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| 型号： | ADP1111ANZ-12 |
| 厂商： | Analog Devices Inc |
| 文件页数： | 7/15页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK BST INV 12V .2A 8DIP |
| 标准包装： | 50 |
| 类型： | 降压（降压），升压（升压），反相 |
| 输出类型： | 固定 |
| 输出数： | 1 |
| 输出电压： | 12V |
| 输入电压： | 2 V ~ 30 V |
| 频率 - 开关： | 72kHz |
| 电流 - 输出： | 200mA |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 通孔 |
| 封装/外壳： | 8-DIP（0.300"，7.62mm） |
| 包装： | 管件 |
| 供应商设备封装： | 8-PDIP |
ADP1111
P L = ( V OUT + V D ? V IN ( MIN ) ) ? ( I OUT )
I PEAK = ? 1 ? e ? = 587 mA
INDUCTOR SELECTION–STEP-UP CONVERTER
In a step-up or boost converter (Figure 18), the inductor must
store enough power to make up the difference between the input
voltage and the output voltage. The power that must be stored
is calculated from the equation:
(Equation 1)
where V D is the diode forward voltage (0.5 V for a 1N5818
Substituting a standard inductor value of 68 μ H with 0.2 Ω dc
resistance will produce a peak switch current of:
6 V ? ? 1.0 Ω ? 7 μ s ?
68 μ H
1.0 Ω ? ?
Once the peak current is known, the inductor energy can be
calculated from Equation 5:
Schottky). Because energy is only stored in the inductor while
the ADP1111 switch is ON, the energy stored in the inductor
on each switching cycle must be equal to or greater than:
E L =
1
2
( 68 μ H ) ? ( 587 mA ) 2 = 11.7 μ J
P L
f OSC
(Equation 2)
Since the inductor energy of 11.7 μ J is greater than the P L /f OSC
requirement of 3.6 μ J , the 68 μ H inductor will work in this
in order for the ADP1111 to regulate the output voltage.
When the internal power switch turns ON, current flow in the
inductor increases at the rate of:
application. By substituting other inductor values into the same
equations, the optimum inductor value can be selected.
When selecting an inductor, the peak current must not exceed
the maximum switch current of 1.5 A. If the equations shown
? 1 ? e L ?
I L ( t ) =
V IN ? ? R ' t ?
R ' ? ?
(Equation 3)
above result in peak currents > 1.5 A, the ADP1110 should be
considered. Since this device has a 70% duty cycle, more energy
is stored in the inductor on each cycle. This results is greater
I L ( t ) =
V IN
where L is in Henrys and R' is the sum of the switch equivalent
resistance (typically 0.8 Ω at +25 ° C) and the dc resistance of
the inductor. In most applications, the voltage drop across the
switch is small compared to V IN so a simpler equation can be
used:
t (Equation 4)
L
Replacing ‘ t ’ in the above equation with the ON time of the
ADP1111 (7 μ s, typical) will define the peak current for a given
inductor value and input voltage. At this point, the inductor
energy can be calculated as follows:
output power.
The peak current must be evaluated for both minimum and
maximum values of input voltage. If the switch current is high
when V IN is at its minimum, the 1.5 A limit may be exceeded at
the maximum value of V IN . In this case, the ADP1111’s current
limit feature can be used to limit switch current. Simply select a
resistor (using Figure 6) that will limit the maximum switch
current to the I PEAK value calculated for the minimum value of
V IN . This will improve efficiency by producing a constant I PEAK
as V IN increases. See the “Limiting the Switch Current” section
of this data sheet for more information.
Note that the switch current limit feature does not protect the
E L = L ? I 2 PEAK
1
2
(Equation 5)
circuit if the output is shorted to ground. In this case, current is
only limited by the dc resistance of the inductor and the forward
P L = ( 12 V + 0.5 V ? 6 V ) ? ( ) = 260 mW
I PEAK =
DC ? ? V IN ? V SW + V D ? ?
As previously mentioned, E L must be greater than P L /f OSC so
that the ADP1111 can deliver the necessary power to the load.
For best efficiency, peak current should be limited to 1 A or
less. Higher switch currents will reduce efficiency because of
increased saturation voltage in the switch. High peak current
also increases output ripple. As a general rule, keep peak current
as low as possible to minimize losses in the switch, inductor and
diode.
In practice, the inductor value is easily selected using the
equations above. For example, consider a supply that will
generate 12 V at 40 mA from a 9 V battery, assuming a 6 V
end-of-life voltage. The inductor power required is, from
Equation 1:
40 mA
On each switching cycle, the inductor must supply:
voltage of the diode.
INDUCTOR SELECTION–STEP-DOWN CONVERTER
The step-down mode of operation is shown in Figure 19.
Unlike the step-up mode, the ADP1111’s power switch does not
saturate when operating in the step-down mode; therefore,
switch current should be limited to 650 mA in this mode. If the
input voltage will vary over a wide range, the I LIM pin can be
used to limit the maximum switch current. Higher switch
current is possible by adding an external switching transistor as
shown in Figure 21.
The first step in selecting the step-down inductor is to calculate
the peak switch current as follows:
2 I OUT ? V OUT + V D ?
(Equation 6)
P L
f OSC
=
260 mW
72 kHz
= 3.6 μ J
where DC = duty cycle (0.5 for the ADP1111)
Since the required inductor power is fairly low in this example,
the peak current can also be low. Assuming a peak current of
500 mA as a starting point, Equation 4 can be rearranged to
recommend an inductor value:
V SW = voltage drop across the switch
V D = diode drop (0.5 V for a 1N5818)
I OUT = output current
V OUT = the output voltage
L =
V IN
I L ( MAX )
t =
6 V
500 mA
7 μ s = 84 μ H
V IN = the minimum input voltage
0
REV. A
–7 –
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| ADP1111ANZ-3.3 | 功能描述:IC REG BUCK BOOST INV 3.3V 8DIP RoHS:是 类别:集成电路 (IC) >> PMIC - 稳压器 - DC DC 开关稳压器 系列:- 标准包装:2,500 系列:- 类型:降压（降压） 输出类型:固定 输出数:1 输出电压:1.2V，1.5V，1.8V，2.5V 输入电压:2.7 V ~ 20 V PWM 型:- 频率 - 开关:- 电流 - 输出:50mA 同步整流器:是 工作温度:-40°C ~ 125°C 安装类型:表面贴装 封装/外壳:10-TFSOP，10-MSOP（0.118"，3.00mm 宽）裸露焊盘 包装:带卷 (TR) 供应商设备封装:10-MSOP 裸露焊盘 |
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