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MIC5219-3.0YM5 参数 Datasheet PDF下载

MIC5219-3.0YM5图片预览
型号: MIC5219-3.0YM5
PDF下载: 下载PDF文件 查看货源
内容描述: 值为500mA的峰值输出LDO稳压器 [500mA-Peak Output LDO Regulator]
分类和应用: 稳压器调节器光电二极管输出元件
文件页数/大小: 14 页 / 385 K
品牌: MICREL [ MICREL SEMICONDUCTOR ]
 浏览型号MIC5219-3.0YM5的Datasheet PDF文件第6页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第7页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第8页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第9页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第11页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第12页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第13页浏览型号MIC5219-3.0YM5的Datasheet PDF文件第14页  
Micrel, Inc.  
MIC5219  
Peak Current Applications  
xBMM, the power MSOP package part. These graphs show  
three typical operating regions at different temperatures. The  
lower the temperature, the larger the operating region. The  
graphs were obtained in a similar way to the graphs for the  
MIC5219-x.xBM5, taking all factors into consideration and  
using two different board layouts, minimum footprint and 1"  
square copper PC board heat sink. (For further discussion  
of PC board heat sink characteristics, refer to “Application  
Hint 17, Designing PC Board Heat Sinks” .)  
TheMIC5219isdesignedforapplicationswherehighstart-up  
currents are demanded from space constrained regulators.  
This device will deliver 500mA start-up current from a SOT-  
23-5 or MM8 package, allowing high power from a very low  
profiledevice.TheMIC5219cansubsequentlyprovideoutput  
current that is only limited by the thermal characteristics of  
the device. You can obtain higher continuous currents from  
the device with the proper design. This is easily proved with  
some thermal calculations.  
Theinformationusedtodeterminethesafeoperatingregions  
can be obtained in a similar manner such as determining  
typical power dissipation, already discussed. Determining  
the maximum power dissipation based on the layout is the  
first step, this is done in the same manner as in the previous  
two sections. Then, a larger power dissipation number multi-  
plied by a set maximum duty cycle would give that maximum  
power dissipation number for the layout. This is best shown  
through an example. If the application calls for 5V at 500mA  
for short pulses, but the only supply voltage available is  
8V, then the duty cycle has to be adjusted to determine an  
average power that does not exceed the maximum power  
dissipation for the layout.  
If we look at a specific example, it may be easier to follow.  
TheMIC5219canbeusedtoprovideupto500mAcontinuous  
output current. First, calculate the maximum power dissipa-  
tion of the device, as was done in the thermal considerations  
section. Worst case thermal resistance (θ = 220°C/W for  
the MIC5219-x.xBM5), will be used for this example.  
JA  
TJ (max ) − TA  
(
)
PD (max ) =  
θJA  
Assuming a 25°C room temperature, we have a maximum  
power dissipation number of  
125 °C 25°C  
(
)
% DC  
ꢀ 100 ꢀ  
ꢀ  
ꢀ  
PD (max ) =  
Avg.P = ꢀ  
ꢀ V – V  
I
+ V I  
IN GND  
(
)
D
IN  
OUT  
OUT  
220 °C /W  
P (max) = 455mW  
D
% DC  
ꢀ  
ꢀ  
ꢀ  
455mW = ꢀ  
ꢀ 100 ꢀ  
(
8V – 5V 500mA + 8V × 20mA  
)
Then we can determine the maximum input voltage for a  
5-voltregulatoroperatingat500mA, usingworstcaseground  
current.  
% Duty Cycle  
ꢀ  
455mW = ꢀ  
ꢀ 100  
ꢀ  
ꢀ1.66W  
ꢀ  
P (max) = 455mW = (V – V  
) I  
+ V I  
IN GND  
D
IN  
OUT OUT  
% Duty Cycle  
100  
I
= 500mA  
= 5V  
OUT  
0.274 =  
V
I
OUT  
% Duty Cycle Max = 27.4%  
= 20mA  
GND  
455mW = (V – 5V) 500mA + V × 20mA  
With an output current of 500mAand a three-volt drop across  
the MIC5219-xxBMM, the maximum duty cycle is 27.4%.  
IN  
IN  
2.995W = 520mA × V  
IN  
Applications also call for a set nominal current output with a  
greateramountofcurrentneededforshortdurations.Thisisa  
trickysituation,butitiseasilyremedied.Calculatetheaverage  
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 MIC5219-3.3BM5 with 5V input voltage as  
our example.  
2.955W  
520mA  
VIN(max ) =  
= 5.683V  
Therefore, to be able to obtain a constant 500mAoutput cur-  
rent from the 5219-5.0BM5 at room temperature, you need  
extremely tight input-output voltage differential, barely above  
the maximum dropout voltage for that current rating.  
You can run the part from larger supply voltages if the proper  
precautions are taken. Varying the duty cycle using the en-  
able pin can increase the power dissipation of the device by  
maintaining a lower average power figure. This is ideal for  
applicationswherehighcurrentisonlyneededinshortbursts.  
Figure1showsthesafeoperatingregionsfortheMIC5219-x.  
xBM5 at three different ambient temperatures and at differ-  
ent output currents. The data used to determine this figure  
assumed a minimum footprint PCB design for minimum heat  
sinking. Figure 2 incorporates the same factors as the first  
figure, but assumes a much better heat sink.A1" square cop-  
per trace on the PC board reduces the thermal resistance of  
thedevice.Thisimprovedthermalresistanceimprovespower  
dissipation and allows for a larger safe operating region.  
P × 50mA = (5V – 3.3V) × 50mA + 5V × 650µA  
D
P × 50mA = 173mW  
D
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, P must be mul-  
D
tiplied by the duty cycle to obtain the actual average power  
dissipation at 50mA.  
Figures3and4showsafeoperatingregionsfortheMIC5219-x.  
June 2009  
10  
M0371-061809