参数资料
| 型号： | A8670EESTR-T |
| 厂商： | Allegro Microsystems Inc |
| 文件页数： | 19/28页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK ADJ 2A 20QFN |
| 标准包装： | 1 |
| 类型： | 降压（降压） |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 可调至 0.6V |
| 输入电压： | 7 V ~ 16 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 200kHz ~ 1MHz |
| 电流 - 输出： | 2A |
| 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： | 20-WFQFN 裸露焊盘 |
| 包装： | 标准包装 |
| 供应商设备封装： | 20-QFN 裸露焊盘（4x4） |
| 其它名称： | 620-1397-6 |
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A8670
Fixed Frequency, 2 A Synchronous Buck Regulator
With Fault Warnings and Power OK
? Recirculation diode losses The recirculation diode losses
(low-side switch) are calculated as:
For this particular solution, a high thermal efficiency board is
required to ensure the junction temperature is kept below 125°C.
P recirc = 0.8 × I OUT × 6 ×10 –9 × f SW
= 0.8 × 2 × 6 ×10 –9 × 500 ×10 3
(35)
It is recommended to use a PCB with four layers. The A8670
should be mounted onto a thermal pad. A number of vias should
= 0.005 W
? Diode transit losses The recirculation diode losses (low-side
switch) are calculated as:
connect the thermal pad to at least one of the internal layers and
the bottom side of the PCB. Both of these layers should be a
ground plane. See the Layout section for more information.
P transit = V IN × I OUT × 3 ×10 –9 × f SW
= 12 × 2 × 3 ×10 –9 × 500 ×10 3
= 0.036 W
? BIAS losses The supply bias losses are calculated as:
(36)
Regulator Efficiency
The overall regulator efficiency can be determined by including
the inductor loss. In the above thermal characteristics example,
P bias = V IN × 7.2 × 10 –3
= 0.086 W
(37)
the inductor resistance, DCR L = 20 m Ω . Therefore the inductor
power loss can be found::
T J – T A
=
V OUT × I OUT
=
2.The total losses in theA8670 can be estimated:
P total = P staticHI + P staticLO + P switch + P recirc + P transit + P bias (38)
= 0.144 + 0.239 + 0.072 + 0.005 + 0.036 + 0.086
= 0.582 W
3. The thermal impedance required for the solution can be found:
R θ JA = (39)
P total
1 25 – 105
0.582
= 34 °C/W
P L = DCR L × I 2OUT
= 0.02 × 2 2
= 0.08 W
The overall regulator efficiency can be found:
η =
( V OUT × I OUT ) + P total + P L
1.2 × 2
(1.2 × 2) + 0.582 + 0.08
= 78. 4 %
Allegro MicroSystems, LLC
115 Northeast Cutoff
Worcester, Massachusetts 01615-0036 U.S.A.
1.508.853.5000; www.allegromicro.com
(40)
(41)
18
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