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I have a 12 V DC power supply and a 9 V DC motor available to me. I want to use the 12 V power supply to power the 9 V motor, and I expected I could do this with voltage division.

I built this circuit, and tested it to confirm that there is 12 V between power and ground, and tested across the 220 k resistor to confirm there was 9 V across it. When I connect the motor across the 220 k resistor, the motor does nothing and when I test across the resistor it now says 0 V. Why does this happen? How can I change my circuit to make the motor spin?

There is a circuit diagram, I was connecting the motor between nodes labelled 'A' and 'B'.enter image description here

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    \$\begingroup\$ Because a voltage divider has far too high an output impedance. \$\endgroup\$
    – Hearth
    Jan 15 at 2:34
  • \$\begingroup\$ The motor current will flow through the 73.4K resistor, increasing the voltage drop across it. If you draw any significant current from the mid-point of a voltage divider, you WILL change the voltage at that point. \$\endgroup\$ Jan 15 at 2:45
  • \$\begingroup\$ Have you learnt KVL yet? The motor coil in small ohms is pretty small, so the best method is to use a FET switch with much lower Ron than the coil resistance and pulse it fast with on 9/12 ratio. Even better if you start by accelerating with a small ratio to reduce the surge current, \$\endgroup\$ Jan 15 at 2:49
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    \$\begingroup\$ Measure the resistance of your motor. \$\endgroup\$
    – Transistor
    Jan 15 at 2:49
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    \$\begingroup\$ KaraLT - Hi, Your question is effectively a duplicate of this previous question: "When would I use a voltage regulator vs voltage divider?" as that explains the problems you are having and why. However it doesn't answer the exact points that you have (it explains the theory so you could apply it yourself, but I don't know if you can do that). If that linked question does allow you to answer your question, then we can close this as a duplicate. As a mod my vote would be binding, so I won't yet vote to close as a duplicate, in case others want to answer. \$\endgroup\$
    – SamGibson
    Jan 15 at 2:58
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A resistor divider does deliver a voltage similar to the one you desire, just like you say and found out. But it also presents a series resistance, too. That series resistance then stands in the way of your motor.

Here's a circuit that is the equivalent to your constructed circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

It's that inserted equivalent resistor that is preventing the motor from running.

The value for the equivalent series resistor (\$R_1\$ above) is just the parallel equivalent of the two resistors you used in the divider. You could redesign the resistor values to get a lower value. But then you'd also be wasting more power and may have to buy physically larger resistors. In general, you don't get to win this game. It's almost always a losing proposition. Especially when discussing motors.

The next step up in complexity is to use a BJT so that the effective resistance is driven down. It looks something like this:

schematic

simulate this circuit

BJTs have something called \$\beta\$ ('beta') that boosts the current compliance (more capable of supplying current to the load.) Put another way, the circuit above with the NPN BJT can be mentally replaced with the first circuit I mentioned above, but where the value of \$R_1\$ is divided by approximately \$\beta\$. And since modern values of \$\beta\$ can be quite large (in the hundreds), the equivalent series resistance is far lower than without the BJT in place.

Doing this allows the actual resistor divider to concentrate on the job of setting the voltage you want without excessively wasting power, while allowing the NPN BJT to pass along a lot more current if the motor requires it. It's slices up the job into two parts, handing off one part to the resistors, just as you already knew about, and handing off the other part -- delivering current that bypasses those resistors -- to the BJT.

(A small note. The design voltage for the resistor divider in this BJT circuit needs to account for the small base-emitter voltage of the BJT -- likely less than one volt -- setting a slightly higher target voltage for the divider so that the desired motor voltage is achieved.)

Will just one BJT be enough? I've no idea. All I've done here is to extend your idea by adding just one part to improve it. Likely, the situation may require still more thought.

You've said nothing much about the motor, itself. If it is a toy motor, then even a relatively small BJT may be able to handle the job. But if it is a treadmill motor, then you are in for a much more complicated circuit.

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