Electrical
Back-EMF in Motors
The counter-voltage a spinning motor makes against its own supply
Back-EMF (counter-electromotive force) is the voltage a rotating motor generates that opposes the supply driving it, in direct obedience to Lenz's law. Because every motor is also a generator, the turning rotor induces a voltage E that subtracts from the applied voltage V, and this counter-voltage rises linearly with speed as E = Ke·ω. Armature current is then set by (V − E)/Ra, so a motor that is spinning fast draws little current, while a stalled motor with E = 0 draws locked-rotor current 5 to 10 times its rated value, limited only by the small winding resistance Ra. The same constant Ke (V·s/rad) equals the torque constant Kt (N·m/A) in SI units, which is why sensing back-EMF gives both rotor speed and — in three-phase machines — rotor position, the foundation of sensorless field-oriented control.
- Governing lawFaraday + Lenz (E opposes supply)
- Speed relationE = Ke·ω (volts)
- CurrentI = (V − E) / Ra
- Constant identityKe [V·s/rad] = Kt [N·m/A]
- Stall current5–10× rated (E = 0)
- SensorlessZero-cross / observer → position
Interactive visualization
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Watch the 60-second explainer
A condensed visual walkthrough — narrated, captioned, under a minute.
Why back-EMF matters
Back-EMF is the single idea that ties a motor's electrical behaviour to its mechanical behaviour. It is not a parasitic side effect — it is the very mechanism by which electrical energy becomes torque. Understanding it explains a long list of otherwise mysterious observations: why an unloaded motor barely draws current, why the same motor pulls a punishing surge the instant you switch it on, why a fan that jams burns out, and how a drone's electronic speed controller knows where the rotor is without a single position sensor.
- Current protection. Inrush and locked-rotor current are governed entirely by back-EMF being zero at start-up; motor protection, soft-starters, and fuse sizing all follow from it.
- Speed self-regulation. A DC or PMSM motor holds speed against load changes automatically because back-EMF is built-in negative feedback — no controller required.
- Sensorless drives. Field-oriented control in appliances, e-bikes, and EV traction inverters estimates rotor position from the back-EMF waveform, eliminating fragile encoders and Hall sensors.
- Regenerative braking. When back-EMF exceeds the supply, the machine pushes current backwards and returns energy to the battery — the same physics run in reverse.
- Efficiency and sizing. Mechanical output power is exactly E·I; a designer picks Ke to place the no-load speed and to trade copper loss against the needed speed range.
How it works, step by step
- Apply voltage, current flows. At the instant of switch-on the rotor is stationary, back-EMF is zero, and current is limited only by winding resistance: I = V/Ra. This is the locked-rotor current.
- Current makes torque. The current in the magnetic field produces torque T = Kt·I. With no back-EMF yet, torque is at its maximum (stall torque), so the rotor accelerates hard.
- Motion induces voltage. As the rotor turns, its conductors sweep through the field and generate a voltage. By Lenz's law this induced EMF opposes the current that produced the motion, so it subtracts from the supply: E = Ke·ω.
- Back-EMF throttles current. The armature loop equation is V = E + I·Ra, so the current becomes I = (V − E)/Ra. As ω climbs, E climbs, (V − E) shrinks, and current falls.
- Equilibrium. The motor keeps accelerating until the torque it makes, Kt·I, exactly equals the load torque plus friction. At that speed current is constant and back-EMF sits just below V — the machine has found its own operating point.
- Load change. Add load, the rotor slows, E drops, (V − E) grows, current and torque rise to meet the new load, and speed settles slightly lower. Remove load and the reverse happens. This is automatic negative feedback.
The governing equation
The steady-state and transient behaviour of a brushed DC motor (and, per-phase, a PMSM in the rotating frame) is captured by the armature voltage loop:
V = E + I·Ra + La·dI/dt, with E = Ke·ω and T = Kt·I
Symbol definitions and units:
- V — applied terminal voltage, volts (V).
- E — back-EMF (counter-EMF), volts (V); zero at standstill, rising with speed.
- I — armature (or phase) current, amps (A).
- Ra — armature/winding resistance, ohms (Ω); often 0.05–5 Ω.
- La — armature inductance, henries (H); sets the electrical time constant τe = La/Ra, typically 0.1–5 ms.
- ω — rotor angular speed, radians per second (rad/s); ω = 2π·n/60 for speed n in rpm.
