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This is a follow up question to my question here about torque and stepper motors.

There Olin explains:

Note also that torque isn't the only criterion for a motor. Power is another important one. For that you have to decide what's the fastest speed you want to be able to pull the mass upwards at. Let's say 2 m/s for sake of example. From above, we know the highest upwards force is 5.8 N.

(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W

After accounting for some losses due to friction and leaving a little margin, the motor should be rated for about 15 W minimum.

I'm unclear about what kind of Watts we're talking about here, is this just Watts in the mechanical sense or in the electrical sense. In other words should I be thinking about needing to deliver 15W as in 15V @ 1A?

I'm also wondering about controlling the amount of power delivered to the motor. In steady state I guess it would be Input Voltage divided by motor coil resistance? But to change the position where the current and voltage are changing the current must be the input voltage / (reactance of the coil plus the coil resistance). But the reactance changes with frequency... so do I have to look at the rise time of the input voltage to understand the frequency involved and then try to calculate from there.

I'm a little lost on that point, but trying to read more and understand.

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  • $\begingroup$ If you put in 15W electrical power into an ideal motor then you will get out 15W of mechanical power. Thats kinda a thermodynamics law (conservation of energy). $\endgroup$ – ratchet freak May 15 '15 at 9:45
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In the text you quoted, I was talking about mechanical power out of the motor. This is what the power rating of a motor, given no other qualifications, means. A "15 W motor" therefore can put out 15 W of mechanical power under the right conditions.

No motor is 100% efficient, so more electrical power than that will need to be supplied to the motor. For example, if the motor is 85% efficient at a particular operating point where it produces 15 W of mechanical power, it will require (15 W)/85% = 17.7 W of electrical power. The remaining 2.7 W that doesn't come out in the form of mechanical power will be dissipated as heat.

Your electrical model of the a motor isn't quite right. Think of it as a resistor in series with a voltage source. Yes, coils are in fact inductive, but the time constant of the inductance with the resistance in the system is usually far less than the mechanical time constant and so isn't relevant.

The resistance is fixed and is the actual DC resistance of the coils. The voltage source is a function of the motor's speed, and always opposes the applied voltage when the motor is spinning in the direction that it is being pushed by the applied voltage. This voltage source represents the motor acting as a generator.

The torque produced by the motor is proportional to the current thru it.

When you first apply a voltage to a stantionary motor, the voltage source is 0 and all the applied voltage is across the resistance. This causes a large current to flow, called the stall current. As the motor speeds up, the voltage source opposes some of the applied voltage. That leaves less votlage across the resitor, which reduces the current, which reduces the torque. Eventually a balance is reached where the current drops to the point where it produces just enough torque to keep the motor going at that speed.

For a unloaded motor, little torque is required to keep it going, and the internal voltage source opposes most of the applied voltage. There is little voltage left across the resistor, so the current drawn by the motor is much lower than the stall current. If the motor is mechanically loaded, then a higher torque is required to keep it going, so the steady state speed will be lower, the internal voltage source lower, more voltage across the resistance, and the current higher.

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