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参数资料
| 型号： | MAX1962EEP+T |
| 厂商： | Maxim Integrated Products |
| 文件页数： | 18/29页 |
| 文件大小： | 0K |
| 描述： | IC REG CTRLR BUCK PWM VM 20-QSOP |
| 产品培训模块： | Lead (SnPb) Finish for COTS Obsolescence Mitigation Program |
| 标准包装： | 2,500 |
| PWM 型： | 电压模式 |
| 输出数： | 1 |
| 频率 - 最大： | 1.12MHz |
| 占空比： | 90% |
| 电源电压： | 2.35 V ~ 5.5 V |
| 降压： | 是 |
| 升压： | 无 |
| 回扫： | 无 |
| 反相： | 无 |
| 倍增器： | 无 |
| 除法器： | 无 |
| Cuk： | 无 |
| 隔离： | 无 |
| 工作温度： | -40°C ~ 85°C |
| 封装/外壳： | 20-SSOP（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
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2.35V to 5.5V, 0.5% Accurate, 1MHz PWM
Step-Down Controllers with Voltage Margining
C C =
5
2 π × R C × f PMOD
Choose the crossover frequency (f C ) in the range f ZESR
< f C < f S /5:
29.3kHz < f C < 200kHz
A small capacitor C F , can also be added from COMP to
GND to provide high-frequency decoupling. C F will add
another high-frequency pole (f PHF ) to the error-amplifier
Select f C = 100kHz, this meets the criteria above, and the
bandwidth is high enough for good transient response.
The power modulator gain at f C is:
response. This pole should be greater than 100 times
the error-amplifier zero frequency to have negligible
impact on the phase margin. This pole should also be
G MOD ( fc ) =
V IN
V RAMP
×
( f PMOD ) 2
f ZESR × f C
less than half the switching frequency for effective
decoupling:
100f ZEA < f PHF < 0.5f S
=
3
0 . 85
×
( 9201 ) 2
29 . 3 k ? × 100 k ?
= 0 . 102
Select a value for f PHF in the range given above, then
solve for C F using the following equation:
Choose R 1 = 8.06k ? , then R 2 = 10k ? (see the Setting
the Output Voltage section):
C =
g mEA FB MOD ( f C )
C F =
1
2 π × R C × f PHF
= 11 k ?
V OUT
× V × G
=
1 . 8
0 . 002 × 0 . 8 × 0 . 102
Below is a numerical example to calculate compensa-
tion values:
V IN = 3.3V
C C =
5
2 π × R C × f PMOD
=
5
2 π × 11 k ? × 9201
= 7863 pF
V RAMP = 0.85V
V OUT = 1.8V
V FB = 0.8V
I OUT(max) = 15A
C O = 2 x 680μF = 1360μF
Select C C = 8200pF (nearest standard capacitor
value).
Select f PHF in the range 100f ZEA < f PHF < 0.5f S .
184kHz < f PHF < 500kHz
Select f PHF = 250kHz, then solve for C F :
ESR = 0.008 ? / 2 = 0.004 ?
L O = 0.22μH
g mEA = 2mS
C F =
1
2 π × R C × f PHF
=
1
2 π × 11 k ? × 250 kHz
= 58 pF
f S = 1MHz
Select the nearest standard capacitor value C F = 56pF.
f PMOD =
=
2 π ×
2 π ×
1
L O × C O
1
0 . 22 × 10 ? 6 × 1360 × 10 ? 6
Summary of feedback divider and compensation com-
ponents:
R 1 = 8.06k ?
R 2 = 10k ?
R C = 11k ?
= 9 . 201 kHz
C C = 8200pF
C F = 56pF
f ZESR =
1
2 π × C O × R ESR
Power MOSFET Selection
When selecting a MOSFET, essential parameters
=
1
2 π × 1360 × 10 ? 6 × 0 . 004
= 29 . 3 kHz
include:
(1) Total gate charge (Q G )
(2) Reverse transfer capacitance (C RSS )
(3) On-resistance (R DS(ON) )
18
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