- Ke — back-EMF constant, volt-seconds per radian (V·s/rad); equivalently V/(rad/s).
- T — electromagnetic torque, newton-metres (N·m).
- Kt — torque constant, newton-metres per amp (N·m/A). In SI, Kt = Ke numerically.
In steady state (dI/dt = 0), rearranging gives the two most useful forms. The current is I = (V − Ke·ω)/Ra, and the speed is ω = (V − I·Ra)/Ke. The no-load speed (I ≈ 0) is ω0 ≈ V/Ke, and the stall torque (ω = 0) is Tstall = Kt·V/Ra. These bracket the linear torque-speed line every brushed and permanent-magnet machine follows.
Worked example: a 24 V servo motor
Take a permanent-magnet DC servo with Ke = 0.05 V·s/rad (so Kt = 0.05 N·m/A), Ra = 0.5 Ω, run from V = 24 V.
- Locked-rotor current. At start-up ω = 0, E = 0, so Istall = V/Ra = 24 / 0.5 = 48 A.
- Stall torque. Tstall = Kt·Istall = 0.05 × 48 = 2.4 N·m.
- No-load speed. ω0 ≈ V/Ke = 24 / 0.05 = 480 rad/s ≈ 4,580 rpm.
- Running point under 0.4 N·m load. I = T/Kt = 0.4/0.05 = 8 A; E = V − I·Ra = 24 − 8×0.5 = 20 V; ω = E/Ke = 20/0.05 = 400 rad/s ≈ 3,820 rpm.
- Current fell from 48 A to 8 A purely because back-EMF climbed from 0 V to 20 V — a 6× reduction with no controller involved.
- Power split at the running point. Mechanical P = E·I = 20 × 8 = 160 W; copper loss = I²·Ra = 8² × 0.5 = 32 W; input = V·I = 192 W (efficiency ≈ 83%, ignoring iron and friction losses).
The table below shows how a real motor's terminal current and back-EMF track speed for this example.
| Operating point | Speed ω (rad/s) | Back-EMF E (V) | Current I (A) | Torque T (N·m) |
|---|---|---|---|---|
| Stall / locked rotor | 0 | 0.0 | 48.0 | 2.40 |
| Heavy load | 200 | 10.0 | 28.0 | 1.40 |
| Rated load | 400 | 20.0 | 8.0 | 0.40 |
| Light load | 460 | 23.0 | 2.0 | 0.10 |
| No load | ~478 | ~23.9 | ~0.2 | ~0.01 |
Ke and Kt: the same constant, two units
The equality Ke = Kt in SI trips up many engineers because datasheets rarely quote them in SI units. The identity comes from energy conservation: electrical power converted to mechanical must equal mechanical power out, E·I = T·ω, and substituting E = Ke·ω and T = Kt·I forces Ke = Kt. The table converts a single physical constant across the units you will actually see.
| Quantity | Symbol | SI unit | Common datasheet unit | Conversion note |
|---|---|---|---|---|
| Back-EMF constant | Ke | V·s/rad | V/krpm | 1 V·s/rad = 104.7 V/krpm |
| Torque constant | Kt | N·m/A | oz·in/A, mN·m/A | 1 N·m/A = 141.6 oz·in/A |
| Identity | Ke = Kt | numerically equal (SI) | hidden by unit choice | from E·I = T·ω |
| Winding resistance | Ra | Ω | Ω (line-to-line for BLDC) | sets stall current V/Ra |
| Electrical time constant | τe = La/Ra | s | ms | current rise, not speed |
Common misconceptions and failure modes
- "Back-EMF is a loss." The opposite — E·I is exactly the mechanical power delivered. The real loss is I²·Ra, and a big back-EMF means small current and low loss.
- "A stalled motor draws its rated current." It draws locked-rotor current, V/Ra, which is typically 5–10× rated. Sustained stall cooks the windings in seconds because there is no motion to convert power to work — all of V·I becomes heat.
- "Back-EMF fights the motor and wastes energy." It opposes the supply, not the motor's purpose; it is the negative feedback that lets a motor hold speed and the term that makes torque possible.
- "Ke and Kt are unrelated numbers." They are the same constant in SI units; a datasheet mismatch is almost always a unit conversion, not a physical difference.
- "Sensorless control works at any speed." Near zero speed E = Ke·ω is too small to measure reliably, so drives start open-loop or with high-frequency signal injection before switching to back-EMF estimation.
