Datasheet

 2018 Microchip Technology Inc. DS20006021A-page 17
MIC5219
EQUATION 4-10:
The power dissipation at 500 mA must also be
calculated.
EQUATION 4-11:
This number must be multiplied by the duty cycle at
which it would be operating: 12.5%.
EQUATION 4-12:
The total power dissipation of the device under these
conditions is the sum of the two power dissipation
figures.
EQUATION 4-13:
The total power dissipation of the regulator is less than
the maximum power dissipation of the SOT23-5
package at room temperature, on a minimum footprint
board and therefore would operate properly.
Multilayer boards with a ground plane, wide traces near
the pads, and large supply-bus lines will have better
thermal conductivity.
For additional heat sink characteristics, please refer to
Application Hint 17, Designing P.C. Board Heat Sinks.
For a full discussion of heat sinking and thermal effects
on voltage regulators, refer to “Regulator Thermals”
section of Microchip’s Designing with Low-Dropout
Voltage Regulators handbook.
4.8 Fixed Regulator Circuits
Figure 4-5 shows a basic MIC5219-x.xYMX
fixed-voltage regulator circuit. A 1μF minimum output
capacitor is required for basic fixed-voltage
applications.
FIGURE 4-5: Low-Noise Fixed Voltage
Regulator.
Figure 4-6 includes the optional 470 pF noise bypass
capacitor between BYP and GND to reduce output
noise. Note that the minimum value of C
OUT
must be
increased when the bypass capacitor is used.
FIGURE 4-6: Ultra-Low Noise Fixed
Voltage Regulator.
4.9 Adjustable Regulator Circuits
Figure 4-7 shows the basic circuit for the MIC5219
adjustable regulator. The output voltage is configured
by selecting values for R1 and R2 using the following
formula.
EQUATION 4-14:
This equation is correct due to the configuration of the
bandgap reference. The bandgap voltage is relative to
the output, as seen in the block diagram. Traditional
regulators normally have the reference voltage relative
to ground and have a different V
OUT
equation.
P
D
50mA 0.875 88.25mW=
P
D
50mA 77.22mW=
P
D
500mA 5V 3.3V–500mA 5V+20mA=
P
D
500mA 950mW=
P
D
0.125 950mW=
P
D
118.75mW=
P
DTOTAL
P
D
50mA P
D
+ 500mA=
P
DTOTAL
77.22mW 118.75mW+=
P
DTOTAL
196mW=
MIC5219-x.x
IN OUT
GND
1μF
V
IN
V
OUT
EN BYP
MIC5219-x.x
IN OUT
GND
470pF
V
IN
EN BYP
2.2μF
V
OUT
V
OUT
1.242V
R2
R1
-------
1+


=