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| 型号： | MAX17067EUA+T |
| 厂商： | Maxim Integrated Products |
| 文件页数： | 9/13页 |
| 文件大小： | 0K |
| 描述： | IC REG BOOST ADJ 0.3A 8UMAX |
| 产品培训模块： | Lead (SnPb) Finish for COTS Obsolescence Mitigation Program |
| 标准包装： | 2,500 |
| 类型： | 升压（升压） |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 1.25 V ~ 18 V |
| 输入电压： | 2.6 V ~ 4 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 640kHz，1.2MHz |
| 电流 - 输出： | 300mA |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-TSSOP，8-MSOP（0.118"，3.00mm 宽） |
| 包装： | 带卷 (TR) |
| 供应商设备封装： | 8-uMAX |
Low-Noise Step-Up DC-DC Converter
The maximum output current, input voltage, output volt-
age, and switching frequency determine the inductor
value. Very high inductance values minimize the cur-
rent ripple and therefore reduce the peak current,
which decreases core losses in the inductor and I 2 R
losses in the entire power path. However, large induc-
Choose an available inductor value from an appropriate
inductor family. Calculate the maximum DC input cur-
rent at the minimum input voltage V IN(MIN) using con-
servation of energy and the expected efficiency at that
operating point ( η MIN ) taken from an appropriate curve
in the Typical Operating Characteristics:
tor values also require more energy storage and more
turns of wire, which increase physical size and can
increase I 2 R losses in the inductor. Low inductance val-
ues decrease the physical size but increase the current
I IN ( DC , MAX ) =
I MAIN(MAX) × V MAIN
V IN ( MIN ) × η MIN
ripple and peak current. Finding the best inductor
involves choosing the best compromise between circuit
Calculate the ripple current at that operating point and
the peak current required for the inductor:
efficiency, inductor size, and cost.
The equations used here include a constant LIR, which
is the ratio of the inductor peak-to-peak ripple current to
I RIPPLE =
V IN ( MIN ) × ( V MAIN ? V IN ( MIN ) )
L × V MAIN × f OSC
I PEAK = I IN ( DC , MAX ) + RIPPLE
the average DC inductor current at the full load current.
The best trade-off between inductor size and circuit
efficiency for step-up regulators generally has an LIR
between 0.3 and 0.5. However, depending on the AC
characteristics of the inductor core material and the
ratio of inductor resistance to other power path resis-
tances, the best LIR can shift up or down. If the induc-
tor resistance is relatively high, more ripple can be
accepted to reduce the number of turns required and
increase the wire diameter. If the inductor resistance is
relatively low, increasing inductance to lower the peak
current can decrease losses throughout the power
path. If extremely thin high-resistance inductors are
I
2
The inductor ’s saturation current rating and the
MAX17067s’ LX current limit (I LIM ) should exceed I PEAK
and the inductor’s DC current rating should exceed
I IN(DC,MAX) . For good efficiency, choose an inductor with
less than 0.1 Ω series resistance.
Considering the application circuit in Figure 4, the maxi-
mum load current (I MAIN(MAX) ) is 250mA with a 9V output
and a typical input voltage of 3.3V. Choosing an LIR of 0.7
and estimating efficiency of 85% at this operating point:
? 3 . 3 V ? ?
? ? ? ?
L = ?
? ≈ 3 . 3 μ H
0 . 25 A × 1 . 2 MHz ? ? 0 . 7 ?
used, as is common for LCD-panel applications, the
best LIR can increase to between 0.5 and 1.0.
Once a physical inductor is chosen, higher and lower
? 9 V
2
9 V ? 3 . 3 V ? ? 0.85 ?
? ?
values of the inductor should be evaluated for efficiency
improvements in typical operating regions.
Using the application’s minimum input voltage (3V) and
estimating efficiency of 80% at that operating point:
Calculate the approximate inductor value using the typ-
ical input voltage (V IN ), the maximum output current
(I MAIN(MAX) ), the expected efficiency ( η TYP ) taken from
I IN ( DC , MAX ) =
0.25 A × 9 V
3 V × 0 . 8
≈ 0 . 94 A
an appropriate curve in the Typical Operating
Characteristics, and an estimate of LIR based on the
The ripple current and the peak current are:
? V IN ? ?
? ? η
? V
TYP ?
L = ?
? ?
?
? ? I
above discussion:
2
V MAIN IN
? V MAIN ? ? MAIN ( MAX ) × f OSC ? ? LIR ?
I RIPPLE =
3 V × ( 9 V ? 3 V )
3 . 3 μ H × 9 V × 1 . 2 MHz
≈ 0 . 51 A
I PEAK = 0 . 94 A +
0.51A
2
≈ 1 . 19 A
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