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SC4525ASETRT датащи(PDF) 14 Page - Semtech Corporation

номер детали SC4525ASETRT
подробное описание детали  28V 3A Step-Down Switching Regulator
PDF  19 Pages
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(3) Place the compensator zero, F
Z, between 0% and
20% of the crossover frequency, F
C.
(4) Use the compensator pole, F
P, to cancel the ESR zero,
F
Z.
(5) Then, the parameters of the compensation network
can be calculated by
where g
m=0.28mA/V is the EA gain of the SC4525A.
Example: Determine the voltage compensator for an
800kHz, 2V to 3.3V/3A converter with 47uF ceramic
output capacitor.
Choose a loop gain crossover frequency of 80kHz, and
place voltage compensator zero and pole at F
Z=6kHz
(20% of F
C), and FP=600kHz. From Equation (9), the
required compensator gain at F
C is
Then the compensator parameters are
Select R
7=3.4k, C5=0.33nF, and C8=0pF for the design.
Compensator parameters for various typical applications
are listed in Table 5. A MathCAD program is also available
upon request for detailed calculation of the compensator
parameters.
CESAT
D
IN
D
O
V
V
V
V
V
D
+
+
=
=
1
V
0
.
1
V
R
R
O
6
4
1
SW
D
O
L
L
F
)
D
1
(
)
V
V
(
I
+
=
D
SW
O
D
O
1
F
I
%
20
)
D
1
(
)
V
V
(
L
+
=
)
D
1
(
D
I
I
O
CIN
_
RMS
=


+
D
=
D
O
SW
L
O
C
F
8
1
ESR
I
V
SW
IN
O
IN
F
V
4
I
C
D
>
,
R
G
R
G
S
CA
PWM
)
/
s
Q
/
s
1
()
/
s
1
(
)
C
R
s
1
(
G
V
V
2
n
2
n
p
O
ESR
PWM
c
o
ω
+
ω
+
ω
+
+
=
7
1
Z
5
R
F
2
1
C
π
=
7
1
P
8
R
F
2
1
C
π
=
,
C
R
1
O
p
ω
,
C
R
1
O
ESR
Z =
ω
k
3
.
22
10
28
.
0
10
R
3
7
20
9
.
15
=
=
nF
45
.
0
10
1
.
22
10
16
2
1
C
3
3
5
=
π
=
pF
12
10
1
.
22
10
600
2
1
C
3
3
8
=
π
=


π
=
O
FB
O
C
S
CA
C
V
V
C
F
2
1
R
G
1
log
20
A
dB
9
.
15
3
.
3
0
.
1
10
22
10
80
2
1
10
1
.
6
28
1
log
20
A
6
3
3
C
=
π
=
m
7
g
10
R
20
C
A
=
CESAT
D
IN
D
O
V
V
V
V
V
D
+
+
=
=
1
V
0
.
1
V
R
R
O
6
4
1
SW
D
O
L
L
F
)
D
1
(
)
V
V
(
I
+
=
D
SW
O
D
O
1
F
I
%
20
)
D
1
(
)
V
V
(
L
+
=
)
D
1
(
D
I
I
O
CIN
_
RMS
=


+
D
=
D
O
SW
L
O
C
F
8
1
ESR
I
V
SW
IN
O
IN
F
V
4
I
C
D
>
,
R
G
R
G
S
CA
PWM
)
/
s
Q
/
s
1
()
/
s
1
(
)
C
R
s
1
(
G
V
V
2
n
2
n
p
O
ESR
PWM
c
o
ω
+
ω
+
ω
+
+
=
7
1
Z
5
R
F
2
1
C
π
=
7
1
P
8
R
F
2
1
C
π
=
,
C
R
1
O
p
ω
,
C
R
1
O
ESR
Z =
ω
k
3
.
22
10
28
.
0
10
R
3
7
20
9
.
15
=
=
nF
45
.
0
10
1
.
22
10
16
2
1
C
3
3
5
=
π
=
pF
12
10
1
.
22
10
600
2
1
C
3
3
8
=
π
=


π
=
O
FB
O
C
S
CA
C
V
V
C
F
2
1
R
G
1
log
20
A
dB
9
.
15
3
.
3
0
.
1
10
22
10
80
2
1
10
1
.
6
28
1
log
20
A
6
3
3
C
=
π
=
m
7
g
10
R
20
C
A
=
CESAT
D
IN
D
O
V
V
V
V
V
D
+
+
=
=
1
V
0
.
1
V
R
R
O
6
4
1
SW
D
O
L
L
F
)
D
1
(
)
V
V
(
I
+
=
D
)
D
1
(
D
I
I
O
CIN
_
RMS
=


