Datasheet

 2018 Microchip Technology Inc. DS20006021A-page 15
MIC5219
EQUATION 4-7:
Therefore, to be able to obtain a constant 500 mA
output current from the MIC5219-5.0YM5 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 enable pin can increase the power dissipation
of the device by maintaining a lower average power
figure. This is ideal for applications where high current
is only needed in short bursts. Figure 4-1 shows the
safe operating regions for the MIC5219-x.xYM5 at
three different ambient temperatures and at different
output currents. The data used to determine this figure
assumed a minimum footprint PCB design for minimum
heat sinking. Figure 4-2 incorporates the same factors
as the first figure, but assumes a much better heat sink.
A 1" square copper trace on the PC board reduces the
thermal resistance of the device. This improved
thermal resistance improves power dissipation and
allows for a larger safe operating region.
Figure 4-3 and Figure 4-4 show safe operating regions
for the MIC5219-x.xYMM, 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.xYM5, 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.
The information used to determine the safe operating
regions 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
multiplied 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 500 mA 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.
EQUATION 4-8:
With an output current of 500 mA and a three volt drop
across the MIC5219-xxYMM, 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
50 mA, but for 12.5% of the time it operates at 500 mA
output, the total power dissipation of the part can be
easily determined. First, calculate the power
dissipation of the device at 50 mA. We will use the
MIC5219-3.3YM5 with 5V input voltage as our
example.
EQUATION 4-9:
P
DMAX
455mW
V
IN
V
OUT
–I
OUT
 V
IN
+ I
GND

==
Where:
I
OUT
= 500 mA
V
OUT
= 5V
I
GND
= 20 mA
455mW V
IN
5V–500mA V
IN
+20mA=
2.995W 520mA V
IN
=
V
IN MAX
2.995W
520mA
------------------
5.683V==
Avg
·
P
D
%DC
100
-------------


V
IN
V
OUT
–I
OUT
 V
IN
+ I
GND
=
455mW
%DC
100
-------------


8V 5V–500mA 8V+20mA=
455mW
%DC
100
-------------


1.66W=
0.274
%DC
100
-------------
=
%DC 27.4% Duty Cycle Max.=
P
D
50mA 5V 3.3V–50mA 5V+ 650A=
P
D
50mA 88.25mW=