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參數(shù)資料
| 型號(hào)： | L6246 |
| 廠(chǎng)商： | STMICROELECTRONICS |
| 元件分類(lèi)： | 運(yùn)動(dòng)控制電子 |
| 英文描述： | VOICE COIL MOTOR CONTROLLER, 3 A, PQFP44 |
| 封裝： | 10 X 10 MM, PLASTIC, QFP-44 |
| 文件頁(yè)數(shù)： | 2/12頁(yè) |
| 文件大?。?/td> | 162K |
| 代理商： | L6246 |
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Using the compensation network of the draw of
pag.8, we have a error amplifier transfer function
of:
VO
VI
=
ZC
R1
=
R2
R1
1
+ scR3
1
+ sc (R3 + R2)
so:
Gmax (DC) =
R2
R1
= 1000 = 60dB
with R1 = 1M
and R2 = 1K
zero =
1
2
π R3C
pole =
1
2
π (R3 + R2) C
Note: Fpole is lower than Fzero
The best choice is to cancel the pole of the load
(at around 1KHz) with the zero of the compensa-
tion.
As can be seen the choice of the pole influence
overall in fixing the gain at high frequency.
The gain at high frequency must be choosen in
order to not create instability problem, because
more higher is this gain and lower is the second
pole that we have at high frequency.
If this pole is taken close to the other that we
have already seen at 130KHz and 210KHz, insta-
bility problems can arise.
Adding together AX | dB and ACERR | dB we ob-
taine the Aloop:
So the choice of the compensation network must
be done in order to fix at the beginning the Gmax
of the error amplifier depending on the ratio
R2
R1
.
To calculate the R3 and C values satisfying the
following system:
1
2
π R3C
=
1
2
π L
RL
+ Rsense
Error amplifier zero equal to load pole
1
2
π (R3 + R2)C
=
Admissible Bandwith
Gloop
=
130KHz
10
8912
= 1.5Hz
This example is for crossing the 0dB one decade
before the first pole of the Power Amplifier
(130KHz), starting with a Gloop max of 79dB.
ERROR AMPL. GAIN
(dB)
40
60
20
1K
10K
100K
D95IN271
100
120
80
1
10
100
1M
10M
OPEN LOOP GAIN
ACERR (dB)
40
60
20
1K
10K
100K
D95IN272
100
120
80
1
10
100
1M
10M
CLOSED LOOP
ACERR
X
DIFFERENTS
POLES
EXAMPLES
X
ACERR
(dB)
20
COMPENSATION AT 3Hz
AX(dB)
19
D95IN273
40
60
ALOOP (dB)
20
40
60
79
10
100
1K
10K
100K
1M
10M
COMPENSATION AT 100Hz
IS STABLE
IS NOT STABLE
L6246
10/12
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