参数资料
型号: MAX1954AEUB+T
厂商: Maxim Integrated Products
文件页数: 15/18页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM CM 10-UMAX
产品培训模块: Lead (SnPb) Finish for COTS
Obsolescence Mitigation Program
标准包装: 2,500
PWM 型: 电流模式
输出数: 1
频率 - 最大: 360kHz
占空比: 93%
电源电压: 3 V ~ 13.2 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: -40°C ~ 85°C
封装/外壳: 10-TFSOP,10-MSOP(0.118",3.00mm 宽)
包装: 带卷 (TR)
Low-Cost, Current-Mode PWM Buck
Controller with Foldback Current Limit
output filtering. For other types of capacitors, due to the
higher capacitance and ESR, the frequency of the zero
created by the capacitance and ESR is lower than the
desired closed-loop crossover frequency. Another
The f C should be much higher than the power modula-
tor pole f PMOD . Also, the crossover frequency should
be less than 1/8th of the switching frequency:
compensation capacitor should be added to cancel
this zero.
The basic regulator loop can be thought of as a power
f pMOD << f C <
f S
8
modulator, output feedback divider, and an error ampli-
fier. The power modulator has DC gain set by g mc x
R LOAD , with a pole and zero pair set by R LOAD , the out-
Therefore, the loop-gain equation at the crossover fre-
quency is:
put capacitor (C OUT ) and its equivalent series resis-
tance (R ESR ). Below are equations that define the
power modulator:
G EA ( fC ) × G MOD ( fC ) ×
V FB
V OUT
= 1
R LOAD × f S × L
R LOAD + f S × L
G MOD = g mc ×
where R LOAD = V OUT / I OUT(MAX) , and g mc = 1 / (A CS x
R DS(ON) ), where A CS is the gain of the current-sense
amplifier and R DS(ON) is the on-resistance of the high-
When f zMOD is greater than f C :
G EA ( fC ) = g mEA × R C and G MOD ( fC ) = g mc × R LOAD ×
then R C is calculated as:
f pMOD
f C
side power MOSFET. A CS is 3.5. The frequencies at
which the pole and zero due to the power modulator
occur are determined as follows:
R C =
V OUT
g mEA × V FB × G MOD ( fC )
f pMOD =
2 π × C OUT × ? LOAD
f zMOD =
C C = LOAD
1
? R × f S × L ?
+ R ESR ?
? R LOAD + f S × L ?
1
2 π × C OUT × R ESR
The feedback voltage-divider used has a gain of G FB =
V FB / V OUT , where V FB is equal to 0.8V. The transcon-
ductance error amplifier has DC gain, G EA(DC) = g m x
R O . The amplifier output resistance (R O ) is typically
10M ? . The C C , R O , and the R C set a dominant pole.
where g mEA = 110μs.
The error-amplifier compensation zero formed by R C
and C C should be set at the modulator pole f pMOD . C C
is calculated by:
R × f S × L × C OUT
( R LOAD + ( f S × L ) ) × R C
If f zMOD is less than 5 x f C , add a second compensa-
tion capacitor, C f , from COMP to GND to cancel the
ESR zero. C f is calculated by:
The R C and the C C set a zero. There is an optional pole
set by C F and R C to cancel the output-capacitor ESR
zero if it occurs before crossover frequency (f C ):
C f =
1
2 π × R C × f zMOD
f pdEA =
1
2 π × C C × ( R O + R C )
As the load current decreases, the modulator pole also
decreases. However, the modulator gain increases
accordingly and the crossover frequency remains
G MOD ( fC ) = G MOD ( DC ) ×
f zEA =
f pEA =
1
2 π × C C × R C
1
2 π × C F × R C
the same.
When f zMOD is less than f C , the power-modulator gain
at f C is:
f pMOD
f zMOD
______________________________________________________________________________________
15
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