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参数资料
| 型号： | MAX16977SAUE/V+ |
| 厂商： | Maxim Integrated Products |
| 文件页数： | 15/18页 |
| 文件大小： | 0K |
| 描述： | IC BUCK SYNC ADJ 2A 16TSSOP |
| 标准包装： | 96 |
| 类型： | 降压（降压） |
| 输出类型： | 两者兼有 |
| 输出数： | 1 |
| 输出电压： | 5V，1 V ~ 10 V |
| 输入电压： | 3.5 V ~ 36 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 1MHz ~ 2.2MHz |
| 电流 - 输出： | 2A |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | * |
| 封装/外壳： | * |
| 包装： | * |
| 供应商设备封装： | * |
MAX16977
36V, 2A, 2.2MHz Step-Down Converter
with Low Operating Current
In a current-mode step-down converter, the output
capacitor, its ESR, and the load resistance introduce a
pole at the following frequency:
The total loop gain as the product of the modulator gain,
the feedback voltage-divider gain, and the error-amplifier
gain at f C should be equal to 1. So:
GAIN MOD(fC) ×
× GAIN EA(fC) =
V FB
f pMOD =
1
2 π × C OUT × R LOAD
1
V OUT
For the case where f zMOD is greater than f C :
The output capacitor and its ESR also introduce a zero at:
GAIN EA(fC) = g m,EA × R C
f zMOD =
1
2 π × ESR × C OUT
GAIN MOD(fC)
= GAIN MOD(DC) ×
f pMOD
f C
GAIN MOD(fC) ×
× g m,EA C =
V FB
R C =
f pdEA =
C C =
C F =
When C OUT is composed of “n” identical capacitors
in parallel, the resulting C OUT = n x C OUT(EACH) and
ESR = ESR (EACH) /n. Note that the capacitor zero for a
parallel combination of alike capacitors is the same as
for an individual capacitor.
The feedback voltage-divider has a gain of GAIN FB =
V FB /V OUT , where V FB is 1V (typ).
The transconductance error amplifier has a DC gain of
GAIN EA(DC) = g m,EA x R OUT,EA , where g m,EA is the
error-amplifier transconductance, which is 900 F S (typ),
and R OUT,EA is the output resistance of the error amplifier.
A dominant pole (f dpEA ) is set by the compensa-
tion capacitor (C C ) and the amplifier output resistance
(R OUT,EA ). A zero (f zEA ) is set by the compensation
resistor (R C ) and the compensation capacitor (C C ).
There is an optional pole (f pEA ) set by C F and R C to
cancel the output capacitor ESR zero if it occurs near
the crossover frequency (f C , where the loop gain equals
1 (0dB)). Thus:
1
2 π × C C × (R OUT,EA + R C )
Therefore:
× R 1
V OUT
Solving for R C :
V OUT
g m,EA × V FB × GAIN MOD(fC)
Set the error-amplifier compensation zero formed by R C
and C C (f zEA ) at the f pMOD . Calculate the value of C C
as follows:
1
2 π × f pMOD × R C
If f zMOD is less than 5 x f C , add a second capacitor,
C F , from COMP to GND and set the compensation pole
formed by R C and C F (f pEA ) at the f zMOD . Calculate the
value of C F as follows:
1
2 π × f zMOD × R C
As the load current decreases, the modulator pole
also decreases; however, the modulator gain increases
f zEA =
1
2 π × C C × R C
accordingly and the crossover frequency remains the
same.
For the case where f zMOD is less than f C :
f pEA =
f pMOD C ≤ SW
<< f
GAIN MOD(fC) MOD(DC) ×
f pMOD
GAIN EA(fC) = g m,EA × R C × zMOD
1
2 π × C F × R C
The loop-gain crossover frequency (f C ) should be set
below 1/5th of the switching frequency and much higher
than the power-modulator pole (f pMOD ):
f
5
Maxim Integrated
The power-modulator gain at f C is:
= GAIN
The error-amplifier gain at f C is:
f
f zMOD
f C
15
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