I studied electromagnetics and with that understanding when I recall the Working of DC Series motor I stumbled upon an inherent negative feedback action in the DC series motor during starting as follows:

Considering the DC Series Motor is started with No Load:

1) At the instant of powering up the motor armature current is low (Transient)

2) Implies the current in the field coil is low

3) So the flux due to field coil in the air gap is low

4) The armature starts rotating because of Lorentz force

5) By faraday's law- the Back Emf is low as the air gap flux is low

6) So the armature current increases as per KVL

7) This, in turn, increases the torque on the armature.

8) However simultaneously the current through field coil also increase(since Ia =If )

9)The FEEDBACK action ----Because of Step 8 the flux increases and thus back EMF induced increases which in turn reduces the armature current. Thus the Torque on armature also reduces and so the speed also inturn reduces!!

Isn't this a negative feedback action? Won't this stabilise the DC series motor at some operating speed?

Why it is said that motor speed increases drastically and reaches un-stability when started with no load?


At startup the speed is low (obviously zero initially) so the back emf term is also small, and the current is largely determined by the motor (plus source) resistance, once it has risen against the inductance of the field, which is typically a lot higher than that of the armature windings. Since both field strength and armature current are high after the starting transient, a large torque is available to accelerate the motor.

As the motor speed rises, the back emf approaches the supply voltage, and the current, and field strength fall. This requires the armature to continue to accelerate - though the available torque to do so is falling - to maintain the emf. Eventually the motor can get to an equilibrium where the torque generated is equal to the friction and windage torque, but this is often a dangerously high value. Once it is at that point, determined by bearing and brush drag plus windage, it'll stay there until something fails. On small motors that's often the commutator that bursts. On larger motors, the commutators tend to be better made than the small molded ones, and it's often the endwindings that will 'fling' - expand outwards outwards until they smash against the field assembly.

On portable power tools, which use a universal series motor, the cooling fan is often sized such that it determines the no load speed as the load it draws increases with speed per the fan laws.

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  • \$\begingroup\$ Eventually, the motor can get to an equilibrium where the torque generated is equal to the friction and windage torque, but this is often dangerously high value. Is it like though the torques generated is equal to friction and windage torque, which is low, because of the machine's inertia the machine takes time to slow down on its own and rotates in that dangerously high speed for a longer time causing issues in the bearing? @Phil G \$\endgroup\$ – VKJ Jul 17 '19 at 17:16
  • \$\begingroup\$ The inertia determines only the rate of acceleration, but plays no part in the eventual balance of torques. Once it is at that point, determined by bearing and brush drag plus windage, it'll stay there until something fails. \$\endgroup\$ – Phil G Jul 17 '19 at 18:00

As you know (because you asked the question), series wound DC motors will continue to speed up until they self-destruct or they reach an equilibrium with their load, like @Phil mentioned. So why does this happen?

Going from statement 5 to statement 6, your logic broke down when you jumped from the present moment "low back-EMF" to calling the resulting high current an increasing trend. Yes it is true that there IS high current when it is first starting (hence using them for high startup torque applications like engine starter motors), but as you know, a second later when the motor physically speeds up, the back-EMF also increases. The increasing back-EMF REDUCES the current in the armature and so statement 8 is reversed: If = Ia, so the field actually weakens. The weaker field means the new back-EMF is lower, so now at this present speed and present current, the motor still hasn't hit its equilibrium where supply voltage = back-EMF (assuming no load), so it speeds up! And this keeps going up and up until the small remaining torque balances whatever friction and windage losses the motor is experiencing.

You ever watched two friends try to one-up each other's stories? That's like how the field and armature enable each other to reach new fascinating and literally unbelievable heights.

*An interesting note about series DC motors is that under certain circumstances, the torque is proportional to the square of the armature current unlike other motors where the field strength is fixed so the torque is directly proportional to current. Here's a good writeup on the various characteristics of series DC motors: http://www.engineeringenotes.com/electrical-engineering/electric-motors/dc-series-motors-characteristics-and-performance-electrical-engineering/36546

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  • \$\begingroup\$ Is this new understanding of the concept, the right one: If the flux is low, the steady-state speed is dangerously high, But when loaded the load continuously tries to reduce the speed of rotation, so the back emf tries to falls and thus part of armature current is always present due to loading. Because of this, some flux is always present (Since Ia=If). So a speed lesser than dangerous speed is enough to produce Back EMF = supply Voltage(steady-state). SO A safer steady-state speed is achieved!! Am I right? @Kent Altobelli \$\endgroup\$ – VKJ Jul 17 '19 at 18:05
  • \$\begingroup\$ Almost! Yes: low field strength = high speed. And yes, if you load the motor, it slows down, but this is not because back-EMF = supply voltage. Driving a load requires a certain amount of torque, and in motors (generally speaking) current = torque. So the load is essentially demanding a certain amount of CURRENT. When you apply a voltage to your spinning armature, some of it is opposed by the back-EMF but the rest is consumed by the resistance of the motor windings (Vsup = Vbemf + Ia*Rmotor). Constant load = stable current = stable field strength = stable back-EMF = stable speed. \$\endgroup\$ – Kent Altobelli Jul 17 '19 at 18:44
  • \$\begingroup\$ To clarify, current does not literally equal torque... Torque = k * flux * Ia, so specifically for a series motor, the high starting current saturates the field coil (so flux is approximately constant) and torque is proportional to current. With lower motor currents, the field coil will likely weaken to the more linear operating range (electromagnet where current is proportional to flux) so the torque will actually increase with the square of current. The previously stated concept is still the same but this is a detail of series motors. \$\endgroup\$ – Kent Altobelli Jul 17 '19 at 19:46

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