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Back-EMF and the motor constants Ke and Kt

A spinning motor is also a generator: it produces a back-EMF E = Ke × ω that opposes the supply, so the armature current is I = (V − E) / R and falls as the motor speeds up. Torque is T = Kt × I, and in SI units Kt and Ke are numerically equal.

Back-EMF: the motor as its own generator

When the rotor of a DC motor turns, its conductors cut the magnetic field and a voltage is induced in them. By Lenz's law this voltage opposes the current that is causing the motion, so it is called the back electromotive force, or back-EMF. It is proportional to speed:

E = Ke × ω

where ω is the angular speed in rad/s and Ke is the voltage constant in V·s/rad. Catalogues often quote it in V per 1000 rpm instead; to convert to V·s/rad, multiply the V per 1000 rpm value by 0.009549.

At standstill the back-EMF is zero. As the rotor accelerates, it rises until it almost balances the supply voltage.

The torque constant Kt

The torque a motor produces is proportional to its armature current:

T = Kt × I

where Kt is the torque constant in N·m/A. Both constants come from the same physical quantities (the field strength and the winding geometry), so in SI units they are the same number:

Kt (N·m/A) = Ke (V·s/rad)

A direct proof comes from energy conservation. Electrical power converted to mechanical power is E × I, and mechanical power is T × ω. So Ke × ω × I = Kt × I × ω, which gives Ke = Kt. This is why a motor with a high torque per amp also generates a high voltage per rad/s, and why you cannot get both a high no-load speed and high torque per amp from the same winding.

The equality holds for permanent-magnet DC motors, for brushless DC motors when the same per-phase or line-to-line definition is used for both constants, and approximately for other machines in their linear region. It breaks if you mix units, for example Kt in N·m/A and Ke in V per 1000 rpm.

The voltage equation and why current falls with speed

Apply Kirchhoff's voltage law to the armature circuit, with supply voltage V, armature resistance R and back-EMF E:

V = I × R + E

so the current is

I = (V − E) / R

At the instant of start the rotor is still, E = 0 and the current is the highest, V / R. As speed rises, E grows, the voltage left across R shrinks and the current falls. The torque falls with it, until the motor settles at the speed where its torque exactly equals the load torque. A heavier load needs more torque, which needs more current, which means a lower back-EMF, so the motor settles at a lower speed.

This is a natural feedback loop. It is what makes DC motors self-regulating, and it is also why a small resistance R and a stalled rotor make a dangerously large current.

No-load speed and stall current

Two extreme points define the whole speed–torque line of a DC motor.

  • Stall (ω = 0): E = 0, so the stall current is I_stall = V / R and the stall torque is T_stall = Kt × V / R.
  • No load (T = 0): the current is zero, so E = V and the ideal no-load speed is ω_0 = V / Ke. A real motor draws a small current to overcome friction and settles a little below this.

Between the two, speed falls linearly with torque:

ω = V / Ke − (R / (Ke × Kt)) × T

The straight line connects ω_0 on the speed axis to T_stall on the torque axis. Raising the voltage moves the line up in parallel. Raising R makes it steeper, meaning that speed drops more under load. Maximum mechanical power occurs at half the no-load speed and half the stall torque, P_max = V² / (4R).

Worked example

Take a permanent-magnet DC motor with V = 24 V, R = 0.5 Ω and Ke = Kt = 0.05 (V·s/rad and N·m/A).

1. Stall current: I_stall = 24 / 0.5 = 48 A, giving a stall torque of 0.05 × 48 = 2.4 N·m. 2. Ideal no-load speed: ω_0 = 24 / 0.05 = 480 rad/s, which is 480 × 60 / (2π) ≈ 4584 rpm. 3. Now load the shaft with 0.5 N·m. The current needed is I = T / Kt = 0.5 / 0.05 = 10 A. 4. The back-EMF must be E = V − I × R = 24 − 10 × 0.5 = 19 V. 5. The speed is ω = E / Ke = 19 / 0.05 = 380 rad/s ≈ 3629 rpm.

Power balance: the supply provides 24 × 10 = 240 W. The shaft receives T × ω = 0.5 × 380 = 190 W. The remaining 50 W is the copper loss I² × R = 100 × 0.5. The efficiency at this point is 190 / 240 ≈ 79%, ignoring friction.

The same motor on 12 V would have half the no-load speed, 2292 rpm, and half the stall current, 24 A.

Practical notes

  • Measuring Ke: spin the motor with a drill or another motor and read the open-circuit voltage; Ke = E / ω.
  • Measuring R: lock the rotor and apply a small voltage; R = V / I. Measure at several rotor positions, because brush contact resistance varies.
  • Supply limits: real drives limit current, so the motor never reaches V / R at start. Without a limit, the starting current can be several times the rated current.
  • Temperature: copper resistance rises about 0.4% per °C, and magnets weaken when hot, so Ke and Kt drop slightly as the motor warms.
  • Regeneration: if the load drives the motor above V / Ke, then E > V, the current reverses and the machine delivers energy back to the supply.
  • Brushless motors follow the same equations per phase, with the inverter doing the commutation. The PWM duty cycle sets the average voltage V.

Try it in the brushed DC motor simulator

Open the brushed DC motor simulator, which starts with a permanent-magnet motor (Field excitation set to Magnet). The Back EMF, Torque and Armature current readouts show E, T and I, and the nameplate has the Armature resistance and Torque constant fields.

  • Set Mechanical load to 0% and raise Supply voltage: the Back EMF readout rises with the speed and stays just below the supply voltage while the Armature current stays small.
  • Increase the Mechanical load slider: the Armature current and Torque readouts rise, and the Back EMF and the speed drop. The Torque readout equals Kt × the armature current.
  • Increase Armature resistance and see the speed fall faster under the same load.
  • Push the Mechanical load slider to its 120% limit with a very low Supply voltage (below about 4 V for the default motor) to reach the Stalled state, where the speed and Back EMF are zero and the current is V / R.

The Field excitation setting lets you switch to Shunt, Series or Compound windings, where the flux depends on the field current and the simple Kt constant no longer applies. The Air-gap flux (rated = 100) readout then shows the change.

Try it in the simulator

Frequently asked questions

What is back-EMF in a motor?
It is the voltage generated in the rotating windings by their motion through the magnetic field. It opposes the supply voltage and is proportional to speed, E = Ke × ω.
Why are Kt and Ke equal in SI units?
Because both come from the same magnetic coupling between current, field and rotor, and because electrical power E × I must equal mechanical power T × ω. Dividing gives Ke = Kt when volts per rad/s and N·m per ampere are used.
Why does motor current decrease as the motor speeds up?
Because the back-EMF grows with speed, leaving less voltage across the winding resistance. Current is (V − E) / R, so it is highest at start and falls as the motor speeds up.
How do I find the stall current and no-load speed?
The stall current is the supply voltage divided by the armature resistance, V / R. The ideal no-load speed is V / Ke in rad/s. A real motor settles slightly below that because of friction.
How do I convert Ke from V per 1000 rpm to V·s/rad?
Multiply the value in V per 1000 rpm by 0.009549. For example, 5.24 V per 1000 rpm is about 0.05 V·s/rad.