DC motor voltage divider with NPN, diode, capacitor, and separate battery
I just pulled a couple of DC motors off the shelf and wired them up to \$6\:\textrm{V}\$ DC and found that the current varied from perhaps \$100\:\textrm{mA}\$ (under little to no load) to about \$500\:\textrm{mA}\$ (when I held my fingers relatively tightly on the shaft as it rotated.) So that gives me a ballpark for the required current.

A typical \$9\:\textrm{V}\$ alkaline battery is from \$1\:\Omega\$ to \$2\:\Omega\$ of parasitic series resistance, depending on its state of charge. In addition, the battery's voltage will itself also vary quite a bit depending on its state of charge. But they are capable of supplying the level of currents I mentioned above. So, from some modest reasoning, I think you may make this work with these batteries.

The following schematic is not for anything but playing around with your motor. I'm going to assume that your battery will typically be providing about \$8.6\:\textrm{V}\$ under some load, when fresh. (I now know you have a TIP120 Darlington transistor.)

This circuit does not support being connected to an I/O port of a microcontroller. It is purely for playing with the motor, directly.







simulate this circuit &ndash; Schematic created using CircuitLab

Use a large valued capacitor for \$C_1\$, consistent with it being able to tolerate at least \$15\:\textrm{V}\$. Small values won't hurt, though. For \$C_2\$ your \$100\:\textrm{nF}\$ should be more than fine.

The point of \$R_1\$ and \$R_2\$ is a voltage divider, not so different from what you already show in your own schematic. The difference here is that the divided voltage is applied to the base of the TIP120. The emitter will "follow" this voltage, dropping about \$1\:\textrm{V}\$ to \$1.2\:\textrm{V}\$ from there. Because of the very high current gain of the TIP120, you can expect that the base current will be relatively small compared to the motor current. So it won't impact the voltage divider's voltage too much. Feel free to adjust \$R_1\$ or \$R_2\$ a little, as needed, to get the right voltage to the motor. But the values I show there should be not too far off.

If you have a potentiometer to add, feel free to wire it into the voltage divider circuit so that you can vary the motor voltage with it.

Use a voltmeter, if you have one, to observe the voltage between the negative terminal of your battery and the emitter of \$Q_1\$.

The idea of the above circuit is foremost to be simple and easy to wire up and where you can vary the value of \$R_1\$ and \$R_2\$ a bit, perhaps. You should be able to arrange things with those resistors in order to get close to what you want, for some time while the battery is "mostly charged."

Please do NOT attempt this with a small, TO-92 packaged BJT. It will simply get hot, very fast, and probably just burn up and fizzle out on you in seconds, smoking a bit after. Not a good thing. You want a larger BJT for this task. The TIP120 you have should be fine, though. I'd probably add some metal to its tab, just so it stays a little cooler. But that's up to you.

\$C_2\$ will help absorb some spikes from the motor. That's the main reason I added it. Value isn't critical. But it should be around the size I mentioned (and that you seem to have), or a bit larger. You could add that reversed diode you had in your circuit if you want. I think \$C_2\$ is fine enough. Up to you, though.

Mostly, I'm hoping you can enjoy playing with this and see how it works for you.