TC2054 Microchip Technology, TC2054 Datasheet - Page 5

• Manufacturer Part Number: TC2054
• Description: 50mA / 100mA / and 150mA CMOS LDOs with Shutdown and Error Output
• Manufacturer: Microchip Technology

4.0
4.1
The amount of power the regulator dissipates is prima-
rily a function of input and output voltage, and output
current.
The following equation is used to calculate worst case
power dissipation:
EQUATION 4-1:
The maximum allowable power dissipation (Equation
4-2) is a function of the maximum ambient temperature
(T
°C) and the thermal resistance from junction-to-air
(θ
approximately 220°C/Watt when mounted on a typical
two layer FR4 dielectric copper clad PC board.
EQUATION 4-2:
©
P
Where:
Where all terms are previously defined
P
V
V
I
JA
A MAX
2002 Microchip Technology Inc.
D
LOAD MAX
D
IN MAX
OUT MIN
). The 5-Pin SOT-23A package has a θ
≈ (V
), the maximum allowable die temperature (125
IN
THERMAL CONSIDERATIONS
Power Dissipation
– V
=
=
=
=
OUT MIN
P
Worst case actual power dissipation
Maximum voltage on V
Minimum regulator output voltage
Maximum output (load) current
D
M A X
)I
=
LOAD MAX
T
---------------------------------- -
J
M AX
θ
–
JA
T
A
M A X
IN
JA
of
Equation 4-1 can be used in conjunction with Equation
4-2 to ensure regulator thermal operation is within lim-
its. For example:
Maximum allowable power dissipation:
In this example, the TC2054 dissipates a maximum of
only 20.7mW; far below the allowable limit of 318mW.
In a similar manner, Equation 4-1 and Equation 4-2 can
be used to calculate maximum current and/or input
voltage limits.
4.2
The primary path of heat conduction out of the package
is via the package leads. Therefore, layouts having a
ground plane, wide traces at the pads, and wide power
supply bus lines combine to lower θ
increase the maximum allowable power dissipation
limit.
Given:
Find
:
Actual power dissipation:
V
V
I
T
1. Actual power dissapation
2. Maximum allowable dissapation
P
TC2054/2055/2186
LOAD MAX
A MAX
IN MAX
OUT MIN
D
Layout Considerations
P
D
= 3.0V ±5%
= 2.7V – 2.5%
= 40mA
= 55°C
≈ (V
= [(3.0 x 1.05) – (2.7 x .975)]40 x 10
= 20.7mW
M A X
=
= 318mW
IN MAX
(
-------------------------- -
125 55
(
------------------------------------- -
T
220
J
M AX
–
– V
θ
–
JA
OUT MIN
T
)
A
M A X
JA
DS21663B-page 5
)
and, therefore,
)I
LOAD MAX
–3