参数资料
型号: MAX15112EVKIT#
厂商: Maxim Integrated Products
文件页数: 19/23页
文件大小: 0K
描述: EVAL KIT STEP-DOWN 12A MAX15112
产品培训模块: Obsolescence Mitigation Program
标准包装: 1
主要目的: DC/DC,步降
输出及类型: 1,非隔离
功率 - 输出: 18W
输出电压: 1.5V
电流 - 输出: 12A
输入电压: 2.7 V ~ 5.5 V
稳压器拓扑结构: 降压
频率 - 开关: 1MHz
板类型: 完全填充
已供物品:
已用 IC / 零件: MAX15112
MAX15112
High-Efficiency, 12A, Current-Mode Synchronous
Step-Down Regulator with Integrated Switches
R LOAD S [(1- D) - 0.5] ?
? 1 +
L × f SW
R1 + R2
R C = × ?
? ?
× 2 π × f CO × C OUT × ? ESR +
?
? 1 K S [(1- D) - 0.5] ?
?
?
K S = 1 + SLOPE SW MC
Figure 3 shows a graphical representation of the asymp-
totic system closed-loop response, including the domi-
nant pole and zero locations.
The loop response’s fourth asymptote (in bold, Figure 3 )
is the one of interest in establishing the desired crossover
frequency (and determining the compensation compo-
nent values). A lower crossover frequency provides for
stable closed-loop operation at the expense of a slower
load and line-transient response. Increasing the cross-
over frequency improves the transient response at the
(potential) cost of system instability. A standard rule of
thumb sets the crossover frequency P 1/5 to 1/10 of the
switching frequency.
Closing the Loop: Designing
the Compensation Circuitry
1) Select the desired crossover frequency. Choose f CO
equal to 1/10th of f SW , or f CO @ 100kHz.
2) Select R C using the transfer-loop’s fourth asymptote
gain equal to unity (assuming f CO > f P1 , f P2 , and f Z1 ).
R C becomes:
? × K
?
?
R2 g M × g MC × R LOAD
? ?
1
+
? R LOAD L × f SW ?
where K S is calculated as:
V × f × L × g
V IN - V OUT
and g M = 1.1mS, g MC = 80A/V, and V SLOPE = 130mV.
dB
1ST ASYMPTOTE
R2 x (R1 + R2) -1 x 10 AVEA(dB)/20 x g MC x R LOAD x {1 + R LOAD x [K S x (1 – D) – 0.5] x (L x f SW ) -1 } -1
2ND ASYMPTOTE
R2 x (R1 + R2) -1 x g M x (2 G C C ) -1 x g MC x R LOAD x {1 + R LOAD x [K S x (1 – D) – 0.5] x (L x f SW ) -1 } -1
GAIN
3RD ASYMPTOTE
R2 x (R1 + R2) -1 x g M x (2 G C C ) -1 x g MC x R LOAD x {1 + R LOAD x [K S x (1 – D) – 0.5] x (L x f SW ) -1 } -1
x (2 G C OUT x {R LOAD-1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1 ) -1
4TH ASYMPTOTE
R2 x (R1 + R2) -1 x g M x R C x g MC x R LOAD x {1 + R LOAD x [K S x (1 – D) – 0.5] x (L x f SW ) -1 } -1
x (2 G C OUT x {R LOAD-1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1 ) -1
3RD POLE
0.5 x f SW
2ND ZERO
(2 G C OUT ESR) -1
UNITY
1ST POLE
[2 G C C (10 AVEA(dB)/20
x g M -1 )] -1
1ST ZERO
(2 G C C R C ) -1
f CO
FREQUENCY
2ND POLE
f PMOD *
NOTE:
R OUT = 10 AVEA(dB)/20 x g M -1
*f PMOD = [2 G C OUT x (ESR + {R LOAD -1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1 )] -1
WHICH FOR
ESR << {R LOAD -1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1
BECOMES
5TH ASYMPTOTE
R2 x (R1 + R2) -1 x g M x R C x g MC x R LOAD x {1 + R LOAD x [K S x (1 – D) – 0.5] x (L x f SW ) -1 } -1
x [(2 G C OUT x {R LOAD-1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1 ) -1 x (0.5 x f SW )2 x (2 G f) -2
6TH ASYMPTOTE
R2 x (R1 + R2) -1 x g M x R C x g MC x R LOAD x {1 + R LOAD x [K S x (1 – D) – 0.5] x (L x f SW ) -1 } -1
x ESR x {R LOAD-1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1 x (0.5 f SW ) 2 x (2 G f) -2
f PMOD = [2 G C OUT x {R LOAD -1 + [K S (1 – D) – 0.5] x (L x f SW ) -1 } -1 ] -1
f PMOD = (2 G C OUT x R LOAD ) -1 + [K S (1 – D) – 0.5] x (2 G C OUT x L x f SW ) -1
Figure 3. Asymptotic Loop Response of Peak Current-Mode Regulator
???????????????????????????????????????????????????????????????? Maxim Integrated Products 19
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