16N78-212P.1001 Portescap Danaher Motion US LLC, 16N78-212P.1001 Datasheet - Page 7

• Manufacturer Part Number: 16N78-212P.1001
• Description: MOTOR BRUSH 16MM 6VDC
• Manufacturer: Portescap Danaher Motion US LLC
• Series: 16N78, Athlonixr
• Type: Brushed DCr

Specifications of 16N78-212P.1001

• Voltage - Rated: 6VDC
• Rpm: 9300 RPM
• Features: -
• Shaft Diameter: 0.059" (1.50mm)
• Shaft Rotational Direction: -
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant
• Other names: 403-1041

The electromechanical properties of motors
with ironless rotors can be described by
means of the following equations:
1. The power supply voltage U
the sum of the voltage drop produced by the
current I in the ohmic resistance R
rotor winding, and the voltage U
the rotor :
U
2. The voltage U
proportional to the angular velocity ω of the
rotor :
U
It should be noted that the following
relationship exists between the angular
velocity ω express in radians per second and
the speed of rotation n express in revolutions
per minute:
ω = 2π n
3. The rotor torque M is proportional to the
rotor current I:
M = k
It may be mentioned here that the rotor torque
M is equal to the sum of the load torque M
supplied by the motor and the friction torque
M
M = M
By substituting the fundamental equations (2)
and (3) into (1), we obtain the characteristics
of torque/angular velocity for the dc motor
Brush DC Working Principles
0
i
f
= k
= I x R
of the motor :
60
E
T
x ω
L
x I
+ M
M
+ U
f
R
i
M
U
I
i
induced in the rotor is
I
0
i
is equal to
induced in
M
of the
U
(1)
(2)
(3)
0
L
with an ironless rotor :
U
By calculating the constant k
dimensions of the motor, the number of turns
per winding, the number of windings, the
diameter of the rotor and the magnetic field
in the air gap, we find for the direct-current
micromotor with an ironless rotor:
M = U
Which means that k = k
The identity k
following energetic considerations:
The electric power P
to the motor must be equal to the sum of the
mechanical power P
the rotor and the dissipated power (according
to Joule’s law) P
P
Moreover, by multiplying equation (1) by I, we
also obtain a formula for the electric power
P
P
The equivalence of the two equations gives
M x ω = U
or U
Quod erat demonstrandum.
Using the above relationships, we may write
the fundamental equations (1) and (2) as
follows:
U
and :
U
I
0
e
e
e
0
0
:
= U
= U
= M x R
ω
= I x R
= M x R
= P
i
ω
= M and k
0
0
i
m
= k
x I = M x ω + I
x I = I
I
+ P
M
k
i
M
+ k x ω
M
v
x I
+ k
+ k x ω
2
x R
E
E
E
= k
x ω
= k
M
v
= I
+ U
T
T
is also apparent from the
= k
2
e
2
x R
i
m
= U
x R
x I
= M x ω produced by
E
M
= k
0
M
:
x I which is supplied
T
E
and k
T
from the
(4)
(5)
(6)
(7)
Graphic
characteristic:
To overcome the friction torque M
the friction of the brushes and bearings, the
motor consumes a no-load current I
gives
M
and:
U
ω
hence:
k = U
Is it therefore perfectly possible to calculate
the motor constant k with the no-load speed
n
resistance R
The starting-current I
follows:
I
It must be remembered that the R
to a great extent on the temperature; in other
words, the resistance of the rotor increases
with the heating of the motor due to the
dissipated power (Joule’s law):
R
Where γ is the temperature coefficient of
copper (γ = 0.004/°C).
As the copper mass of the coils is
comparatively small, it heats very quickly
d
0
M
0
0
, the no-load current I
= U
f
= I
= 2π x n
= R
n
n
= k x I
0
R
0
60
M
0
0
0
M
x R
- I
M0
L
ω
(1 + γ x ∆T)
0
0
0
M
x R
+ k x ω
0
M
M
U
.
0
express
0
where
d
I
is calculated as
0
“speed-torque”
M
and the rotor
L
I
M
M
L
depends
f
due to
0
. This
M
(8)