L292 STMicroelectronics, L292 Datasheet - Page 10

• Manufacturer Part Number: L292
• Description: IC DC MOTOR DRIVER MULTIWATT15V
• Manufacturer: STMicroelectronics
• Type: Driverr

Specifications of L292

• Applications: DC Motor Driver
• Number Of Outputs: 1
• Current - Output: ±2A
• Voltage - Supply: 18 V ~ 36 V
• Operating Temperature: -40°C ~ 150°C
• Mounting Type: Through Hole
• Package / Case: Multiwatt-15 (Horizontal, Bent and Staggered Leads)
• Operating Temperature Classification: Automotive
• Motor Controller Type: DC Motor Driver
• Package Type: MULTIWATT V
• Operating Supply Voltage (max): 36V
• Operating Supply Voltage: 18 V to 36 V
• Mounting Style: Through Hole
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Voltage - Load: -
• Lead Free Status / Rohs Status: Compliant
• Other names: 497-1388-5

L292
Figure 8. Small Signal Step Response (normalized amplitude vs. t / R
V
I
t = 100 s/div. with V
It is possible to verify that the L292 works in "closed-loop" conditions during the entire motor current rise-
time: the voltage at pin 7 inverting input of the error amplifier) is locked to the reference voltage VR,
present at the non-inverting input of the same amplifier.
The previous linear analysis is correct for this example.
Descresing the
(6), the maximum value of
b) Large signal reponse
The large step signal response is limited by slew-rate and inductive load.
In this case, during the rise-time of the motor current, The L292 works is open-loop condition.
CLOSED LOOP SYSTEM BANDWIDTH.
A good choice for x is the value 1 / Ö2. In this case :
The module of the transfer function is :
The cutoff frequency is derived by the expression (9) by putting
10/13
I
----- s
V
M
I
-----
V
M
7
M
I
I
= 100 mA/div.
= 200 mV/div.
=
=
0.044
-------------- -
R
0.044
-------------- -
s
R
s
--------------------------------------------------------------------------------------------------------------------- -
--------------------------------------------------------------------- -
1
1
+
value, the rise-time of the current decreases. But for a good stability, from relationship
+
2s R
2 R
I
= 1.5 Vp.
1
F
C
+
F
F
C
2 1
sR
+
F
is:
F
2s
2
C
+
+
2
F
min
R
1
2
F
R
2
C
F
T
=
2
1 2
F
C
=
------------- -
2
2
–
4
F
1
-------------- -
R
2
0.9
F
2
R
C
F
(phase margin = 45°)
F
C
F
2
f
+
T
1
=
---------------------- -
2 R
0.9
I
-----
V
F
M
C
I
F
= 0.707 ·
F
C
F
).
0.044
-------------- -
R
s
(-3 dB), from which:
(8)
(9)