参数资料
型号: MAX5099ATJ+
厂商: Maxim Integrated Products
文件页数: 22/27页
文件大小: 0K
描述: IC REG BUCK SYNC ADJ DL 32TQFN
产品培训模块: Lead (SnPb) Finish for COTS
Obsolescence Mitigation Program
标准包装: 60
类型: 降压(降压)
输出类型: 可调式
输出数: 2
输出电压: 0.8 V ~ 17.1 V
输入电压: 4.5 V ~ 19 V
PWM 型: 电压模式
频率 - 开关: 200kHz ~ 2.2MHz
电流 - 输出: 1A,2A
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: 32-WFQFN 裸露焊盘
包装: 管件
供应商设备封装: 32-TQFN-EP(5x5)
Dual, 2.2MHz, Automotive Synchronous Buck
Converter with 80V Load-Dump Protection
Procedure 2 (See Figure 4)
If the output capacitor used is a low-ESR ceramic type,
the ESR frequency is usually far away from the targeted
unity crossover frequency (f C ). In this case, Type III
V OUT
C CF
compensation is recommended. Type III compensation
provides two-pole zero pairs. The locations of the zero
R I
R1
R F
C F
and poles should be such that the phase margin peaks
around f C . It is also important to place the two zeros at
or below the double pole to avoid the conditional stabil-
ity issue.
C I
R2
FB_
V REF
-
g M
+
COMP_
1) Select a crossover frequency:
f SW ≤ SW
f LC =
3) Place a zero f Z 1 =
at 0 . 75 × f LC .
C CF =
f
20
2) Calculate the LC double-pole frequency, f LC :
1
2 π × L OUT × C OUT
1
2 π × R F × C F
Figure 4. Type III Compensation Network
7) Place a second pole at 1/2 the switching frequency.
C F
( 2 π × 0 . 5 × f SW × R F × C F ) ? 1
Boost Converter Compensation
where
C F =
1
2 π × 0 . 75 × f LC × R F
The boost converter compensation gets complicated
due to the presence of a right-half-plane zero
f ZERO,RHP . The right-half-plane zero causes a drop in
phase while adding positive (+1) slope to the gain
and R F ≥ 10k Ω .
4) Calculate C I for a target unity crossover frequency, f C .
curve. It is important to drop the gain significantly below
unity before the RHP frequency. Use the following pro-
cedure to calculate the compensation components:
C I =
2 π × f C × L OUT × C OUT × V OSC
V IN × R F
1) Calculate the LC double-pole frequency, f LC , and
the right-half-plane-zero frequency.
5) Place a pole f P 1 =
1
2 π × R I × C I
at f ZERO , ESR .
f LC =
1 ? D
2 π × L OUT × C OUT
R I =
1
2 π × f ZERO , ESR × C I
f ZERO , RHP =
( 1 ? D ) 2 R ( MIN )
2 π × L OUT
6) Place a second zero, f Z2 , at 0.2 x f C or at f LC ,
whichever is lower.
where
D = 1 ?
R ( MIN ) =
R 1 =
1
2 π × f Z 2 × C I
? R I
V IN
V OUT
V OUT
I OUT ( MAX )
Target the unity-gain crossover frequency for:
f C ≤
f ZERO,RHP
5
22
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