DC Motor Calculator (Back-EMF, Kv and Torque)
DC Motor Calculator (Back-EMF, Kv and Torque)
The brushed permanent-magnet motor from first principles: back-EMF from speed, current from what is left of the supply, torque from current, and where that leaves mechanical power, copper loss and efficiency — plus the stall and no-load points and the straight line between them, with Kv and Kt shown to be the same constant.
DC motor
12 V across a motor with 1.2 Ω of winding, Kv = 480 rpm/V and a no-load current of 0.25 A, running at 4,200 rpm
Three equations, and everything that follows
E·I = T·ω ⟹ Ke = Kt in SI Kv = 60 / (2π·Kt) rpm/V
stall: I = V/Ra, T = Kt(V/Ra − I₀) no load: ω₀ = (V − I₀Ra)/Ke
maximum power at ω₀/2, Pmax = Tstall·ω₀/4
maximum efficiency at I = √(I₀V/Ra), ηmax = (1 − √(I₀Ra/V))²
- Ke = Kt
- one constant with two names and two unit systems: volt-seconds per radian and newton metres per amp are the same thing, because volts times amps and newton metres times radians per second are both watts
- I 0
- no-load current. Modelled here as a constant friction torque Kt·I₀, which is a simplification: on a real motor part of it is iron loss and rises with speed
- R a
- winding resistance including the brushes. It rises about 0.4% per °C, so a hot motor has a lower stall torque and a steeper speed–torque line than a cold one
Worked example
12 V across a motor with 1.2 Ω of winding, Kv = 480 rpm/V and a no-load current of 0.25 A, running at 4,200 rpm
Kt = 60 ÷ (2π × 480) = 19.894 mN·m/A — and the back-EMF constant is the same number, 0.019894 V·s/rad
ω = 2π × 4,200 ÷ 60 = 439.823 rad/s, so the back-EMF is E = Keω = 8.7500 V — exactly 4,200 ÷ 480, which is what a speed constant in rpm per volt means
That leaves 12 − 8.7500 = 3.2500 V across 1.2 Ω, so I = 2.7083 A
T = Kt(I − I₀) = 0.019894 × (2.7083 − 0.25) = 48.91 mN·m, which is 48.91 mN·m or 6.93 oz·in
Shaft power is Tω = 21.51 W; the supply is delivering 32.5 W, of which 8.8021 W is heating the winding and 2.1875 W is friction. Efficiency 66.2%
The ends of the line: stall at 10.0 A and 194 mN·m, no load at 5,616 rpm. Maximum power is 28.52 W at 2,808 rpm, and maximum efficiency — 70.9% — at 1.581 A, which is √(I₀V/Ra), the geometric mean of the no-load and stall currents
The same motor, right across its range
| Operating point | Speed | Current | Torque | Shaft power | Efficiency |
|---|---|---|---|---|---|
| Stall | 0 rpm | 10.00 A | 194 mN·m | 0 W | 0% |
| Maximum power | 2,808 rpm | 5.13 A | 96.99 mN·m | 28.52 W | 46.4% |
| The worked example | 4,200 rpm | 2.71 A | 48.91 mN·m | 21.51 W | 66.2% |
| Maximum efficiency | 4,849 rpm | 1.58 A | 26.48 mN·m | 13.45 W | 70.9% |
| No load | 5,616 rpm | 0.25 A | 0 N·m | 0 W | 0% |
One constant, two names, and a straight line
A brushed permanent-magnet motor is about as simple as a machine gets. The magnets make a fixed field; the commutator keeps the armature current always in the part of the winding where it produces useful torque. Two proportionalities follow. Spin the armature and it generates a voltage proportional to speed — the back-EMF. Push current through it and it produces a torque proportional to that current. And the two constants of proportionality are, in SI units, the same number.
Why Ke equals Kt. Not a coincidence and not an approximation. The electrical power the motor converts is E·I, and the mechanical power it produces is T·ω. Those are the same power, so Keω·I = KtI·ω and the constants cancel. A motor with a torque constant of 0.02 N·m/A has a back-EMF constant of 0.02 V·s/rad, exactly. The familiar rpm-per-volt figure is the reciprocal with a unit conversion: Kv = 60/(2πKt), so 0.02 N·m/A is 477 rpm/V. Anyone quoting a motor’s Kv has also told you its torque constant.
Why the speed–torque curve is a straight line. The supply splits between the resistive drop and the back-EMF: V = IRa + Keω. Solve for I, multiply by Kt, and the torque falls linearly with speed. Two points fix the line. At stall there is no back-EMF, so the current is V/Ra — which is enormous, and is the number that decides your driver and your fuse. At no load the back-EMF has risen until only the no-load current flows. Everything in between is on the line joining them, and maximum mechanical power is exactly halfway along it, at Tstallω₀/4.
