参数资料
型号: SC4525ASETRT
厂商: Semtech
文件页数: 14/19页
文件大小: 0K
描述: IC REG BUCK 3A 8SOIC
标准包装: 1
类型: 降压(降压)
输出数: 1
输入电压: 8 V ~ 28 V
PWM 型: 电流模式
频率 - 开关: 300kHz ~ 1.3MHz
电流 - 输出: 3A
同步整流器:
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 8-SOIC(0.154",3.90mm Width)裸露焊盘
包装: 标准包装
供应商设备封装: 8-SOIC-EP
产品目录页面: 1358 (CN2011-ZH PDF)
其它名称: SC4525ASEDKR
2 π ? 16 ? 10 ? 22 . 1 ? 10 3
C 5 =
1
3
= 0 . 45 nF
SC4525A
C 8 =
1
2 π ? 600 ? 10 3 ? 22 . 1 ? 10 3
= 12 pF
V o G PWM ( 1 + s R ESR C O )
(3) Place the compensator zero, F Z  , 2 between  0% and
V c ( 1 + s / ω p ) ( 1 + s / ω n Q + s / ω 2 )
G PWM ≈
ω p ≈
ω Z =
F Z . R 1
(5) Then, the parameters of the compensation network R
G CA ? R S
1 conduction loss P C , the switching loss P SW , and bootstrap
circuit P loss P BST, P can + P be SW estimated P as follows:
TOTAL =
+ P BST + Q
ApplicationsInformation(Cont.)
=
20% of the crossover frequency, F C .
(4) Use the compensator pole, F P  , to cancel the ESR zero,
, ,
R C O
can be calculated by
Thermal Considerations
For the power transistor inside the SC4525A, the
,
ESR C O C
10 20
P SW =
? t S ? V IN ? I O ? F SW
P BST = D ? V BST ?
R 7 =
C 5 =
C 8 =
A C
g m
1
2 π F Z 1 R 7
1
2 π F P 1 R 7
P C = D ? V CESAT ? I O
1
2
I O
40
P Q = V IN ? 2 mA
( 0)
where g m =0.28mA/V is the EA gain of the SC4525A.
Example : Determine the voltage compensator for an
V ( 1 ? D ) ? D ? I O D
where P BST = is the BST V supply voltage and t S is the equivalent
switching time of the NPN transistor (see Table 4).
Input Voltage
800kHz,  2V to 3.3V/3A converter with 47uF ceramic
output capacitor.
P 1 ~ 1 . 3 ? I 2O ? R DC IND = 1
Table ( 4. . Typical ) switching time
Load Current
1A 2A
3A
Choose a loop gain crossover frequency of 80kHz, and
place voltage compensator zero 1 and pole ? at F Z  = 6kHz
?
V FB
(20% of F = C ), ? ? ? 20 1 ? log P  ? ? 1 =600kHz. 2 π From Equation (9), the
1 28V 25.3ns 28ns 31ns
and F G R V FB ? ? ?
A C
?
required compensator F C gain V at ? ? F C is
A C = ? 20 ? log ? ? ? CA ? S F C C O V O ?
P TOTAL = P C + P SW + P BST In P addition, the quiescent current loss is
A C = ? A 20 C ? = log ? ? 20 ? log ? ? 3 ?
? 1 ? 1 1 1 . 0 ? 1 1 . 0 ?
? 28 ? 4 . 1 ? 10 28 2 ? π 4 ? . 1 ? ? 10 3 ? 3 ? 47 2 10 ? ? 6 80 3 . ? 3 10 3 ? 47 ? 10 ? 6
? π
?
3 . 3 ?
P C = D ? V CESAT ? I O
10
The total power loss of the SC4525A is therefore
R 7 =
= 31 . 19 8 k
0 . 28 ? 10 ? 3 10 20
P SW =
? t S ? V IN ? I O ? F SW
R 7 = 1 = 31 . 8 k 2
P TOTAL = P C + P SW + P BST + P Q
C 5 = 0 . 28 ? 10 ? 3 = 0 . 31 nF
C 5 =
= P 0 BST . 31 = nF ? V BST ?
1 I O
1 D
2 π ? 600 2 10 3 ? 16 . ? 4 10 3 ? 31 . 4 ? 10
? π ? 31 ? 10
40
The temperature rise ? of the SC4525A P is Q the V product of the
= IN ? 2 mA
P C = D ? V CESAT I O
total power dissipation (Equation ( 2)) and q JA (36 o C/W),
C 8 = G
= 8 . 5 pF
1
2 π ? 600 ? 10 2 ? 31 . 4 ? 10 3 P D = ( 1 ? D ) ? V D ? I O
1
which is the thermal impedance from junction to ambient
P SW =
? t S ? V IN ? I O ?
for the SOIC-8 2 EDP package. F SW
V o PWM ( 1 + s R ESR C O ) 3
=
( 1 + s / ω p ) ( 1 + s / ω n Q + s / ω n )
V c
I
G PWM ≈ ( 1 1 + , s R ESR C O ω ) Z P = IND 1 = ( 1 , . 1 ~ 1 . 3 ) ? I O ? R DC It is not BST recommended O to operate the SC4525A above
G PWM ≈ V o R , ω p P = D ? V BST ?
V c CA ( S 1 + s / ω p ) ( 1 + O s / ω n Q + s / ω ESR n 2 )
upon request for detailed calculation of the compensator
frequency = may ? D need D to I O be reduced to meet the thermal
10
) ? V ?
P D
( 1
R 7 =
parameters.
1 requirement.
R
1
G CA ? R S
G PWM ≈ , ω p ≈ , ω Z = ,
P IND = ( 1 . 1 ~ 1 . 3 ) ? I 2O ? R DC
C 5 = R C O R ESR C O
12V 12.5ns 15.3ns 18ns
24V 22ns 25ns 28ns
? ?
? G CA R S 2 π C O O
? ? ? = 19 dB ? ? = 19 dB
80 10 ?
P Q = V IN ? 2mA (  )
19
Then the compensator parameters are
20
1
2 π ? 16 ? 10 3 ? 31 . 4 ? 10 3 ( 2)
C 8 = 3 = 8 . 5 pF 3
2
Select R 7 =3 .4k, C 5 =0.33nF, and C 8 = 0pF for the design.
2
? G = R 2 O 40 R R C
A C
Compensator parameters C for various typical applications  25 o C junction temperature. In the applications with high
are listed in Table 5. A MathCAD program is also available input voltage and high output current, the switching
20
g m
2 π F Z 1 R 7
2 R π F P 1 = R 7 10
g m
C 8 =
1 A C
7
20
C 5 =
1
 4
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