op-amp vs direct control

Question

I'm experimenting with small motors (just building useless toys so far) and I'm wondering what benefit there would be to using an op-amp to control a motor (top circuit) vs just controlling the motor with a pot directly (bottom circuit).

^{simulate this circuit – Schematic created using CircuitLab}

EDIT: please ignore the exact component values; 100 ohm and TL081 is just the circuitlab defaults that I didn't change -- I'm just thinking about the general setup of active vs. passive components for the control.

Answer 1

Referring to the schematics in the question, there are unlikely to be any advantages to using an op-amp, and quite possibly a number of disadvantages.

First, your op-amp schematic is drawn without feedback and with the negative input grounded. Due to the high DC gain of the op-amp, the output will most probably be saturated at either the positive or negative rail (even if you can adjust the potentiometer to just the right position, the resistor's temperature coefficient will probably make the circuit too unstable to be usable).

Second, although the op-amp can output a voltage close to the supply rails, the potentiometer circuit allows for slightly higher voltage.

That said, there are ways you could use an op-amp as a buffer to drive the motor, which would reduce the fluctuations in the motor voltage due to load on the motor. This is because the back-emf produced by the motor will fall as the motor is loaded, and thus the current drawn will increase. Without the op-amp, this changes the effective resistance of one leg of your potentiometer voltage divider.

Answer 2

A few problems with the op amp circuit:

• Depending on your motor, its highly likely that the op amp cannot source enough current to drive it.
• The op amp is running open loop. The output will almost always be saturated at the positive rail because the inverting pin is pulled to common. You may have a few tenths of a millivolt of control, depending on the open loop gain of the op amp.
• \$100\Omega\$ is generally a low resistor value to be using with op amps. Use something in the \$k\Omega\$ range.

The potentiometer solution isn't very good either. Putting a low impedance load on the wiper of a pot like that will result in nonlinear taper. Only high impedances should be driven from the wiper of a pot.

Answer 3

Some points:

• In the non-op-amp circuit, as you turn the motor near maximum voltage, a lot of current flows through a smaller and smaller segment of the pot. So it's a good idea to think of something better, indeed.

• 100 ohm resistors to ground are wasteful of energy; you don't want on battery power. If you have the op-amp in place, the potentiometer value can be a lot larger, like 10K or more. The amplifier isolates the impedance of the pot from the load, so not having to use a tiny potentiometer is one benefit.

• Not all op-amps can go to the negative rail. (Certainly not the TL081 that you have nominally stuck in the schematic). A better part for this would be an op-amp designed for single voltage supply. You need that for turning the potentiometer all the way down to 0V. But ...

• Not many op-amps can drive such tiny load impedances. You can use the op-amp for voltage gain, but then follow it with a power transistor output stage to deliver the current to the load. Or else you need a power amplifier IC.

• Since the goal here is to drive a motor with a variable voltage DC power supply, it would be better to build ... a variable voltage DC power supply! For instance with an LM317 voltage regulator. Right tool for the right job.

• A much better idea is pulse width modulation to control current to a motor, especially on battery power. Pulses generated by an oscillator (built with, e.g. a 555 timer IC) rapidly turn a MOSFET on and off, which passes current to the motor. Why don't you build a PWM circuit using a 555 timer. Google potentiometer PWM 555 timer motor.