MIC5216 Datasheet, PDF (7/12 Pages) Micrel Semiconductor – 500mA-Peak Output LDO Regulator Preliminary Information

MIC5216
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_JA = 220°C/W for
the MIC5216-x.xBM5), will be used for this example.
( ) _{PD(MAX) =}
T_J(MAX) – T_A
θ_JA
Assuming room temperature, we have a maximum power
dissipation number of
P_D(MAX) =
(125°C – 25°C)
220°C/W
P_D = 455mW
Then we can determine the maximum input voltage for a five-
volt regulator operating at 500mA, using worst case ground
current.
P_D(max) = 455mW = (V_IN – V_OUT) I_OUT + V_IN I_GND
I_OUT = 500mA
V_OUT = 5V
I_GND = 20mA
455mW = (V_IN – 5V) 500mA + V_IN × 20mA
2.995W = 520mA × V_IN
V_IN(max) =
2.955W
520mA
= 5.683V
Therefore, to be able to obtain a constant 500mA output
current from the 5216-5.0BM5 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
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 1 shows the safe operating regions for the
MIC5216-x.xBM5 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 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.
Figures 3 and 4 show safe operating regions for the MIC5216-
x.xBMM, the power MSOP package part. These graphs
show three typical operating regions at different tempera-
tures. The lower the temperature, the larger the operating
region. The graphs were obtained in a similar way to the
graphs for the MIC5216-x.xBM5, taking all factors into con-
sideration 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”.
Micrel
The information used to determine the safe operating regions
can be obtained in a similar manner to that used in determin-
ing typical power dissipation, already discussed. Determin-
ing 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 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.
( ) ^Avg.P_D ^{=  %10D0C V}_IN ^{– V}_OUT ^I_OUT ^{+ V}_IN ^I_GND
455mW = ^{ %10D0C} (8V – 5V) 500mA + 8V × 20mA
455mW = ^{ % Du1t0y0Cycle} 1.66W
0.274 = ^{% Duty Cycle}
100
% 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.
P_D × 50mA = (5V – 3.3V) × 50mA + 5V × 650µA
P_D × 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, P_D must be
multiplied by the duty cycle to obtain the actual average
power dissipation at 50mA.
P_D × 50mA = 0.875 × 173mW
P_D × 50mA = 151mW
The power dissipation at 500mA must also be calculated.
P_D × 500mA = (5V – 3.3V) 500mA + 5V × 20mA
P_D × 500mA = 950mW
This number must be multiplied by the duty cycle at which it
would be operating, 12.5%.
P_D × = 0.125 × 950mW
P_D × = 119mW
January 2000
7
MIC5216