L6238S View Datasheet(PDF) - STMicroelectronics
Part Name
Description
MFG CO.
'L6238S' PDF : 31 Pages View PDF
L6238S
Figure 5-4
Lmtr
Rmtr
+
Eg
-
D95IN321
The relation between these variables is given by:
V
=
Lmtr
dimtr
dt
Rmtr
imtr
+
Eg
(5.1.2)
where:
V = Applied Voltage
imtr = Motor Current
Lmtr = Total inductance of the motor
windings
Rmtr = Resistance in series with the motor
Eg = The internally generated voltage of
the motor, proportional to the motor
velocity
Since:
Eg
= KEω
(5.1.3)
The above equations can be combined to form
the basic electrical equation for a motor:
V
=
Lmtr
dimtr
dt
Rmtr
imtr
+
KEω
(5.1.4)
Figure 5.5 is a simplified electrical equivalent of
the output stage of the L6238S along with the
model of the motor during the time that the Out-
put Drives are conducting.
Figure 5-5
UPPER
Rdson
Lmtr
24/31
D95IN322
Rmtr
+
KEW
-
LOWER
Rdson
Rsense
The additional resistance associated with the out-
put stage and sensing resistor are also in series
with the motor. If we let Rs equal the total series
resistence:
Rs = 2*RdsON + Rmtr + Rsense
(5.1.5)
then (5.1.4) becomes:
V
=
Lmtr
dimtr
dt
Rs
imtr
+
Eg
(5.1.6)
Figure 5-6
Lmtr
Rmtr
-
KEW
+
LOWER
Rdson
LOWER
Rdson
D95IN323
Figure 5-6 is an equivalent circuit of the output
stage during the Constant-OFF period. During the
OFF time the lower driver for the particular phase
beign driven remains ON.
The internally generated voltage forces the path
of current though the motor, its series resistance,
the RdsON of the Lower Driver and finally through
the opposite lower driver.
PWM Example (Refer to Figure 5-7)
The following is an example on how to select the
timing parameters.
Given:
DCStart Current
= 1.25A
Ripple Current
= 100mA
Duty Cycle
= 50%
Motor Interface (L)
= 880µH
Total Series Resistance (Rs)
= 4.8Ω
If the worst case start current is 1.25A and the
duty cycle is 50%, then the Peak Current, It will
be:
it
=
1.25
+
0.1
2
it = 1.30A
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