- "Adding inductance stops the inrush surge." Inductance only slows how fast current rises (τe = La/Ra, a few milliseconds); the steady stall current is still V/Ra. You limit stall current with current control, resistance, or a soft-starter, not the winding L.
- "Overhauling loads never damage the drive." When an external load spins the motor faster than V/Ke, back-EMF exceeds the supply and current reverses (regeneration); without a brake resistor or a battery to absorb it, the DC bus can over-volt and destroy the inverter.
Frequently asked questions
What is back-EMF in a motor?
Back-EMF is the voltage a spinning motor generates that opposes the supply driving it. Any conductor moving through a magnetic field induces a voltage (Faraday's law), and by Lenz's law that induced voltage points against the current that created the motion. A motor is therefore also a generator: as the rotor turns, the windings produce a counter-electromotive force E that subtracts from the applied voltage. E is proportional to speed, E = Ke·ω, so it is zero at standstill and largest at full speed.
Why does a stalled motor draw so much current?
At standstill the rotor is not moving, so back-EMF is zero. The only thing limiting current is the winding resistance Ra, so I = V/Ra — the locked-rotor or stall current. Because Ra is deliberately small (often a fraction of an ohm) to keep the motor efficient, this current is typically 5 to 10 times the rated running current. A 12 V motor with 0.5 ohm winding resistance draws 24 A stalled versus perhaps 2 to 3 A at speed. That is why blocked fans, jammed actuators, and hard-started motors trip breakers, blow fuses, and overheat windings.
What is the back-EMF constant Ke and how does it relate to the torque constant Kt?
Ke is the back-EMF constant, the proportionality between speed and generated voltage: E = Ke·ω, with SI units of volt-seconds per radian (V·s/rad), equivalently V/(rad/s). Kt is the torque constant, torque per amp: T = Kt·I, in newton-metres per amp (N·m/A). In a lossless machine energy balance forces them to be numerically equal in SI units: Ke (V·s/rad) = Kt (N·m/A). They are the same physical constant seen from the electrical and mechanical sides. Datasheets often list Ke in V/krpm and Kt in N·m/A or oz·in/A, which hides the equality behind unit conversions.
How does back-EMF limit motor current at speed?
The armature current obeys I = (V − E)/Ra. As the motor accelerates, E = Ke·ω grows and the numerator (V − E) shrinks, so current falls. The motor settles at the speed where the torque it makes, T = Kt·I, exactly balances the load torque. A lightly loaded motor spins up until E is almost equal to V and draws only enough current to overcome friction. This self-regulation is automatic and needs no controller — back-EMF is the negative feedback built into the machine itself.
How is back-EMF used for sensorless speed and position control?
Because E = Ke·ω, measuring the generated voltage gives rotor speed directly with no tachometer. In a three-phase brushless motor, the back-EMF of each phase is a shifted waveform (trapezoidal in BLDC, sinusoidal in PMSM). Sensing the zero-crossings of the un-energised phase, or estimating the back-EMF vector with an observer, reveals rotor position for commutation. This is sensorless control — the heart of field-oriented control (FOC) drives in drones, appliances, and EVs. Its weakness is low speed: at near-zero speed back-EMF is too small to sense, so drives use open-loop or high-frequency injection to start.
Does back-EMF make a motor more or less efficient?
Back-EMF is where useful electrical power turns into mechanical power. The power delivered to the shaft is P_mech = E·I, while the power lost as heat is I²·Ra. A high Ke means E is large and current is small for a given load, so copper losses I²·Ra fall — high-Ke, low-current designs are efficient. But a high Ke also lowers the no-load speed for a given supply, because the motor can only spin until E reaches V. Designers trade off Ke against the required speed range; field weakening deliberately reduces effective flux to push speed above the base point where E would otherwise equal V.
What is the difference between back-EMF and the voltage drop across the winding resistance?
They are two different terms in the same loop equation, V = E + I·Ra + L·dI/dt. The I·Ra term is an ohmic drop that always dissipates as heat and exists even in a stationary motor. Back-EMF E is a motional voltage that only appears when the rotor turns and represents energy converted to mechanical work, not heat. At stall E = 0 and the whole supply falls across I·Ra (pure loss); at full no-load speed E ≈ V, I·Ra is tiny, and almost all voltage is back-EMF. The L·dI/dt term is the transient from winding inductance during current changes.