Where the efficiency peak is, and why it is nowhere near the power peak. At half the no-load speed, half the supply voltage is being dropped across the winding, so about half the input power is heating copper and the efficiency is about 50%. Peak efficiency is much further up, at a current of √(I₀V/Ra) — the geometric mean of the no-load and stall currents — where the fixed friction loss and the current-squared copper loss are balanced. The peak value is (1 − √(I₀Ra/V))², a pleasingly compact result that says everything: low resistance, low no-load current and high voltage are what make an efficient motor, and the three trade against each other.
What this model leaves out, which is a lot. There is no iron loss that rises with speed — the no-load current is treated as a fixed friction torque, which is only roughly right. There is no brush voltage drop, which on a real brushed motor is one to two volts and is a large fraction of a 6 V or 12 V supply. There is no armature inductance, so nothing here says anything about how fast the current can change or what a PWM drive does to it. There is no armature reaction and no magnet weakening with temperature. And Ra is treated as constant when it rises about 0.4% per degree, so a motor that has been working hard has a lower stall torque than the cold figure above. Treat the page as the shape of the behaviour, accurate to a few per cent on a well-characterised motor, and reach for measured curves when the last few per cent matter.
Also: I₀ is not zero. The textbook model has no friction and reaches 100% efficiency at zero load, which is nonsense. A small brushed motor draws a few per cent of its stall current running free, and that current alone sets the peak efficiency through the expression above. If you set I₀ to zero here the page will happily give you the textbook answer; it is not the motor you have.
For the general relation between power, torque and speed in any machine, the motor power, torque and speed calculator — it uses the same N·m and rpm conventions as this page. For a brushless motor, the same constants apply to the machine but the switching stage has losses of its own, which the BLDC inverter loss calculator covers. For turning the shaft torque here into something useful at a slower shaft, the gear ratio calculator.
Frequently asked questions
What is Kv on a DC motor?
The speed constant: how many rpm the motor turns per volt of back-EMF. A 480 rpm/V motor on 12 V spins at about 5,760 rpm with no load, a little less once the no-load current’s resistive drop is taken off. It is the reciprocal of the torque constant with a unit conversion: Kt = 60/(2π·Kv) N·m/A.
Is Kv the same as Kt?
Yes, in the sense that either one gives you the other exactly. In SI units the back-EMF constant in volt-seconds per radian and the torque constant in newton metres per amp are the same number, because electrical power converted equals mechanical power produced. Kv in rpm per volt is 60/(2π) divided by that number, so a 0.02 N·m/A motor is 477 rpm/V.
How do I calculate DC motor torque from current?
T = Kt × (I − I₀). Subtracting the no-load current matters: it is the current that goes into spinning the motor itself rather than into the shaft. A motor drawing 2 A with a 0.25 A no-load current and Kt of 0.02 N·m/A gives 0.035 N·m at the shaft, not 0.04.
What is the stall current of a DC motor?
Supply voltage divided by winding resistance, because at zero speed there is no back-EMF to oppose it. It is usually ten to fifty times the running current, it is what your driver and fuse have to survive, and it turns the entire input into heat in the winding — which is why a stalled motor has minutes, not hours.
At what speed is a DC motor most efficient?
At a current of √(I₀·V/Ra), which is well above the no-load speed end of the range and nowhere near the maximum-power point. The peak efficiency itself is (1 − √(I₀Ra/V))². Maximum power is at half the no-load speed, where the efficiency is only about 50%.
Why does my motor not reach the speed this page predicts?
Usually the supply. The stall current is large, so a battery or a bench supply with any source resistance sags under it, and the voltage at the terminals is less than the voltage you set. After that: brush drop, which this model ignores and which is a whole volt or two; a winding that is hotter and therefore more resistive than when you measured it; and iron loss that rises with speed while the model treats the no-load torque as constant.
Related calculators
References
- Hughes A, Drury B. Electric Motors and Drives: Fundamentals, Types and Applications, 5th ed., Elsevier, 2019, ISBN 9780081026151, Chapter 3 Conventional D.C. Motors — the armature circuit equation, the identity of the back-EMF and torque constants, the speed–torque characteristic and the effect of armature resistance on its slope.
- maxon motor ag, maxon DC motor and maxon EC motor — Key information (the explanatory section of the maxon catalogue). States that the speed constant and the torque constant are not independent of one another, defines the no-load speed n₀ and the stall torque MH as the ends of a straight speed–torque line with a gradient proportional to 1/k², and defines efficiency as mechanical power delivered over electrical power consumed. Its equations are typeset as images in the PDF, so the constant 60/2π used here was re-derived rather than copied.
- Chapman SJ. Electric Machinery Fundamentals, 5th ed., McGraw-Hill, 2011, Chapter 8 — the separately excited and permanent-magnet DC machine, including the magnetisation and armature-reaction effects that the linear model on this page deliberately leaves out.