Here’s how the stepper drive voltage affects the current rise, using that kludge to sync the scope on one of those motors with L=2.6 mH and R=2.2 Ω. The peak winding current is 1 A, so the first step current-limits at 200 mA.

At 9 V:

Current Rise - 9 V 1A 3 RPS

At 18 V:

Current Rise - 18 V 1A 3 RPS

Knowing the rise time and current change, you can calculate the actual voltage across the inductor using:

VL = L di/dt

With 9 V drive the motor sees:

4.4 V = 2.6 mH x 220 mA / 130 us

With 18 V drive the motor sees:

14 V = 2.6 mH x 240 mA / 45 us

So, in round numbers, the driver MOSFETs, winding resistance, and all the crappy solderless breadboard connections soak up about 4 V of the available supply voltage. There’s some back EMF in there, too, but I haven’t measured that part of the puzzle yet.

The motor is turning at 3 rev/s in 1/8 microstepping mode, so each microstep is:

200 us = 1/(3 rev/s x 1600 step/rev)