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参数资料
| 型号： | MAX16976AEE/V+T |
| 厂商： | Maxim Integrated Products |
| 文件页数： | 15/17页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK 5V/ADJ 0.6A 16QSOP |
| 产品培训模块： | Lead (SnPb) Finish for COTS Obsolescence Mitigation Program |
| 标准包装： | 2,500 |
| 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 ~ 28 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 220kHz ~ 1MHz |
| 电流 - 输出： | 600mA |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 16-SSOP（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
| 供应商设备封装： | 16-QSOP |
MAX16976
28V, 600mA Automotive Step-Down Converter
with Low Operating Current
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 O
R OUT,EA , where g m,EA is the error-amplifier transcon-
ductance, which is 1000 F S (typ), and R OUT,EA is the
output resistance of the 50M I error amplifier.
Therefore:
GAIN MOD(fC) ×
Solving for R C :
V FB
V OUT
× g m,EA × R C = 1
f dpEA =
f zEA =
f pEA =
R C =
C C =
C F =
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 resis-
tor (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 )
1
2 π × C C × R C
1
2 π × C F × R C
The loop-gain crossover frequency (f C ) is set below
1/5th the switching frequency and much higher than the
power-modulator pole (f pMOD ):
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 lower 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
accordingly and the crossover frequency remains the
same. For the case where f zMOD is less than f C :
f pMOD C ≤ SW
<< f
f
5
The total loop gain as the product of the modulator gain,
The power-modulator gain at f C is:
GAIN MOD(f C ) = GAIN MOD(dc) ×
f pMOD
f zMOD
GAIN MOD(fC) ×
× GAIN EA(f ) = 1
V OUT
GAIN EA(f ) m,EA C × zMOD
= g
× R
f C
thefeedbackvoltage-dividergain,andtheerror-amplifier
gain at f C is equal to 1. So:
V FB
C
The error-amplifier gain at f C is:
C
f
× g m,EA C × = 1
f zMOD
For the case where f zMOD is greater than f C :
GAIN EA(fC) = g m,EA × R C
Therefore:
GAIN MOD(f C ) ×
V FB
V OUT
× R
f C
R C =
GAIN MOD(fC) = GAIN MOD(dc) ×
f pMOD
f C
Solving for R C :
V OUT × f C
g m,EA × V FB × GAIN MOD(f C ) × f zMOD
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