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Data and formules about little electric motors

This page provides data and formules for those little low-voltage DC electric motors that can be found inside toys, home appliances like electric razors, cassette players and CD players or electric radio-command planes, gliders and cars. They can be bought in model kit or electronic components shops and can be powered with little batteries from 1.5 Volt up to 12 Volt. (Do not confound them with AC motors intended to be plugged on the high-voltage AC network. The AC home network is dangerous.) Please note the first formules are wrong. See the formules send in by Olof Lind.

Feed the motor with a constant current


This is the approximated graph of the mechanical power delivered, versus turning speed. The motor is connected to a power source with a constant voltage. e.g. 3 Volts.

The changes of speed are not due to changes in the electric force connected to the motor, but to mechanical strains (weights to lift, friction, ...)

When a motor is not braked, it will turn at its maximum speed (left-most part of the graph). It will deliver no mechanical power.

It will deliver the highest amount of mechanical power when the braking is such that the motor turns at one half its maximum speed (middle of the graph).

For example: if you want to put a propeller on a motor and make it blow the biggest possible quantity of air, you must dimension the propeller so it will brake the motor and make it turn at half its maximum speed.

The left part of the graph

The approximate justification of the shape of the left, rising, side of the graph is as follows:

Why does the rotor of an electric motor turn? Because it is an electromagnet, and it is alternatively attracted and repelled by the permanent magnet placed near it (the stator), inside the motor.

At low speed, the force acting between the two magnets is independent of the turning speed. It is not dependent of the speed at which they approach each other, or leave.

According to Newton, energy E is proportional to the amount of force F effected on a certain distance d:

E  =  F d

The faster the motor turns, the more distance it's rotor will cover (on it's circular path). And it is submitted to the same amount of force, whatever that speed or distance.

(If you compress a big spring between two walls, it won't deliver any energy just by pushing on the walls. But if the walls give in to the force of the spring, and move a few centimeters apart, you will have to deliver some energy to compress the spring again (and put a brick between the spring and one of the walls). Each time the walls move apart, you have to re-compress the spring. The total amount of energy finally delivered will be function of the total distance moved by the walls. No matter the time taken to make the movement.)

Thus, the faster the motor turns, the more energy it will deliver in a certain amount a time. Thus, the higher the power delivered.

(Another way of looking at it is to consider you want to lift a weight using a lever, and the person who has to pull on the lever tells you it will pull with one given force, but at any speed you want. The longer the lever, the more weight will be lifted in the same amount of time (and the fastest the person moves).)

Note that the fact the motor delivers no mechanical energy when still, does not mean it's not consuming electric energy. The slower the motor turns, the more electric energy it consumes. But that electric energy is turned into heat by the Coulomb effect in the coils of the rotor.

The right part of the graph

The right, falling, part of the graph is due to the appearance of an electromagnetic phenomenon: the electric counter-force:

When an inductance moves past a magnet, an electric force (tension, voltage) appears at it's ends. The fastest the movement, the greater the force. The rotor of our motor is such an inductance. And the electric force created is opposite to that delivered by the electric generator that makes the rotor turn. When the speed of the rotor is such that the electric force created compensates the force of the generator, no more current can flow trough the electromagnets of the rotor, and thus the rotor can't turn faster.

Another way to say it: the motor is also a dynamo. When the dynamo reaches a given turning speed it produces an electric tension equal and opposite to the tension of the electric power source of the motor, so no more current can flow through the motor/dynamo. (Should the motor be forced to turn even faster by an external mechanical force, it would force current back towards the power source.)

(Mechanic, aerodynamic, and other electromagnetic phenomenon also trade in to slow the motor down, but we will neglect them.)


If v is the average linear speed of the rotor, and Fm the average force acting between the rotor and the stator, the mechanical power given by the rotor is:

Pm  =  v Fm         (Newton)

The electric force Fec induced in the coils of the rotor by the stator (dynamo effect) will be:

Fec  =  Cst1 v         (Faraday)

(Cst1 is a constant depending of the characteristics and size of the motor.)

The force Fm being roughly proportional to the current I flowing trough the coils of the rotor, if R is the electric resistance of the coils, and Feg the electric force of the generator:

I  =  (Feg - Fec) / R         (Ohm)

Fm  =  Cst1 I         (Oërsted)


Pm  =  v Cst1 (Feg - Cst1 v) / R

Pm  =   (Cst1 / R) (Feg v - Cst1 v2)

As the normal maximum turning speed vm of a little motor is often given in the documentation, or can at least easily be measured with an oscilloscope connected to the power source, it is useful to express Cst1 in function of it.

Indeed, power is zeroed at maximum speed, so:

Feg vm  =   Cst1 vm2

Cst1  =  Feg/vm


Pm  =   Feg2 / (R vm) (v - v2 / vm)

With the maximum at vm / 2.

A very good question is that of the electric power consumption Pe of the motor versus its turning speed v. It is maximum when the motor is blocked (current  =  Imax), and decreases linearly to be zero at maximum speed.

Imax  =   Feg / R

Pe  =  Feg ( -(Imax / vm) v + Imax)

Pe  =   Feg2 / R (1 - v / vm)

This is the graph of the electric power consumption versus speed. The amount of mechanical power produced is given by the dotted line:

This is very useful because it allows to measure easily some key data. The electric tension being constant, the electric power used by the motor can be almost directly known by simply measuring the current (power = tension x current). To know if a motor is producing its maximum amount of mechanical energy, divide the current that flows through it when it is turning by the current when it is blockaded. The result should be 1/2. To know the importance of the mechanical friction inside the motor and how far from its ideal maximum speed the motor is turning, measure the current when it is blockaded and when it turns freely, and use the upper graph (trough a motor with no friction, turning freely, there should flow virtually no current). (To estimate the power lost due to friction at any speed, roughly consider it decreases linearly towards zero at null speed.)

