SC4524CSETRT Semtech, SC4524CSETRT Datasheet - Page 14

• Manufacturer Part Number: SC4524CSETRT
• Description: IC BUCK ADJ 2A 8SOIC
• Manufacturer: Semtech
• Series: -r
• Type: Step-Down (Buck), PWM - Current Moder

Specifications of SC4524CSETRT

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1 V ~ 26.88 V
• Current - Output: 2A
• Frequency - Switching: 300kHz ~ 1.3MHz
• Voltage - Input: 3 V ~ 28 V
• Operating Temperature: -40°C ~ 105°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (0.154", 3.90mm Width) Exposed Pad
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant
• Other names: SC4524CSETR

Applications Information (Cont.)
integrator pole (-90deg) and the dominant pole (-90deg).
The high frequency pole nulls the ESR zero and attenuates
high frequency noise.
Therefore, the procedure of the voltage loop design for
the SC4524C can be summarized as:
(1) Plot the converter gain, i.e. control to feedback transfer
function.
(2) Select the open loop crossover frequency, F
10% and 20% of the switching frequency. At F
required compensator gain, A
ceramic output capacitors, the ESR zero is neglected and
the required compensator gain at F
(3) Place the compensator zero, F
20% of the crossover frequency, F
(4) Use the compensator pole, F
F
(5) Then, the parameters of the compensation network
can be calculated by
Z
.
-60
-60
-30
-30
60
60
30
30
0
0
1K
1K
R
R
C
C
C
C
A
A
A
A
R
R
C
C
C
C
G
G
R
R
C
C
C
C
V
V
A
A
A
A
V
V
V
V
V
V
7
7
o
o
Figure 8. Bode plots for voltage loop design
C
C
C
C
5
5
8
8
c
c
7
7
7
7
C
C
C
C
5
5
8
8
o
o
c
c
PWM
PWM
5
5
8
8
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
. 0
. 0
1 (
1 (
−
−
−
−
10
10
2
2
2
2
. 0
. 0
1 (
1 (
−
−
−
−
2
2
2
2
2
2
2
2
≈
≈
g
g
⋅ π
⋅ π
⋅ π
⋅ π
⋅ π
⋅ π
20
20
20
20
28
28
⋅ π
⋅ π
π
π
π
π
20
20
20
20
28
28
+
+
G
G
+
+
10
10
m
m
10
10
A
A
20
20
F
F
F
F
Fp
Fp
C
C
CA
CA
16
16
600
600
1
1
1
1
/ s
/ s
16
16
600
600
/ s
/ s
1 Z
1 Z
P
P
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
1
1
⋅
⋅
R
R
15
15
15
15
log
log
log
log
10
10
20
20
log
log
log
log
10
10
20
20
⋅
⋅
10K
10K
R
R
R
R
G
G
9 .
9 .
G
G
ω
ω
9 .
9 .
R
R
⋅
⋅
ω
ω
⋅
⋅
10
10
7
7
PWM
PWM
10
10
7
7
PWM
PWM
⋅
⋅
p
p
S
S
⋅
⋅
p
p


 
 








−
−


 
 








−
−
10
10
1 ( )
1 ( )
10
10
1 ( )
1 ( )
3
3
3
3
,
,
G
G
28
28
Fz1
Fz1
G
G
28
28
3
3
1
1
3
3
1
1
1 (
1 (
=
=
CA
CA
1 (
1 (
1
1
=
=
CA
CA
1
1
3
3
1
1
⋅
⋅
3
3
+
+
1
1
⋅
⋅
+
+
FREQUENCY (Hz)
FREQUENCY (Hz)
⋅
⋅
22
22
22
22
⋅
⋅
22
22
22
22
R
R
R
R
⋅
⋅
+
+
⋅
⋅
+
+
6
6
/ s
/ s
6
6
/ s
/ s
22
22
22
22
S
S
S
S
1 .
1 .
R s
R s
1 .
1 .
R s
R s
.
.
1 .
1 .
1
1
.
.
1 .
1 .
1
1
k 3
k 3
⋅
⋅
k 3
k 3
C
⋅
⋅
ω
ω
ω
ω
1 .
1 .
. In typical applications with
1 .
1 .
2
2
⋅
⋅
Fc
Fc
100K
100K
ESR
ESR
2
2
⋅
⋅
ESR
ESR
⋅
⋅
⋅
⋅
n
n
10
10
n
n
10
10
π
π
ω
ω
10
10
π
π
10
10
P1
Q
Q
⋅
⋅
Q
Q
⋅
⋅
F
F
10
10
F
F
p
p
C
C
10
10
1
1
C
C
, to cancel the ESR zero,
1
1
C
+
+
C
C
−
−
+
+
C
C
−
−
.
3
3
≈
≈
O
O
3
3
O
O
3
3
C
C
Z1
3
3
C
C
C
s
s
Fp1
Fp1
s
s
)
)
3
3
Fz
Fz
)
)
3
3
R
R
=
=
can be estimated by
O
O
, between 10% and
=
=
⋅
⋅
O
O
2
2
⋅
⋅
2
2
2
2
1
1
C
C
=
=
2
2
=
=
. 0
. 0
/
/
. 0
. 0
⋅
⋅
/
/
⋅
⋅
⋅ π
⋅ π
O
O
12
12
⋅ π
⋅ π
12
12
V
V
ω
ω
Fsw/2
Fsw/2
ω
ω
V
V
V
V
45
45
V
V
45
45
FB
FB
,
,
FB
FB
O
O
2
2
n
n
1M
1M
2
2
n
n
O
O
80
80
80
80
pF
pF
pF
pF
)
)
)
)
nF
nF
nF
nF


