January 2000

7

MIC5216

MIC5216

Micrel

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

the MIC5216-x.xBM5), will be used for this example.

P

=

T

– T

D(MAX)

J(MAX)

A

JA

()

θ

Assuming room temperature, we have a maximum power

dissipation number of

P

=

125 C – 25 C

C/W

D(MAX)

°°

()

°

220

P

455mW

Then we can determine the maximum input voltage for a five-

volt regulator operating at 500mA, using worst case ground

current.

P

I

V

I

455mW = (V

2.995W = 520mA

× V

IN

V

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”.

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 =

% DC

100

V

– V

I

V

I

DIN

OUT

OUT

IN GND







+

455mW =

% DC

100

8V – 5V 500mA

8V

20mA







+×

455mW =

% Duty Cycle

100

1.66W









0.274 =

% Duty Cycle

100

% Duty Cycle Max =

7.4%

2

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

P

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

multiplied by the duty cycle to obtain the actual average

power dissipation at 50mA.

P

P

The power dissipation at 500mA must also be calculated.

P

P

This number must be multiplied by the duty cycle at which it

would be operating, 12.5%.

P

P