And finally, the efficiency Y, being the result of the division of the amount of produced mechanic power by the amount of consumed electric power:

Y  =  v / vm

This is the graph of the efficiency versus speed:

It is a linear function too. One can easily see that the maximum efficiency is achieved at the maximum speed (vm). But it is of no use, as there is no energy delivered. For common purposes, the fact that a motor delivers its maximum mechanical energy with an efficiency in the average of 50%, is very satisfactory. (50% of the consumed electric energy is transformed into useable mechanical power, and 50% into loosen heat). Higher efficiencies, obtained by using bigger motors or generators (to get or use the same amount of mechanical energy), are only interesting for static power plants, heavy industry, or some few space applications.

The graph above is theory. It is based on mathematical calculations that do not involve parameters like friction inside the motor. Olof Lind send in the text below that adds the necessary precisions:

A motor running in the no-load condition cannot have the efficiency 100%: the output power is zero and there is still an input power to cover for the friction losses of the motor. A motor giving zero Watt output power will of course have the efficiency 0 percent. The true efficiency curve is an "unsymmetric parabola" (not a straight line) and at some point between the zero-load condition and the blocked motor condition there will be a point of maximum efficiency. 

The most astonishing fact regarding the PM motor with strong ceramic magnet is that virtually all the motor properties can be derived from two torque figures: The starting torque (blocked motor) and the friction torque (mainly caused by the brushes and the bearings). This can be seen from following formulae. During my time as a motor designer (35 years ago) this statement was confirmed by many measurements and I can assure that it is true.

Making the simplifications that:
Then the current and the speed of the motor can be plotted against the output torque as straight lines. We assume that the motor is fed from a power source giving constant voltage independent on the load. The symbol n represents speed (rpm) and T is the torque:

The right vertical axis represents the motor running freely (no external load). The prolonged current line meets the torque axis at Tfr, the point representing the friction torque at no load (bottom end of the left axis). The speed line reaches the torque axis at the starting torque point Tst (zero speed, blocked motor). The no-load speed of the motor is n0. At the speed n0s, the counter-emf is exactly balancing the input voltage.
At some point (n, T) the output power of the motor is

Pout = T x 2πn/60  Watt

At the ideal no-load speed (no friction) the induced voltage counterbalances the power source voltage:

Eind = C x p x n0s x Φp/60 

Where p is the number of poles, Φp is the magnetic flux of the pole  and C is a constant.
Thus, the input power in this point is

Pin = C x p x I x n0s x Φp/60

Furthermore, at this point (n, T) the torque of the armature is

T + Tfr = C x p x I x Φp/2 π

and  thus:

I = 2 π (T+Tfr)/(p x C x Φp)

Thus, by combination of the above, the input power is:

Pin = 2 π x n0s (T+Tfr)/60

Thus the efficiency =

Pout/Pin = (n x T) / n0s (T+Tfr)

The astonishing fact here is that the efficiency of a PM motor in any point of work along its characteristic is given by the friction torque and the starting (blocked condition) torque and nothing else - that´s it.  With these two values given, it does not matter how much copper or length or diameter you put into the motor design; the efficiency at a torque load T is still the same. When the motor is running at no load, the torque T is = 0 and the efficiency is 0. When the motor is blocked, the speed  n is 0 and the efficiency is 0.

Working on with the formulae, you will find that the maximum efficiency will appear at the speed:

nmaxeff = n0 (1 + Tfr/Tst – ((Tfr/Tst)2 + Tfr/Tst)0,5)

and the maximum efficiency is:

efficiencymax = (1- (Tfr/(Tfr+Tst))0,5 )2

As you can easily observe from the formulae, the basic thing to do when you are trying to achieve a high motor efficiency is to keep the friction (including the air friction or the armature) as low as possible. Not very astonishing!

Feed the motor with a constant current

Ic will be the symbol for the constant current. See above for the significance of the other symbols.

Mechanical power delivered by the motor:

Pm  =  Cst1 Ic v

This is the graph of the mechanical power produced with a constant current, and that of the power produced with constant voltage (dotted line):

(The current when the motor is blocked is the same in both cases.)

Electrical power consumed:

Pe  =  Cst1 Ic v + Ic2 R


Y  =  1 / (1 + (Ic R) / (Cst1 v))

The graph of the efficiency achieved with a constant current, and that of the efficiency with constant voltage (dotted line):

(Again, the current when the motor is blocked is the same in both cases.)

Note that in normal circumstances, when the motor turns at a certain speed, it probably won't have to give it's full power. So the efficiency will be increased by the fact that the current or the voltage are lower than the maximum. Conversely, the efficiency will fall down during burst modes.

Attention: the maximum speed shown on the graphs is the same vm as for a motor fed with a constant voltage. Don't forget that with a constant current the speed can be increased above vm. We didn't show it on the graphs. (Little DC- current motors get mechanical problems when pushed too far above their vm.)

To allow the average current to be constant, the average voltage must increase with the speed.

This is the graph of the voltage versus speed:

It is calculated by considering that the power source has to override the normal resistance of the coils AND the electric counter-force Fec:

Fec  =  Cst1 v

Feg  =  Ic R + Fec


PHYSICS / Foundations and Applications
Robert M. Esberg / Lawrence S. Lerner
Mc Graw Hill
ISBN 0-07-066268-1

Joseph Kane / Morton Sternheim
ISBN 2-7296-0098-1

The graphs were produced by programs, and converted to a standard vectorial format thanks to software written by M. Frédéric Cloth.

Eric Brasseur  -  1994  till November 2004