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Characteristics of Direct Current Motors

For the drive of metal kit models mostly low cost direct current motors are used, whose electrical and mechanical specifications are more or less unknown.
When data of the idle speed and -current frequently are available the nominal rating often is missing, and informations about engine power, torque and efficiency generally is the exception.
But especially the knowledge of the torque at different load cases can make sense to evaluate the qualification of a motor.

For the specification of a motor only a few data are necessary from which all other characteristics are calculable. The basis for this is the balance of engine power:

U · I = I² · R + c · I · 2 · p · n

and consequential

U = I · R + c · 2 · p · n

U = Supply voltage in V (Volt)
I = Supply current in A (Ampere)
R = Ohmic resistance of the armature winding in Ohm
c = Engine constant in Vs
n = Revolutions per second in 1/s

The product U · I is the power consumption, which can be measured with a simple multimeter.
I² · R describes the electrical power loss of the armature winding.
c · I · 2 · p · n is the mechanical power, which consists of the mechanical loss (friction, aerodynamic resistance etc.) and the mechanical output power.

The rotation speed n must be available for the calculations and if necessary it has to be measured. For this several ways exist, e. g. commercial measuring devices and simple magnetical or optical methods by using an oscilloscope; also by evaluation of the audio frequency spectrum with a Fourier analysis on a PC the rotation speed can be determined.

To calculate the still unknown items R and c the current consumption I and the rotation speed n at two different load cases has to be provided, e. g. at idle and at nominal load.
The insertion of this 2 pairs of variates into the balance of engine power results into 2 equations with 2 unknowns, from which R and c can be educed.

Example: A motor shall have the following data

Idle:	Rated voltage U = 6 V, rotation speed n = 93 1/s (5600 1/min), current I = 0,5 A
Nominal load :	Rated voltage U = 6 V, rotation speed n = 75 1/s (4500 1/min), current I = 1,7 A

The appropriate system of equations is:

6V · 0,5A = (0,5A)² · R + c · 0,5A · 2 · p · 93 1/s
6V · 1,7A = (1,7A)² · R + c · 1,7A · 2 · p · 75 1/s

resp. over the current

6V = 0,5A · R + c · 2 · p · 93 1/s
6V = 1,7A · R + c · 2 · p · 75 1/s

The calculated values for R and c

R = 0,9078 Ohm c = 9,457 · 10^-3 Vs
at last complete the balance of engine power:

U · I = I² · 0,9078 Ohm + 9,457 · 10^-3 Vs · I · 2 · p · n

The next step is the calculation of the mechanical output power.
It is obtained after subtracting the mechanical power loss from the total mechanical power:

P_{mech out} = P_{mech total} - P_{mech loss}
P_{mech out} = (c · I · 2 · p · n) - (c · I_idle · 2 · p · n) = c · 2 · p · n · (I - I_idle)
This equation includes the dependence of the mechanical loss on the rotation speed:
at idle it is maximum, and at a blocked engine shaft it is zero, because the armature doesn´t move anymore; here a linear reduction between idle and blocked engine shaft is assumed.

For the in this example assumed nominal load the mechanical output power is given to:

P_{mech out} = 9,457 · 10^-3 Vs · 2 · p · 75 1/s · (1,7 A - 0,5 A)

P_{mech out} = 5,348 W

The torque results from the correlation:

M = P / w = P / 2 · p · n = c · I

M = Torque in Nm bzw. Ws
P = Power in W
n = Revolutions per second in 1/s
w = Angular frequency 2 · p · n in 1/s

Also at this the loss has to be subtracted from the total torque to get the output torque:

M_out = M_total - M_loss
M_out = 9,458 · 10^-3 Vs · (1,7 A - 0,5 A)

M_out = 1,135 Ncm
Of practical importance is the locked rotor torque.
It is also given from M = c · (I - I_idle), where as current I the max. value I = U / R = 6 V / 0,9078 Ohm = 6,609 A is to be inserted:

M_max = 5,78 Ncm

Furthermore from the existing data still the efficiency can be determined.
It is the ration of mechanical output power to the electrical power consumption

h = P_{mech out} / P_electrical = 5,348 W / 10,2 W

h = 52,4%

So far the characteristics describe the performance of a motor in only one operating point, in fact under the assumed nominal load condition.
But sometimes also the behaviour in the complete working range is of interest. Moreover it is to check, whether the assumed nominal rate really corresponds to the max. efficiency operation.

Correlation between rotation speed and current consumption

This function can be educed by transposing the balance of engine power to the rotation speed n:

n = (U - I · R) / (c · 2 · p)

n = f (I)

Correlation between power, torque and current consumption

The function of the output power results from insertion of the above equation into the formula for the mechanical output power:

P_{mech out} = (U - R · I) · (I - I_idle)
The function of the torque follows from

M_out = c · (I - I_idle)

P_{mech out} = f (I), M_out = f (I)

Thus the max. mechanical output power is not delivered at the assumed nominal load I = 1,7 A, but at approx. I = 3,55 A; in this case the power reaches approx. 8,5 W.

Correlation between efficiency and current consumption

This function is the ratio of mechanical output power and electrical power consumption:

h = (U - R · I) · (I - I_idle) / (U · I)

h = f (I)

Similar to the output power also the efficiency has a maximum at one current, but this currents are not identical.

To get the maximum of a mathematical function generally the methods of the differential calculus are used.
Differentiation of a function results into it´s gradient function. Because in a maximum the gradient is zero, the gradient function also is to set to zero, and from this the to the maximum corresponding x-value (in this case the current) can be calculated.

I _{h max} = Ö(U · I_idle / R)

I _{h max} = 1,818 A
This shows that the assumed nominal load I = 1,7 A is not exactly the operation with max. efficiency.
The other characteristics for max. efficiency can be determined to

n _{h max} = 73,2 1/s = 4392 1/min

P _{mech out h max} = 5,733 W

M _{out h max} = 1,247Ncm

h _max = 52,56%
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