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參數(shù)資料
| 型號： | L6258EA |
| 廠商： | STMICROELECTRONICS |
| 元件分類： | 運動控制電子 |
| 英文描述： | STEPPER MOTOR CONTROLLER, 2 A, PDSO36 |
| 封裝： | POWER, SO-36 |
| 文件頁數(shù)： | 3/23頁 |
| 文件大小： | 214K |
| 代理商： | L6258EA |
11/23
L6258EA
4
PWM CURRENT CONTROL LOOP
4.1 Open Loop Transfer Function Analysis
Block diagram : refer to Fig. 6.
Figure 8. Application data:
these data refer to a typical application, and will be used as an example during the analysis of the stability of the
current control loop.
The block diagram shows the schematics of the L6258EA internal current control loop working in PWM mode;
the current into the load is a function of the input control voltage VDAC , and the relation between the two vari-
ables is given by the following formula:
Iload RS GS = VDAC Gin
where:
VDAC
is the control voltage defining the load current value
Gin
is the gain of the input transconductance amplifier ( 1/Ra )
Gs
is the gain of the sense transconductance amplifier ( 1/Rb )
Rs
is the resistor connected in series to the output to sense the load current
In this configuration the input voltage is compared with the feedback voltage coming from the sense resistor,
then the difference between this two signals is amplified by the error amplifier in order to have an error signal
controlling the duty cycle of the output stage keeping the load current under control.
It is clear that to have a good performance of the current control loop, the error amplifier must have an high DC
gain and a large bandwidth .
Gain and bandwidth must be chosen depending on many parameters of the application, like the characteristics
of the load, power supply etc..., and most important is the stability of the system that must always be guaran-
teed.
To have a very flexible system and to have the possibility to adapt the system to any application, the error am-
plifier must be compensated using an RC network connected between the output and the negative input of the
same.
For the evaluation of the stability of the system, we have to consider the open loop gain of the current control
loop:
VS = 24V
Gs transconductance gain = 1/Rb
LL = 12mH
Gin transconductance gain = 1/Ra
RL = 12
Ampl. of the Tria_0_180 ref. = 1.6V (peak to peak)
RS = 0.33
Ra = 40K
RC = to be calculated
Rb = 20K
CC = to be calculated
Vr = Internal reference equal to VDD/2 (Typ. 2.5V)
I
LOAD
R
S
1
R
b
-------
V
DAC
1
R
a
-------
=
I
LOAD
V
DAC
R
b
R
a
R
s
------------------
0.5
V
DAC
R
S
--------------- A
()
==
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