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AT5560
Preliminary Product Information
VCM Driver for Mobile Phone
7F, No.9,PARK AVENUE. II, Science-Based Industrial Park, Hsinchu 300,Taiwan, R.O.C.
Tel: 886-3-563-0878 Fax: 886-3-563-0879 WWW:
http://www.aimtron.com.tw
2006/08/24 REV:0.4 Email:
service@aimtron.com.tw
6
Application Information
z
AT5560 provides constant current which can be evaluated by following equation.
RNF
10
V
×
I
IN
=
.
(1)
It’s obviously that the accuracy value of constant current not only depends on the resistance, RNF,
between IM pin and GND, but also the input voltage, V
IN
. In order to get the suitable and stable input
voltage level, VIN, the PWM signal must be low-pass filtered by R1, R2, and C1. The input voltage should
be calculated by following equation.
R2
V
×
+
where D% means duty ratio of PWM signal, and V
pp
is the amplitude of PWM signal. By setting different
value of duty ratio of PWM signal, the corresponding voltage at IM pin we can get, then the constant
current also be set simultaneously.
-
PP
PWM
IN
V
D%
R2
R1
R2
+
V
R2
R1
×
×
=
=
, (2)
z
As Fig 2 shows, in order to get the suitable and stable input voltage level, V
IN
, the PWM signal must be
low-pass filtered by R1, R2, and C1. The values of R1, R2, and C1 will determine -3dB frequency and DC
value of V
IN
. The -3dB frequency can be got by
(
)
C1
×
R1//R2
1
ω
3dB
=
. (3)
For accuracy and stable value of constant current, the input voltage, V
IN
, should be a DC value. So, it is
suggested that the frequency of PWM signal must be larger than 100 times of -3dB frequency.
PWM
3dB
ω
100
1
ω
=
(4)
z
As above condition, to choose the R1, R2, and C1 is necessary for DC value of input voltage
corresponding PWM signal. The following procedure will explain how to design the R1, R2, and C1.
For PWM signal frequency, f
PWM
=100KHz, the sense resistance, RNF=1.5
Ω
, the desired maximum
output constant current is 70mA when duty ratio of PWM signal is 95%. The Vpp=2.8V, choosing R1, R2,
and C1.
1
f
3dB
×
×
If choosing C1=0.1uF, the (R1//R2)=1.59K
Ω
.
And owing to equation (1) and (2), therefore,
10
V
D%
R2
R1
-
PP
×
+
Thus, R1=4.03K
Ω
, R=2.63 K
Ω
can be get.
(
)
1KHz
f
100
1
C1
R1//R2
2
π
PWM
=
×
=
=
394
.
8
×
%
95
5
×
07
.
V
R2
IN
≈
×
=
=
.