HCPL3150300E AVAGO TECH, HCPL3150300E Datasheet - Page 16

• Manufacturer Part Number: HCPL3150300E
• Manufacturer: AVAGO TECH

Figure 28a. Thermal Model.
The value of 4.25 mA for I
the previous equation was
obtained by derating the I
of 5 mA (which occurs at -40 C)
to I
Since P
than P
increased to reduce the HCPL-
3150 power dissipation.
P
For Qg = 500 nC, from Figure
27, a value of E
gives a Rg = 41
Thermal Model
(HCPL-3150)
The steady state thermal model
for the HCPL-3150 is shown in
Figure 28a. The thermal
resistance values given in this
model can be used to calculate
the temperatures at each node for
a given operating condition. As
shown by the model, all heat
generated flows through
which raises the case temperature
T
depends on the conditions of the
LC
O(SWITCHING MAX)
C
E
= 391°C/W
accordingly. The value of
CC
= P
= 154 mW - 85 mW
= 69 mW
SW(MAX)
max at 90 C (see Figure 7).
O(MAX)
O(MAX)
O
T
JE
for this case is greater
= –––––––––––––––
= ––––––– = 3.45 J
LD
, Rg must be
- P
69 mW
= 439°C/W
20 kHz
P
T
SW
O(SWITCHINGMAX)
C
O(BIAS)
= 3.45 J
.
CA
T
A
= 83°C/W*
T
f
DC
JD
= 119°C/W
CC
CA
CC
max
in
CA
board design and is, therefore,
determined by the designer. The
value of
obtained from thermal measure-
ments using a 2.5 x 2.5 inch PC
board, with small traces (no
ground plane), a single HCPL-
3150 soldered into the center of
the board and still air. The
absolute maximum power
dissipation derating specifications
assume a
From the thermal mode in Figure
28a the LED and detector IC
junction temperatures can be
expressed as:
T
+ P
T
+ P
Inserting the values for
T
T
T
T
JE
JD
DC
JE
JD
T
DC
LC
LD
CA
CA
JD
JE
C
D
+ P
+ P
= P
D
= P
= P
= P
shown in Figure 28 gives:
will depend on the board design and the placement of the part.
= LED junction temperature
= detector IC junction temperature
= case temperature measured at the center of the package bottom
= LED-to-case thermal resistance
= LED-to-detector thermal resistance
= detector-to-case thermal resistance
= case-to-ambient thermal resistance
(
(
E
E
D
D
DC
E
E
–––––––––––––––– +
(
(49 C/W +
(104 C/W +
||(
(230 C/W +
(49 C/W +
(
LC
CA
––––––––––––––– +
CA
LC
+
LC
= 83 C/W was
value of 83 C/W.
LD
||(
LC
+
LC
DC
+
LD
DC
+
DC
LC
+
DC
+
) +
CA
CA
LD
CA
CA
DC
) + T
)
LD
LC
) + T
)
) +
CA
and
) + T
A
CA
A
CA
CA
)
)
)
+ T
A
A
16
For example, given P
P
= 83 C/W:
T
T
T
125 C based on the board layout
and part placement (
to the application.
JE
JD
O
JE
= 250 mW, T
= P
= P
= 45 mW 313 C/W + 250 m
= 45 mW 132C/W + 250 m
and T
E
132 C/W + 70 C = 117 C
E
187 C/W + 70 C = 123 C
313 C/W + P
132 C/W + P
JD
should be limited to
A
= 70 C and
D
D
E
CA
132 C/W + T
187 C/W + T
= 45 mW,
) specific
W
W
CA
A
A