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MIC3232 Datasheet(PDF) 14 Page - Micrel Semiconductor

Part # MIC3232
Description  Constant Current Boost Controller for Driving High Power LEDs
PDF  19 Pages
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Manufacturer  MICREL [Micrel Semiconductor]
Direct Link  http://www.micrel.com
Logo MICREL - Micrel Semiconductor

MIC3232 Datasheet(HTML) 14 Page - Micrel Semiconductor

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Micrel, Inc.
MIC3230/1/2
November 2009
14
M9999-113009-B
(see the current waveforms in Figure 5).
It can be difficult to find large inductor values with high
saturation currents in a surface mount package. Due to
this, the percentage of the ripple current may be limited
by the available inductor. It is recommended to operate
in the continuous conduction mode. The selection of L
described here is for continuous conduction mode.
Eq. (13)
PP
in
IN
I
T
D
V
L
_
×
×
=
Using the nominal values, we get:
H
A
s
V
L
μ
μ
43
31
.
0
2
56
.
0
12
=
×
×
=
Select the next higher standard inductor value of 47µH.
Going back and calculating the actual ripple current
gives:
Eq. (13a)
PP
nom
nom
IN
PP
in
A
uh
us
v
L
T
D
V
I
29
.
0
47
2
56
.
0
12
_
_
=
×
×
=
×
×
=
The average input current is different than the RMS input
current because of the ripple current. If the ripple current
is low, then the average input current nearly equals the
RMS input current. In the case where the average input
current is different than the RMS, Equation 10 shows the
following:
Eq. (13b)
() (
)
12
2
_
2
max
_
_
max
_
_
PP
IN
RMS
IN
AVE
IN
I
I
I
−
=
() (
)
A
I
AVE
IN
64
.
1
12
/
29
.
0
64
.
1
2
2
max
_
_
≈
−
=
The Maximum Peak input current IL_PK can found using
equation 11:
A
I
I
I
PP
L
AVE
IN
PK
L
78
.
1
5
.
0
max
_
_
max
_
_
max
_
_
=
×
+
=
The saturation current (ISAT) at the highest operating
temperature of the inductor must be rated higher than
this.
The power dissipated in the inductor is:
Eq. (13c)
DCR
I
P
RMS
in
INDUCTOR
×
=
2
max
_
_
Current Limit and Slope Compensation
Having calculated the IL_pk above, We can set the current
limit 20% above this maximum value:
A
A
I
Limit
pk
L
9
.
1
6
.
1
2
.
1
_
=
×
=
The internal current limit comparator reference is set at
0.45V, therefore when
45
.
0
_
=
PIN
IS
V
, the IC enters
current limit.
Eq. (14)
(
)
PK
A
Vcs
V
PK
+
=
45
.
0
Where
PK
A
V
is the peak of the
A
V waveform and
PK
Vcs
is the peak of the Vcs waveform
Eq. (14a)
CS
pk
L
SLC
RAMP
R
I
D
R
I
Limit
×
+
×
×
=
_
45
.
0
To calculate the value of the slope compensation resistance,
RSLC, we can use Equation 5:
(
)
SW
CS
IN
OUT
SLC
F
A
L
R
V
V
R
MIN
MAX
×
×
×
−
=
μ
250
First we must calculate RCS, which is given below in
Equation 15:
Eq. (15)
()
Limit
pk
L
SW
MIN
MAX
CS
I
F
L
D
VIN
VOUT
R
_
max
45
.
0
+
×
×
−
=
Therefore;
() (
)
Ω
=
+
×
×
−
=
m
A
kHz
H
v
v
RCS
179
9
.
1
500
47
50
.
0
8
28
45
.
0
μ
Using a standard value 150mΩ resistor for RCS, we obtain
the following for RSLC:
(
)
Ω
=
×
×
Ω
×
−
=
511
500
250
47
150
8
28
kHz
A
H
m
RSLC
μ
μ
Use the next higher standard value if this not a standard
value. In this example 511Ω is a standard value.
Check: Because we must use a standard value for Rcs and
RSLC;
Limit
pk
L
I _
may be set at a different level (if the
calculated value isn’t a standard value) and we must
calculate the actual
Limit
pk
L
I _
value (remember
Limit
pk
L
I _
is
the same as
Limit
pk
in
I _
).
Rearranging Equation 14a to solve for
Limit
pk
L
I _
:
CS
SLC
RAMP
pk
in
R
D
R
I
I
Limit
)
45
.
0
(
_
×
×
−
=
A
ua
I
Limit
actual
in
34
.
2
150
.
)
75
.
0
511
250
45
.
0
(
_
=
×
×
−
=
This is higher than the initial
A
I
PK
L
9
.
1
2
.
1
max
_
_
=
×
limit
because we have to use standard values for RCS and for
RSLC. If
Limit
actual
in
I _
is too high than use a higher value for
RCS. The calculated value of RCS for a 1.9A current limit was
179mΩ. In this example, we have chosen a lower value
which results in a higher current limit. If we use a higher
standard value the current limit will have a lower value. The
designer does not have the same choices for small valued
resistors as with larger valued resistors. The choices differ
from resistor manufacturers. If too large a current sense
resistor is selected, the maximum output power may not be
able to be achieved at low input line voltage levels. Make
sure the inductor will not saturate at the actual current limit
Limit
actual
in
I _
.
Perform a check at IIN=2.34Apk.
(
)
V
m
A
A
V
PIN
IS
45
.
0
150
34
.
2
511
78
.
0
250
_
=
Ω
×
+
Ω
×
×
=
μ



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