MIC5190
Micrel
The V
IN
(min) to V
OUT
 ratio and current will determine the
maximum R
DSON
 required. For example, for a 1.8V (?%) to
1.5V conversion at 5A of load current, dropout voltage can be
calculated as follows (using V
IN
(min)):
R
V    V
I
R
1 71V  1 5V
5A
R
m
DSON
IN
OUT
OUT
DSON
DSON
=

(
)
=

(
)
=
.
.
42  &
Running the N-Channel in dropout will seriously affect tran-
sient response and PSRR (power supply ripple rejection). For
this reason, we want to select a MOSFET that has lower than
42m& for our example application.
Size is another important consideration. Most importantly,
the design must be able to handle the amount of power being
dissipated.
The amount of power dissipated can be calculated as follows
(using V
IN
(max)):
P
D
 = (V
IN
  V
OUT
) ?I
OUT
P
D
 = (1.89V  1.5V) ?5A
P
D
 = 1.95W
Now that we know the amount of power we will be dissipating,
we will need to know the maximum ambient air temperature.
For our case were going to assume a maximum of 65癈
ambient temperature. Different MOSFETs have different
maximum operating junction temperatures. Most MOSFETs
are rated to 150癈, while others are rated as high as 175癈.
In this case, were going to limit our maximum junction
temperature to 125癈. The MIC5190 has no internal thermal
protection for the MOSFET so it is important that the design
provides margin for the maximum junction temperature. Our
design will maintain better than 125癈 junction temperature
with 1.95W of power dissipation at an ambient temperature of
65癈. Our thermal resistance calculates as follows:
So our package must have a thermal resistance less than
31癈 /W. Table 1. shows a good approximation of power
dissipation and package recommendation.
Package
Power Dissipation
TSOP-6
<850mW
TSSOP-8
<950mW
TSSOP-8
<1W
PowerPAK"1212-8
<1.1W
SO-8
<1.125W
PowerPAK" SO-8 D-Pack
<1.4W
TO-220/TO-263 (D
2
Pack)
>1.4W
Table 1. Power Dissipation and
Package Recommendation
In our example, our power dissipation is greater than
1.4W, so well choose a TO-263 (D
2
Pack) N-Channel
MOSFET. ?/DIV>
JA
 is calculated as follows.
?/DIV>
JA
 = ?/DIV>
JC
 + ?/DIV>
CS
 + ?/DIV>
SA
Where ?/DIV>
JC
 is the junction-to-case resistance, ?/DIV>
CS
 is the
case-to-sink resistance and the ?/DIV>
SA
 is the sink-to-ambient
air resistance.
In the D
2
 package weve selected, the ?/DIV>
JC
 is 2癈/W. The
?/DIV>
CS
, assuming we are using the PCB as the heat sink, can
be approximated to 0.2癈/W. This allows us to calculate
the minimum ?/DIV>
SA
:
?/DIV>
SA
= ?/DIV>
JA
 ?/DIV>
CS
  ?/DIV>
JC
?/DIV>
SA
= 31癈/W  0.2癈/W  2癈/W
?/DIV>
SA
= 28.8癈/W
Referring to Application Hint 17, Designing PCB Heat
Sinks, the minimum amount of copper area for a D
2
Pack
at 28.8癈/W is 2750mm
2
 (or 0.426in
2
 ). The solid line
denotes convection heating only (2 oz. copper) and the
dotted line shows thermal resistance with 250LFM air-
flow. The copper area can be significantly reduced by
increasing airflow or by adding external heat sinks.
Figure 8. PC Board Heat Sink
Another important characteristic is the amount of gate
capacitance. Large gate capacitance can reduce tran-
sient performance by reducing the ability of the MIC5190
to slew the gate. It is recommended that the MOSFET
used has an input capacitance <10nF (C
ISS
).
?/DIV>
?/DIV>
?/DIV>
JA
J
J
D
JA
JA
T max   T ambient
P
125 C  65 C
1.95W
C  W
=
(   )

(
)
=
?nbsp    ?/DIV>
=   ?/DIV>
31   /
    PC Board Heat Sink
Thermal Resistance vs. Area
December 2005
9
M9999-120105
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