mic3230 Micrel Semiconductor, mic3230 Datasheet - Page 14

• Manufacturer Part Number: mic3230
• Description: Constant Current Boost Controller For Driving High Power Leds
• Manufacturer: Micrel Semiconductor

Iout is the same as ILED
Selecting the inductor current (peak-to-peak), I
between 20% to 50% of I
obtain:
(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)
Using the nominal values, we get:
Select the next higher standard inductor value of 47µH.
Going back and calculating the actual ripple current
gives:
Eq. (13a)
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)
The Maximum Peak input current I
equation 11:
The saturation current (I
temperature of the inductor must be rated higher than
this.
The power dissipated in the inductor is:
Eq. (13c)
Micrel, Inc.
January 2009
I
in
I
L
_
_
I
IN
PP
I
PK
IN
I
_
IN
_
_
_
RMS
I
nom
in
_
max
RMS
P
AVE
INDUCTOR
_
PP
_
=
_
nom
=
_
min
=
max
I
L
I
0
V
IN
L
IN
4 .
IN
=
=
=
=
_
_
I
_
V
AVE
V
V
12
in
=
AVE
nom
OUT
OUT
IN
=
I
_
V
in
rms
_
I
eff
×
(
eff
×
×
L
_
in
. 1
max
_
_
. 0
_
max
D
D
. 0
PP
_
nom
IN_RMS_nom
64
min
×
31
nom
×
_
SAT
56
RMS
×
V
V
nom
=
A
)
+
T
IN
2
IN
×
×
×
×
) at the highest operating
I
I
0
_
T
−
OUT
2
(
_
OUT
I
_
IN
max
5 .
μ
nom
(
=
max
=
s
. 0
_
12
×
0
=
29
_
RMS
_
, in this case 40%, we
4 .
min
nom
v
I
2
43
L_PK
)
×
L
2
×
*
μ
_
47
_
. 0
=
. 0
/
H
max
DCR
PP
=
56
12
uh
. 0
78
. 0
can found using
_
48
×
) (
2
max
78
≈
2
−
=
us
A
. 1
A
. 0
_
I
64
IN
=
=
_
rms
31
L_PP
. 0
12
. 1
_
rms
A
PP
29
A
78
P
A
)
, to be
2
PP
−
A
P
14
Current Limit and Slope Compensation
Having calculated the I
limit 20% above this maximum value:
The internal current limit comparator reference is set at
0.45V, therefore when
current limit.
Eq. (14)
Where
Eq. (14a)
To calculate the value of the slope compensation
resistance, R
First we must calculate R
Equation 15:
Eq. (15)
Therefore;
Using a standard value 150mΩ resistor for R
obtain the following for R
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 R
calculated value isn’t a standard value) and we must
calculate
Rearranging Equation 14a to solve for
This is higher than the initial
limit because we have to use standard values for R
Vcs
I
L
_
I
PK
pk
in
Limit
SLC;
_
is the peak of the Vcs waveform
V
actual
A
is the same as
I
I
PK
R
in
L
R
R
CS
the
_
is the peak of the
SLC
_
CS
Limit
SLC
pk
pk
=
=
Limit
Limit
I
R
(
, we can use Equation 5:
=
VOUT
=
L
. 0
. 0
(
SLC
actual
_
28
47
(
45
47
45
pk
. 0
=
v
may be set at a different level (if the
μ
μ
Limit
=
45
=
MAX
−
H
(
(
H
=
28
. 0
(
8
V
L_pk
I
×
v
×
(
OUT
−
RAMP
L
=
45
V
500
) (
−
I
SLC
−
250
L
V
CS
×
. 0
in _
250
1
×
A
VIN
8
×
I
F
above, We can set the current
2 .
MAX
IS
PK
45
L
−
, which is given below in
)
SW
. 0
250
kHz
:
×
×
μ
_
_
pk
I
×
50
.
MIN
A
ua
+
. 0
150
RAMP
−
150
PIN
R
pk
1
V
Limit
V
μ
45
×
Vcs
)
6 .
SLC
IN
A
Limit
R
A
)
×
+
500
1
m
×
A
MIN
×
=
1
CS
).
511
2 .
D
waveform and
Ω
×
F
9 .
=
PK
max
SW
. 0
)
×
kHz
×
×
D
A
1
R
R
45
)
×
I
9 .
value
+
=
L
CS
+
SLC
I
. 0
179
A
L
_
=
I
M9999-011409-A
I
, the IC enters
L
L
75
_
PK
511
_
_
m
pk
MIC3230/1/2
pk
pk
×
)
Ω
_
Limit
Limit
Ω
Limit
D
CS
=
max
(remember
)
, we
. 2
×
:
34
R
=
CS
A
1
9 .
CS
A