参数资料
型号: MIC5216-5.0BMM
厂商: MICREL INC
元件分类: 固定正电压单路输出LDO稳压器
英文描述: 500mA-Peak Output LDO Regulator Preliminary Information
中文描述: 5 V FIXED POSITIVE LDO REGULATOR, 0.6 V DROPOUT, PDSO8
封装: MSOP-8
文件页数: 2/13页
文件大小: 288K
代理商: MIC5216-5.0BMM
Micrel, Inc.
MIC5216
March 2007
10
M9999-032307
1.66W
100
Cycle
%Duty
455mW
=
=
100
Cycle
%Duty
274
.
0
% Duty Cycle Max = 27.4%
With an output current of 500mA and a three-volt drop
across the MIC5216-xxBMM, the maximum duty cycle is
27.4%.
Applications also call for a set nominal current output
with a greater amount of current needed for short
durations. This is a tricky situation, but it is easily
remedied. Calculate the average power dissipation for
each current section, then add the two numbers giving
the total power dissipation for the regulator. For
example, if the regulator is operating normally at 50mA,
but for 12.5% of the time it operates at 500mA output,
the total power dissipation of the part can be easily
determined. First, calculate the power dissipation of the
device at 50mA. We will use the MIC5216-3.3BM5 with
5V input voltage as our example.
PD × 50mA = (5V – 3.3V) × 50mA + 5V × 650A
PD × 50mA = 173mW
However, this is continuous power dissipation, the actual
on-time for the device at 50mA is (100%-12.5%) or
87.5% of the time, or 87.5% duty cycle. Therefore, PD
must be multiplied by the duty cycle to obtain the actual
average power dissipation at 50mA.
PD × 50mA = 0.875 × 173mW
PD × 50mA = 151mW
The power dissipation at 500mA must also be
calculated.
PD × 500mA = (5V – 3.3V) 500mA + 5V × 20mA
PD × 500mA = 950mW
This number must be multiplied by the duty cycle at
which it would be operating, 12.5%.
PD × = 0.125mA × 950mW
PD × = 119mW
a. 25
°C Ambient
b. 50
°C Ambient
c. 85
°C Ambient
Figure 1. MIC5216-x.xBM5 (SOT-23-5) on Minimum Recommended Footprint
a. 25
°C Ambient
b. 50
°C Ambient
c. 85
°C Ambient
Figure 2. MIC5216-x.xBM5 (SOT-23-5) on 1-inch
2 Copper Cladding
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