IDT82V3280PFG IDT, Integrated Device Technology Inc, IDT82V3280PFG Datasheet - Page 149
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IDT82V3280PFG
Manufacturer Part Number
IDT82V3280PFG
Description
IC PLL WAN SE STRATUM 2 100-TQFP
Manufacturer
IDT, Integrated Device Technology Inc
Type
PLL Clock Generatorr
Datasheet
1.IDT82V3280DQG8.pdf
(171 pages)
Specifications of IDT82V3280PFG
Input
CMOS, LVDS, PECL, TTL
Output
CMOS, LVDS, PECL, TTL
Frequency - Max
622.08MHz
Voltage - Supply
3 V ~ 3.6 V
Operating Temperature
-40°C ~ 85°C
Mounting Type
Surface Mount
Package / Case
100-TQFP, 100-VQFP
Frequency-max
622.08MHz
Function
Wan PLL
Operating Temperature (max)
85C
Operating Temperature (min)
-40C
Package Type
TQFP
Pin Count
100
Mounting
Surface Mount
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Other names
82V3280PFG
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
IDT82V3280PFG
Manufacturer:
IDT, Integrated Device Technology Inc
Quantity:
10 000
Company:
Part Number:
IDT82V3280PFG8
Manufacturer:
IDT, Integrated Device Technology Inc
Quantity:
10 000
Table 43: Power Consumption and Maximum Junction Temperature
8
+85°C. To ensure the functionality and reliability of the device, the maxi-
mum junction temperature T
applications, the device will consume more power and a thermal solution
should be provided to ensure the junction temperature T
exceed the T
8.1
geographical center of the chip where the device's electrical circuits are.
It can be calculated as follows:
be used. The θ
in the loads.
ments.
8.2
Table 44: Thermal Data
Thermal Management
TQFP/EQG100
IDT82V3280
*note: Simulated with 3 x 3 array of thermal vias.
TQFP/PN100
TQFP/EQG100
TQFP/EQG100
The device operates over the industry temperature range -40°C ~
Junction temperature T
Where:
In order to calculate junction temperature, an appropriate θ
Power consumption is the core power excluding the power dissipated
Package
Assume:
TQFP/PN100
Package
Equation 1: T
θ
T
T
P = Device Power Consumption
T
θ
s)
P = 1.9W
A
A
JA
j
JA
= Junction Temperature
= 85°C
= Ambient Temperature
= 18.9°C/W (TQFP/EQG100 Soldered & when airfow rate is 0 m/
= Junction-to-Ambient Thermal Resistance of the Package
THERMAL MANAGEMENT
JUNCTION TEMPERATURE
EXAMPLE OF JUNCTION TEMPERATURE
CALCULATION
jmax
Table 43
JA
Consumption (W)
.
is shown in
Pin Count Thermal Pad
j
Power
= T
1.9
1.9
100
100
100
provides power consumption in special environ-
A
+ P X
j
is the temperature of package typically at the
jmax
Table
Yes/Soldered*
θ
Yes/Exposed
JA
should not exceed 125°C. In some
Operating
44:
Voltage
No
(V)
3.6
3.6
T
A
θ
85
85
(°C)
JC
11.0
16.1
16.1
(°C/W)
Temperature (°C)
Maximum
Junction
j
125
125
does not
JA
θ
JB
must
34.2
34.2
1.3
(°C/W)
149
temperature of 125°C so no extra heat enhancement is required.
might exceed the maximum junction temperature of 125°C and an exter-
nal thermal solution such as a heatsink is required.
8.3
attached. θ
resistance, as the heat flowing from the die junction to ambient goes
through the package and the heatsink. θ
be selected to ensure the junction temperature does not exceed the
maximum junction temperature. According to Equation 1 and 2,
below or equal to 5.0°C/W is used in such operation environment, the
junction temperature will not exceed the maximum junction temperature.
The junction temperature T
The junction temperature of 120.9°C is below the maximum junction
In some operation environments, the calculated junction temperature
A heatsink is expanding the surface area of the device to which it is
Where:
θ
Assume:
θ
That is, if a heatsink and heatsink attachment whose
θ
CH
CH
CH
35.8
39.3
18.9
+ θ
+ θ
0
T
Equation 2:
θ
θ
θ
Equation 3:
T
T
P = 1.9W
θ
θ
+ θ
A
j
JC
CH
HA
j
JC
CH
= T
= 125°C (T
HA
= 85°C
HA
HA
JA
+
= Junction-to-Case Thermal Resistance
= 16.1°C/W (TQFP/EQG100)
= Case-to-Heatsink Thermal Resistance
= Heatsink-to-Ambient Thermal Resistance
HEATSINK EVALUATION
A
θ
can be calculated as follows:
is now a combination of device case and heat-sink thermal
determines which heatsink and heatsink attachment can
can be calculated as follows:
+ P X
HA
36.2
31.1
14.6
= (125°C - 85°C ) / 1.9W - 16.1°C/W = 5.0°C/W
1
θ
θ
jmax
θ
JA
CH
JA
+
=
)
= 85°C + 1.9W X 18.9°C/W = 120.9°C
θ
θ
θ
JA
HA
JC
(°C/W) Air Flow in m/s
34.3
29.5
13.5
+
= (T
j
2
θ
can be calculated as follows:
CH
j
- T
+
A
θ
) / P -
HA
JA
33.5
28.6
12.9
3
θ
can be calculated as follows:
JC
December 9, 2008
32.9
27.9
12.6
4
θ
CH
WAN PLL
+
32.6
27.4
12.4
θ
5
HA
is