J = 0.0283 oz-in-s
2
= 2 * 10
-4
kg . m
2
L = 0.004H
Then the corresponding time constants are
T
m
= 0.04 sec
and
T
e
= 0.002 sec
Assuming that the amplifier gain is K
v
= 4, the resulting transfer function is
P/V = 40/[s(0.04s+1)(0.002s+1)]
Current Drive
The current drive generates a current I, which is proportional to the input voltage, V, with a gain of K
a
. The
resulting transfer function in this case is
P/V = K
a
K
t
/ Js
2
where Kt and J are as defined previously. For example, a current amplifier with K
a
= 2 A/V with the motor
described by the previous example will have the transfer function:
P/V = 1000/s
2
[rad/V]
If the motor is a DC brushless motor, it is driven by an amplifier that performs the commutation. The combined
transfer function of motor amplifier combination is the same as that of a similar brush motor, as described by the
previous equations.
Velocity Loop
The motor driver system may include a velocity loop where the motor velocity is sensed by a tachometer and is fed
back to the amplifier. Such a system is illustrated in Figure 10.5. Note that the transfer function between the input
voltage V and the velocity ω is:
ω /V = [K
a
K
t
/Js]/[1+K
a
K
t
K
g
/Js] = 1/[K
g
(sT
1
+1)]
where the velocity time constant, T
1
, equals
T
1
= J/K
a
K
t
K
g
This leads to the transfer function
P/V = 1/[K
g
s(sT
1
+1)]
Figure 10.5: Elements of velocity loops
Chapter 10 Theory of Operation ▫ 174
DMC-40x0 User Manual
K
a
Kt/Js
K
g
V