Data and formules about little electric motors
This page provides data and formules for those little lowvoltage DC
electric motors that can be found inside toys, home appliances like
electric razors, cassette players and CD players
or electric radiocommand 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 highvoltage AC network. The AC home network
is dangerous.) Please note the first formules are wrong. See the
formules send in by Olof Lind.
Characteristics
Feed the motor with a constant current
Characteristics
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 (leftmost 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 recompress
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 counterforce:
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.)
formules
If v is the average linear speed of the rotor, and F
_{m} the
average force acting between the rotor and the stator, the mechanical
power given by the rotor is:
P_{m} = v F_{m}
(Newton)
The electric force F
_{ec} induced in the coils of the rotor by
the stator (dynamo effect) will be:
F_{ec} = Cst_{1} v
(Faraday)
(Cst
_{1} is a constant depending of the
characteristics and size of the motor.)
The force F_{m} being roughly proportional to the current I
flowing trough the coils of the rotor, if R is the electric resistance
of the coils, and F_{eg} the electric force of the generator:
I = (F_{eg}  F_{ec}) / R
(Ohm)
F_{m} = Cst_{1} I
(OÃ«rsted)
Thus:
P_{m} = v Cst_{1} (F_{eg}
 Cst_{1} v) / R
P_{m} =
(Cst_{1} / R) (F_{eg} v  Cst_{1} v^{2})
As the normal maximum turning speed v
_{m} 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 Cst
_{1} in function of it.
Indeed, power is zeroed at maximum speed, so:
F_{eg} v_{m} =
Cst_{1} v_{m}^{2}
Cst_{1} = F_{eg}/v_{m}
So:
P_{m} =
F_{eg}^{2} / (R vm) (v  v^{2} / v_{m})
With the maximum at v
_{m} / 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 = I_{max}), and
decreases linearly to be zero at maximum speed.
I_{max} =
F_{eg} / R
Pe = F_{eg} ( (I_{max} / v_{m})
v +
I_{max})
Pe =
F_{eg}^{2} / R (1  v / v_{m})
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 / v_{m}
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 (v
_{m}). 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 noload
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 zeroload
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:
 The internal friction of the motor (from bearing, brushes etc) is
independent of the rpm (rotation per minute).
 The magnetic reaction from the armature can be neglected.
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 T
_{fr}, 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 T
_{st}
(zero speed, blocked motor). The noload speed of the motor is n
_{0}.
At the speed n
_{0s}, the counteremf is exactly balancing the
input voltage.
At some point (n, T) the output power of the motor is
P_{out} = T x 2πn/60 Watt
At the ideal noload speed (no friction) the induced voltage
counterbalances the power source voltage:
E_{ind} = C x p x n_{0s} x Φ_{p}/60
Where p is the number of poles, Φ
_{p} is the magnetic flux of
the pole and C is a constant.
Thus, the input power in this point is
P_{in} = C x p x I x n_{0s} x Φ_{p}/60
Furthermore, at this point (n, T) the torque of the armature is
T + T_{fr} = C x p x I x Φ_{p}/2 π
and thus:
I = 2 π (T+T_{fr})/(p x C x Φ_{p})
Thus, by combination of the above, the input power is:
P_{in} = 2 π x n_{0s}
(T+T_{fr})/60
Thus the efficiency =
P
_{out}/P
_{in} = (n
x T) / n
_{0s} (T+T
_{fr})
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:
n_{maxeff} = n_{0} (1
+ T_{fr}/T_{st} – ((T_{fr}/T_{st})^{2}
+ T_{fr}/T_{st})^{0,5})
and the maximum efficiency is:
efficiency_{max} = (1 (T_{fr}/(T_{fr}+T_{st}))^{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!
I
_{c} will be the symbol for the constant current. See
above for the significance of the other symbols.
Mechanical power delivered by the motor:
P_{m} = Cst_{1} I_{c} 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 = Cst_{1} I_{c} v +
I_{c}^{2} R
Efficiency:
Y = 1 / (1 + (I_{c} R) / (Cst_{1}
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 v_{m}
as for a motor fed with a constant voltage. Don't forget that with a
constant current the speed can be increased above v_{m}. We
didn't show it on the graphs. (Little DC
current motors get mechanical problems when pushed too far above their v_{m}.)
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 counterforce F
_{ec}:
F_{ec} = Cst_{1} v
F_{eg} = I_{c} R + F_{ec}
Bibliography
PHYSICS / Foundations and Applications
Robert M. Esberg / Lawrence S. Lerner
Mc Graw Hill
ISBN 0070662681
PHYSIQUE
Joseph Kane / Morton Sternheim
InterEditions
ISBN 2729600981
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