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Suppose I have a DC motor and I put a load on it (in fact, the loading part isn't quite necessary for my question) but I have a load on the motor shaft and I connect it to a power supply. I can calculate what the final rotational speed of the motor would be given all the parameters (voltage of power supply, armature resistance, ke, etc) and I know that the torque of the motor would eventually equal to the loading torque (if loaded, 0 ideally if not loaded) and the torque would get driven down with time due to the back emf from the motor, but my question is this: what would be the start up torque of the motor? If there was no load on the motor, then a finite start up torque would be needed to accelerate the motor shaft due to its rotational inertia, but what would this value be?

I can derive a torque speed characteristic for the motor because I have all its parameters, so would the torque just on start up be the same as the motor's stall torque (because both lie at omega = 0 on the torque-speed characteristic curve)? The same goes for loading: what would be the torque just as the motor starts (not just to start the motor)? I have a section here in my lecture notes that's implying the startup torque is equal to the loading torque but that makes no sense to me: as the motor shaft accelerates, the back emf would drive the current (and therefore torque) down, so surely the torque on startup should be greater than the load torque (otherwise it'd be impossible for the motor shaft to accelerate). I guess my question is, is there a way to calculate the torque on startup using DC analysis and the specifications of the motor? Thanks for all your help.

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  • \$\begingroup\$ The fact that the motor is at standstill does not imply zero torque. The static load could be X [Nm] and the airgap torque could cancel it out if it were also X [Nm]. Many loads have the load torque proportional to the square of the shaft speed. Hence, the load torque is zero at standstill. This means that your DC motor can develop any torque equal to \$K_t I\$ up to rated winding current \$I_{rated}\$. The startup torque of the motor would be equal to \$V_{dc}/R * K_t\$ [Nm]. If the motor windings are disconnected and the speed is zero, this implies zero load torque. \$\endgroup\$ – SunnyBoyNY Oct 21 '15 at 1:11
  • \$\begingroup\$ Assume it's pretty much the same as stall torque, unless you have means of limiting the current at startup, such as a soft start circuit. \$\endgroup\$ – Brian Drummond Oct 21 '15 at 13:22
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I have a section here in my lecture notes that's implying the startup torque is equal to the loading torque but that makes no sense to me.

And you are right. The airgap torque has to be higher than the load torque in order for the machine to accelerate. Alternatively, the airgap torque has to be lower than the load torque for the machine to accelerate in the backward direction.

Just the fact that the motor is at standstill does not imply zero torque. It implies that the load torque is equal to the airgap torque, zero or non-zero.

The static load could be X [Nm] and the airgap torque could cancel it out if it were also X [Nm]. Many loads have the load torque proportional to the square of the shaft speed. Hence, the load torque is zero at standstill. This means that your DC motor can develop any torque equal to \$K_t I_{winding}\$ up to rated winding current \$I_{rated}\$. The startup torque of the motor would be equal to \$T_{start} = V_{dc}/R_{winding} * K_t\$ [Nm]. Note that the current \$I_{winding}\$ would likely be higher than the rated current if not actively controlled.

If the motor windings are initially disconnected or shorted (provided non-zero winding resistance) and the shaft speed is zero, then such state implies zero load torque.

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With no load coupled to the motor, the torque applied to accelerate the motor's inertia is the torque of the torque-speed capability characteristic. The current will be whatever current corresponds to that curve. If that is too much current, it is necessary to limit the current and reduce the torque by reducing the voltage or putting some resistance in series with the motor.

With a load coupled to the motor, the torque applied to accelerate the motor and load inertia is the torque of the motor's torque-speed characteristic minus the torque required to drive the load at a given speed without accelerating the inertia. That is the load's torque-speed requirement characteristic.

Here is an illustration for an induction motor. The shape of the curve for a DC motor is different, but the principle is the same.

Added detail: The motor produces the torque given by its characteristic curve whenever it is not operating at the intersection of the motor curve and the load curve. The difference between the torque given by the motor curve and the torque given by the load curve is applied to accelerate (or decelerate) the load to the intersection of the two curves. That is the normal operating point. The diagram has been revised to show that more clearly (I hope).

enter image description here

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  • \$\begingroup\$ Thank you for your response, Charlie. Could you clarify what you mean by "the torque of the torque-speed capability characteristic"? \$\endgroup\$ – AtticusFinch95 Oct 21 '15 at 0:58
  • \$\begingroup\$ The torque at any given speed is the torque on the motor characteristic curve for that speed. I will try to make a better illustration. \$\endgroup\$ – Charles Cowie Oct 21 '15 at 1:03
  • \$\begingroup\$ I see. So, just on start up for (say no load), the electromagnetic torque developed within the motor would be the torque on the torque-speed characteristic corresponding to 0 rotational speed (the same value as the stall torque once the motor reaches its maximum rotational speed)? \$\endgroup\$ – AtticusFinch95 Oct 21 '15 at 1:05

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