+
D
=
D
O
SW
L
O
C
F
8
1
ESR
I
V
SW
IN
O
IN
F
V
4
I
C
D
>
,
R
G
R
G
S
CA
PWM
)
/
s
Q
/
s
1
()
/
s
1
(
)
C
R
s
1
(
G
V
V
2
n
2
n
p
O
ESR
PWM
c
o
ω
+
ω
+
ω
+
+
=
7
1
Z
5
R
F
2
1
C
π
=
7
1
P
8
R
F
2
1
C
π
=
,
C
R
1
O
p
ω
,
C
R
1
O
ESR
Z =
ω


π
=
O
FB
O
C
S
CA
C
V
V
C
F
2
1
R
G
1
log
20
A
m
7
g
10
R
20
C
A
=
SW
O
D
O
1
F
I
%
35
)
D
1
(
)
V
V
(
L
+
=
dB
19
3
.
3
0
.
1
10
47
10
80
2
1
10
1
.
4
28
1
log
20
A
6
3
3
C
=
π
=
k
8
.
31
10
28
.
0
10
R
3
7
20
19
=
=
nF
31
.
0
10
4
.
31
10
16
2
1
C
3
3
5
=
π
=
pF
5
.
8
10
4
.
31
10
600
2
1
C
3
3
8
=
π
=
CESAT
D
IN
D
O
V
V
V
V
V
D
+
+
=
=
1
V
0
.
1
V
R
R
O
6
4
1
SW
D
O
L
L
F
)
D
1
(
)
V
V
(
I
+
=
D
)
D
1
(
D
I
I
O
CIN
_
RMS
=


+
D
=
D
O
SW
L
O
C
F
8
1
ESR
I
V
SW
IN
O
IN
F
V
4
I
C
D
>
,
R
G
R
G
S
CA
PWM
)
/
s
Q
/
s
1
()
/
s
1
(
)
C
R
s
1
(
G
V
V
2
n
2
n
p
O
ESR
PWM
c
o
ω
+
ω
+
ω
+
+
=
7
1
Z
5
R
F
2
1
C
π
=
7
1
P
8
R
F
2
1
C
π
=
,
C
R
1
O
p
ω
,
C
R
1
O
ESR
Z =
ω


π
=
O
FB
O
C
S
CA
C
V
V
C
F
2
1
R
G
1
log
20
A
m
7
g
10
R
20
C
A
=
SW
O
D
O
1
F
I
%
35
)
D
1
(
)
V
V
(
L
+
=
dB
19
3
.
3
0
.
1
10
47
10
80
2
1
10
1
.
4
28
1
log
20
A
6
3
3
C
=
π
=
k
8
.
31
10
28
.
0
10
R
3
7
20
19
=
=
nF
31
.
0
10
4
.
31
10
16
2
1
C
3
3
5
=
π
=
pF
5
.
8
10
4
.
31
10
600
2
1
C
3
3
8
=
π
=
CESAT
D
IN
D
O
V
V
V
V
V
D
+
+
=
=
1
V
0
.
1
V
R
R
O
6
4
1
SW
D
O
L
L
F
)
D
1
(
)
V
V
(
I
+
=
D
)
D
1
(
D
I
I
O
CIN
_
RMS
=


+
D
=
D
O
SW
L
O
C
F
8
1
ESR
I
V
SW
IN
O
IN
F
V
4
I
C
D
>
,
R
G
R
G
S
CA
PWM
)
/
s
Q
/
s
1
()
/
s
1
(
)
C
R
s
1
(
G
V
V
2
n
2
n
p
O
ESR
PWM
c
o
ω
+
ω
+
ω
+
+
=
7
1
Z
5
R
F
2
1
C
π
=
7
1
P
8
R
F
2
1
C
π
=
,
C
R
1
O
p
ω
,
C
R
1
O
ESR
Z =
ω


π
=
O
FB
O
C
S
CA
C
V
V
C
F
2
1
R
G
1
log
20
A
m
7
g
10
R
20
C
A
=
SW
O
D
O
1
F
I
%
35
)
D
1
(
)
V
V
(
L
+
=
dB
19
3
.
3
0
.
1
10
47
10
80
2
1
10
1
.
4
28
1
log
20
A
6
3
3
C
=
π
=
k
8
.
31
10
28
.
0
10
R
3
7
20
19
=
=
nF
31
.
0
10
4
.
31
10
16
2
1
C
3
3
5
=
π
=
pF
5
.
8
10
4
.
31
10
600
2
1
C
3
3
8
=
π
=
CESAT
D
IN
D
O
V
V
V
V
V
D
+
+
=
=
1
V
0
.
1
V
R
R
O
6
4
1
SW
D
O
L
L
F
)
D
1
(
)
V
V
(
I
+
=
D
)
D
1
(
D
I
I
O
CIN
_
RMS
=


