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Brush DC Working Principles
through the effect of the rotor current,
required for obtaining a speed of rotation n
particularly in the event of slow or repeated
starting. The torque M d produced by the
starting-current I d is obtained as follows:
for a given load torque M L (angular velocity ω
= n x 2π/60). By introducing equation (10) into
(6) we obtain:
Using equation (10) we first calculate the
current which is supplied to the motor under
these conditions:
(
)
M d = I d x k - M f = (I d - I 0 )k
(9)
U 0 = M L +I 0 R M +kx ω
k
(13)
I = M L + I 0 = 0.008 + 0.012
k 0.0232
By applying equation (1), we can calculate the
angular velocity ω produced under a voltage
U 0 with a load torque M i . We first determine
= 0.357A
Practical examples of calculations
Please note that the International System of Equation (11) gives the angular velocity ω:
the current required for obtaining the torque
M=M L +M f :
I=M L +M f
Units (S.I.) is used throughout.
1. Let us suppose that, for a Portescap ?
ω = U 0 ? I x R M = 9 ? 0.357 x 10.2
k 0.0232
= 231 rad/s
k
Since M f = I 0
k
motor 23D21-216E, we wish to calculate the
motor constant k, the starting current I d and and the speed of rotation n:
the starting torque M d at a rotor temperature n = 60 ω = 2200 rpm
of 40°C. With a power supply voltage of 12V,
2π
we may also write
I=M L +I 0
(10) the no-load speed is n 0 is 4900 rpm ( ω 0 = 513
rad/s), the no-load current I 0 = 12 mA and the Thus the motor reaches a speed of 2200 rpm
k
resistance R M0 = 9.5 Ω at 22°C.
and draws a current of 357 mA.
For the angular velocity ω , we obtain the By introducing the values ω 0 , I 0 , R M0 and U 0 into
3. Let us now calculate the torque M at a
relationship
ω =U 0 ?IxR M
k
= U 0 ? R M (M L + M f )
the equation (8), we obtain the motor constant
(11) k for the motor 23D21-216E:
k = 12 ? 0.012 x 9.5 = 0.0232 Vs
15
given speed of rotation n of 3000 rpm ( ω = 314
rad/s) and a power supply voltage U 0 of 15V;
equation (12) gives the value of the current:
I=U 0 ?kx ω =I d ?k x ω
k k
2
R
M
R M
Before calculating the starting-current, we
In which the temperature dependence of the must calculate the rotor resistance at 40°C.
rotor resistance R M must again be considered; With ?T = 18°C and R M0 = 9.5Ω, we obtain
in other words, the value of R M at the working R M = (1 + 0.004 x 18) = 9.5 x 1.07
= 1.18 ? 0.0232 x 314 = 0.466A
10.2
and the torque load M L :
temperature of the rotor must be calculated.
= 10.2Ω
M L = k(I ? I 0 )
On the other hand, with the eqation (6), we
can calculate the current I and the load The starting-current I d at a rotor temperature
torque M L for a given angular velocity ω and a of 40°C becomes
= 0.0232 (0.466 ? 0.012)
= 0.0105 Nm
(M L = 10.5 mNm)
given voltage U 0 :
I d = U 0 = 12 = 1.18A
I=U 0 ?kx ω= I d ?k ω
(12)
R
M
10.2
4. Lastly, let us determine the power supply
R M
R M
and the starting-torque M d , according to
voltage U 0 required for obtaining a speed
rotation n of 4000 rpm ( ω = 419 rad/s) with
M = (I ? I )k ? k ω
Andwithequation(10)
M L = (I ? I 0 )k
We get the value of M L :
2
L 0
R M
equation (9), is
M d = k(I d ? I 0 ) = 0.0232 (1.18 ? 0.012)
= 0.027 Nm
2. Let us ask the following question: what is
the speed of rotation n attained by the motor
a load torque of M L of 0.008 Nm, the rotor
temperature again being 40°C (R M = 10.2Ω).
As we have already calculated, the current I .
necessary for a torque of 0.008 Nm is 0.357 A
U 0 =IxR M +kx ω
with a load torque of 0.008 Nm and a power
The problem which most often arises is that supply voltage of 9V at a rotor temperature
= 0.357 x 10.2 + 0.0232 x 419
= 13.4 volt
of determining the power supply voltage U
0
of 40°C?
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