresee.

You want 1.25V, which is exactly 1/4th of 5V, so if the total resistance is R, then Rd (for R-pull-down) needs to be 1/4th of R in the divider, so that 1/4th of the source signal falls across it.

So Ru (for R-pull-up) needs to be the rest, or 3/4th of R.

Which means Ru = 3*Rd.

So 1.1k and 3.3k would work.



But funny thing:

Did you notice me mentioning that resistor earlier?

That pulls the CTRL pin to 1.25V all by itself, no worries. So why not try this:







simulate this circuit

Whenever your PWM signal is high, the transistor is on, pulls the CTRL low to 0V. Whenever it it low, it lets the pin go, and the internal resistor will pull it up to 1.25V.

You may see a small RC-curve bump at the peak current, created by the 50k resistor and trace and transistor capacitance, but it should be very small (possibly even unnoticeable) in almost all normal use cases.

Of course, that will invert your PWM effect, but in an MCU (at least the Atmels I know from experience) that's very easy to flip.