 
 




 
 


⋅
⋅
⋅
⋅
C
10
10
10
10
, between
C
, find the
(9)
1
1
1
1
3
3
3
3
⋅
⋅
⋅
⋅
ω
ω
22
22
22
22
Z
Z
10M
10M
=
=
⋅
⋅
⋅
⋅
10
10
10
10
R
R
ESR
ESR
−
−
−
−
1
1
6
6
6
6
C
C
where g
Example: Determine the voltage compensator for an
800kHz, 12V to 3.3V/2A converter with 22uF ceramic
output capacitor.
Choose a loop gain crossover frequency of 80kHz, and
place voltage compensator zero and pole at F
(20% of F
required compensator gain at F
Then the compensator parameters are
Select R
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.
Thermal Considerations
For the power transistor inside the SC4524C, the
conduction loss P
circuit loss P
where V
⋅
⋅
⋅
⋅
O
O
1
1
3
3
1
1
3
3
,
,
0 .
0 .
3 .
3 .
0 .
0 .
3 .
3 .












P
P
P
P
P
P
P
P
P
P
P
P
A
A
R
R
C
C
C
C
=
=
=
=
C
C
BST
BST
D
D
IND
IND
TOTAL
TOTAL
SW
SW
7
BST
C
C
m
5
5
7
7
8
8
=12.4k, C
15
15
15
15
=0.3mA/V is the EA gain of the SC4524C.
=
=
=
=
is the BST supply voltage and t
=
=
C
=
=
=
=
D
D
1 (
1 (
20
20
), and F
9 .
9 .
0.3
0.3
2
2
9 .
9 .
2
2
BST,
=
=
1
1
1 (
1 (
2
2
D
D
⋅
⋅
10
10
log
log
dB
dB
−
−
dB
dB
V
V
P
P
1 .
1 .
⋅
⋅
⋅
⋅
16
16
can be estimated as follows:
600
600
11.4
11.4
CESAT
CESAT
10
10
C
C
) D
) D
t
t
V
V
20
20
18.5
18.5
S
S
BST
BST
~
~
+
+
C
5
10
10
⋅
⋅
⋅
⋅
3
3
=1nF, and C
, the switching loss P
1
1
P
P
V
V
10
10
V
V
5.5
5.5
3
3
1
1
1
1
⋅
⋅
⋅
⋅
D
D
3 .
3 .
P1
SW
SW
IN
IN
1
1
I
I
3
3
40
40
1
1
O
O
=600kHz. From Equation (9), the
I
I
1
1
⋅
⋅
10
10
I )
I )
O
O
2
2
⋅
⋅
I
I
2
2
1
1
⋅
⋅
+
+
I
I
O
O
4 .
4 .
4 .
4 .
3
3
2
2
O
O
2
2
O
O
P
P
k
k
4 .
4 .
⋅
⋅
2
2
BST
BST
⋅
⋅
10
10
F
F
R
R
10
10
SW
SW
80
80
3
3
DC
DC
+
+
3
3
8
10
10
=22pF for the design.
C
P
P
1
1
3
3
0.8nF
0.8nF
is
Q
Q
22
22
2
2
1
1
10
10
pF
pF
P
P
6
6
Q
Q
SW
1.0
1.0
3.3
3.3
=
=
S
is the equivalent
, and bootstrap
V
V
IN
IN
11.4dB
11.4dB
⋅
⋅
2
2
mA
mA
(10)
Z1
=16kHz
14