Here is a schematic of the ULN2003 being used to drive the 28-byj48 stepper motor (unipolar).
Here is a schematic of the internals of the ULN2003:
So basically, power (5V) is delivered to the stepper by the red line (connected to the COM of ULN2003). Drive signals from the microcontroller goes into the IN1, IN2...IN4 pins of the ULN2003, which allows current to flow through the respective channels of the Darlington. This is the current that flows through the motor and allows it to turn. This current goes from the OUT pin of the ULN2003 and flows to the GND pin of the ULN2003.
The OUT pin is essentially the collector, and the emitter is essentially the ground according to the ULN2003 schematic. There is a free wheeling diode at the COM and its purpose is to allow current to flow back to motor in a cycle once current gets shut off in the channel. This is because the motor has inductance and current cannot drop to zero instantaneously.
Did I understand it correctly?
Thanks