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参数资料
| 型号： | MAX8588ETM+T |
| 厂商： | Maxim Integrated Products |
| 文件页数： | 25/31页 |
| 文件大小： | 0K |
| 描述： | IC PMIC HI EFF LOW IQ 48-TQFN |
| 产品培训模块： | Lead (SnPb) Finish for COTS Obsolescence Mitigation Program |
| 标准包装： | 2,500 |
| 应用： | 处理器 |
| 电源电压： | 2.6 V ~ 5.5 V |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 48-WFQFN 裸露焊盘 |
| 供应商设备封装： | 48-TQFN-EP（6x6） |
| 包装： | 带卷 (TR) |
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High-Efficiency, Low-I Q PMIC with
Dynamic Core for PDAs and Smartphones
If the capacitor has significant ESR, the output ripple
component due to capacitor ESR is:
Table 4. Compensation Parameters
V RIPPLE(ESR) = I L(PEAK) x ESR
Output capacitor specifics are also discussed in the
Compensation and Stability section.
Compensation and Stability
The relevant characteristics for REG1, REG2, and
PARAMETER
Error-Amplifier
Transconductance, gmEA
Current-Sense Amp
Transresistance, R CS
REG1
87μS
0.5V/A
REG2
87μS
0.75V/A
REG3
68μS
1.25V/A
REG3 compensation are:
1) Transconductance (from FB_ to CC_), gm EA
Table 5. Typical Compensation Values
2) Current-sense amplifier transresistance, R CS
3) Feedback regulation voltage, V FB (1.25V)
4) Step-down output voltage, V OUT , in V
5) Output load equivalent resistance, R LOAD = V OUT /
I LOAD
The key steps for step-down compensation are:
1) Set the compensation RC zero to cancel the R LOAD
C OUT pole.
2) Set the loop crossover at or below approximately
1/10th the switching frequency.
COMPONENT OR
PARAMETER
V OUT
Output Current
Inductor
Load-Step Droop
Loop Crossover Freq (fC)
C C
R C
C OUT
REG1
3.3V
1300mA
3.3μH
3%
100kHz
330pF
240k ?
22μF
REG2
2.5V
900mA
6.8μH
3%
100kHz
270pF
240k ?
22μF
REG3
1.3V
500mA
10μH
3%
100kHz
330pF
240k ?
22μF
For example, with V IN(MAX) = 5V, V OUT = 2.5V for
REG2, and I OUT = 800mA, then R LOAD = 3.125 ? . For
REG2, R CS = 0.75V/A and gm EA = 87μS.
Choose the crossover frequency, f C ≤ f OSC / 10.
Choose 100kHz. Then calculate the value of the com-
pensation capacitor, C C :
C C = (V FB / V OUT ) x (R LOAD / R CS ) x (gm / (2 π x f C ))
= (1.25 / 2.5) x (3.125 / 0.75) x (87 x 10 -6 / (6.28
x 100,000)) = 289pF
Choose 330pF, the next highest standard value.
Now select the compensation resistor, R C , so transient-
droop requirements are met. As an example, if 3% tran-
sient droop is allowed for the desired load step, the
input to the error amplifier moves 0.03 x 1.25V, or
37.5mV. The error-amplifier output drives 37.5mV x
gm EA , or I EAO = 37.5mV x 87μS = 3.26μA across R C to
provide transient gain. Find the value of R C that allows
the required load-step swing from:
R C = R CS x I IND(PK) / I EAO
where I IND(PK) is the peak inductor current. In a step-
down DC-DC converter, if L IDEAL is used, output cur-
rent relates to inductor current by:
I IND(PK) = 1.25 x I OUT
So for an 800mA output load step with V IN = 3.6V and
V OUT = 2.5V:
R C = R CS x I IND(PK) / I EAO = (0.75V/A) x
(1.25 x 0.8A) / 3.26μA = 230k ?
We choose 240k ? . Note that the inductor does not limit
the response in this case since it can ramp at (V IN -
V OUT ) / L, or (3.6 - 2.5) / 3.3μH = 242mA/μs.
The output-filter capacitor is then selected so that the
C OUT R LOAD pole cancels the R C C C zero:
C OUT x R LOAD = R C x C C
For the example:
R LOAD = V OUT x I LOAD = 2.5V / 0.8A =
3.125 ?
C OUT = R C x C C / R LOAD = 240k ? x 330pF /
3.125 ? = 25μF
We choose 22μF.
Recalculate R C using the selected C OUT .
R C = C OUT x R LOAD / C C = 208k ?
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