參數(shù)資料
型號: L6258E
廠商: 意法半導體
元件分類: 基準電壓源/電流源
英文描述: PWM CONTROLLED - HIGH CURRENT DMOS UNIVERSAL MOTOR DRIVER
中文描述: PWM控制-大電流DMOS的通用型電機驅(qū)動
文件頁數(shù): 11/24頁
文件大?。?/td> 216K
代理商: L6258E
11/24
L6258E
4
PWM CURRENT CONTROL LOOP
4.1 Open Loop Transfer Function Analysis
Block diagram : refer to Fig. 6.
Table 7. 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 L6258E internal current control loop working in PWM mode; the
current into the load is a function of the input control voltage V
DAC
, and the relation between the two variables
is given by the following formula:
I
load
· R
S
· G
S
= V
DAC
· G
in
where:
V
DAC
G
in
G
s
R
s
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:
is the control voltage defining the load current value
is the gain of the input transconductance amplifier ( 1/Ra )
is the gain of the sense transconductance amplifier ( 1/Rb )
is the resistor connected in series to the output to sense the load current
V
S
= 24V
Gs transconductance gain = 1/Rb
L
L
= 12mH
Gin transconductance gain = 1/Ra
R
L
= 12
Ampl. of the Tria_0_180 ref. = 1.6V (peak to peak)
R
S
= 0.33
R
a
= 40K
R
C
= to be calculated
R
b
= 20K
C
C
= to be calculated
V
r
= Internal reference equal to V
DD
/2 (Typ. 2.5V)
I
LOAD
R
S
R
b
------
V
DAC
R
a
------
=
I
LOAD
V
DAC
R
R
a
R
s
------------------
0.5
V
R
S
--------------
A
( )
=
=
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