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| 型号： | LT1316IS8#PBF |
| 厂商： | Linear Technology |
| 文件页数： | 8/16页 |
| 文件大小： | 0K |
| 描述： | IC REG BST FLYBK ADJ 0.5A 8SOIC |
| 标准包装： | 100 |
| 类型： | 升压，反激式 |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 1.23 V ~ 30 V |
| 输入电压： | 1.5 V ~ 12 V |
| 电流 - 输出： | 500mA |
| 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-SOIC（0.154"，3.90mm 宽） |
| 包装： | 管件 |
| 供应商设备封装： | 8-SOIC |
LT1316
APPLICATIO N S I N FOR M ATIO N
2(I OUT ) 2(10mA)
Discontinuous Mode Operation
A boost converter with a high V OUT :V IN ratio operates with
2. I PEAK =
= = 58mA
1 – DC 1 – 0.654
) )
V OUT – V IN + V D
)
)
= 5 – 2 + 0.4 2 μ s
a high duty cycle in continuous mode. For duty cycles
exceeding the LT1316’s guaranteed minimum specifica-
tion of 0.73, the circuit will need to be designed for
discontinuous operation. Additionally, very low peak cur-
rent limiting below 50mA may necessitate operating in this
mode unless high inductance values are acceptable. When
operating in discontinuous mode, a different equation
governs available output power. For each switch cycle, the
inductor current ramps down to zero, completely releas-
3. Find L
L=
0.4(I PEAK )
0.4(58mA)
= 293 μ H
t OFF
)
)
Where f = I PEAK (L)
V IN – V SAT
) )
V IN
)
)
=
330 μ H
) )
I PEAK ? L
) ) ) )
= 28 – 3.3 + 0.4 = 0.89
V OUT – V IN + V D
V OUT – V SAT + V D
) ) ) )
= 5 – 2 + 0.4 = 0.654
V OUT – V IN + V D
V OUT – V SAT + V D
(t ON2 )(V IN – V SAT ) 2
= 52 μ H
(3.4 μ s 2 )(3.3 – 0.2) 2
2(0.196W)(3.4 μ s + 2 μ s)
ing the stored energy. Energy stored in the inductor at any
time is equal to 1/2 LI 2 . Because this energy is released
each cycle, the equation for maximum power out is:
P OUT(MAX) = 1/2L(I PEAK2 )f
1
+ t OFF
When designing for very low peak currents (< 50mA), the
inductor size needs to be large enough so that on-time is
a least 1 μ s. On-time can be calculated by the equation:
On-Time =
(V IN – V SAT )
where V SAT = 0.2V.
Also, at these low current levels, current overshoot due to
power transistor turn-off delay will be a significant portion
of peak current. Increasing inductor size will keep this to
a minimum.
Design Example 1
Requirements: V IN = 2V, V OUT = 5V and I LOAD = 10mA.
1. Find duty cycle
DC =
5 – 0.2 + 0.4
Because duty cycle is less than the LT1316 minimum
specification (0.73), the circuit can be designed for
continuous operation.
8
4. Find R SET resistor
Overshoot = 300ns
L
2
= 1.8mA
Find R SET from Figure 3 for 58mA – 1.8mA = 56.2mA
R SET ≈ 47k
Design Example 2
Requirements: V IN = 3.3V, V OUT = 28V and I LOAD = 5mA.
1. Find duty cycle:
DC =
28 – 0.2 + 0.4
Because duty cycle exceeds LT1316 minimum specifi-
cation of 73%, the circuit must be designed for discon-
tinuous operation.
2. Find P OUT(MAX)
Multiply P OUT by 1.4 to give a safe operating margin
P OUT(MAX) = P OUT (1.4) = (5mA)(28V)(1.4) = 0.196W
3. Set the on-time to the data sheet minimum of 3.4 μ s and
find L
L=
2P OUT(MAX) (t ON + t OFF )
=
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