+
D
=
D
O
SW
L
O
C
F
8
1
ESR
I
V
SW
IN
O
IN
F
V
4
I
C
D
>
,
R
G
R
G
S
CA
PWM
)
/
s
Q
/
s
1
()
/
s
1
(
)
C
R
s
1
(
G
V
V
2
n
2
n
p
O
ESR
PWM
c
o
ω
+
ω
+
ω
+
+
=
7
1
Z
5
R
F
2
1
C
π
=
7
1
P
8
R
F
2
1
C
π
=
,
C
R
1
O
p
ω
,
C
R
1
O
ESR
Z =
ω


π
=
O
FB
O
C
S
CA
C
V
V
C
F
2
1
R
G
1
log
20
A
m
7
g
10
R
20
C
A
=
SW
O
D
O
1
F
I
%
35
)
D
1
(
)
V
V
(
L
+
=
dB
19
3
.
3
0
.
1
10
47
10
80
2
1
10
1
.
4
28
1
log
20
A
6
3
3
C
=
π
=
k
8
.
31
10
28
.
0
10
R
3
7
20
19
=
=
nF
31
.
0
10
4
.
31
10
16
2
1
C
3
3
5
=
π
=
pF
5
.
8
10
4
.
31
10
600
2
1
C
3
3
8
=
π
=
Thermal Considerations
For the power transistor inside the SC4525A, the
conduction loss P
C, the switching loss PSW, and bootstrap
circuit loss P
BST, can be estimated as follows:
(0)
whereV
BST is the BST supply voltage and tS is the equivalent
switching time of the NPN transistor (see Table 4).
Table 4. Typical switching time
In addition, the quiescent current loss is
()
The total power loss of the SC4525A is therefore
(2)
The temperature rise of the SC4525A is the product of the
total power dissipation (Equation (2)) and q
JA (36
o
C/W),
which is the thermal impedance from junction to ambient
for the SOIC-8 EDP package.
It is not recommended to operate the SC4525A above
25oC junction temperature. In the applications with high
input voltage and high output current, the switching
frequency may need to be reduced to meet the thermal
requirement.
O
CESAT
C
I
V
D
P
=
40
I
V
D
P
O
BST
BST
=
DC
2
O
IND
R
I
)
3
.
1
~
1
.
1
(
P
=
O
D
D
I
V
)
D
1
(
P
=
SW
O
IN
S
SW
F
I
V
t
2
1
P
=
Q
BST
SW
C
TOTAL
P
P
P
P
P
+
+
+
=
mA
2
V
P
IN
Q
=
O
CESAT
C
I
V
D
P
=
40
I
V
D
P
O
BST
BST
=
DC
2
O
IND
R
I
)
3
.
1
~
1
.
1
(
P
=
O
D
D
I
V
)
D
1
(
P
=
SW
O
IN
S
SW
F
I
V
t
2
1
P
=
Q
BST
SW
C
TOTAL
P
P
P
P
P
+
+
+
=
mA
2
V
P
IN
Q
=
1A
2A
3A
12V
12.5ns
15.3ns
18ns
24V
22ns
25ns
28ns
28V
25.3ns
28ns
31ns
Load Current
Input Voltage
1A
2A
3A
12V
12.5ns
15.3ns
18ns
24V
22ns
25ns
28ns
28V
25.3ns
28ns
31ns
Load Current
Input Voltage
O
CESAT
C
I
V
D
P
=
40
I
V
D
P
O
BST
BST
=
DC
2
O
IND
R
I
)
3
.
1
~
1
.
1
(
P
=
O
D
D
I
V
)
D
1
(
P
=
SW
O
IN
S
SW
F
I
V
t
2
1
P
=
Q
BST
SW
C
TOTAL
P
P
P
P
P
+
+
+
=
mA
2
V
P
IN
Q
=
O
CESAT
C
I
V
D
P
=
40
I
V
D
P
O
BST
BST
=
DC
2
O
IND
R
I
)
3
.
1
~
1
.
1
(
P
=
O
D
D
I
V
)
D
1
(
P
=
SW
O
IN
S
SW
F
I
V
t
2
1
P
=
Q
BST
SW
C
TOTAL
P
P
P
P
P
+
+
+
=
mA
2
V
P
IN
Q
=
O
CESAT
C
I
V
D
P
=
40
I
V
D
P
O
BST
BST
=
DC
2
O
IND
R
I
)
3
.
1
~
1
.
1
(
P
=
O
D
D
I
V
)
D
1
(
P
=
SW
O
IN
S
SW
F
I
V
t
2
1
P
=
Q
BST
SW
C
TOTAL
P
P
P
P
P
+
+
+
=
mA
2
V
P
IN
Q
=
O
CESAT
C
I
V
D
P
=
40
I
V
D
P
O
BST
BST
=
DC
2
O
IND
R
I
)
3
.
1
~
1
.
1
(
P
=
O
D
D
I
V
)
D
1
(
P
=
SW
O
IN
S
SW
F
I
V
t
2
1
P
=
Q
BST
SW
C
TOTAL
P
P
P
P
P
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