Replacing TOLL with ThinPAK

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NXTY_Fujioka
Level 4
Level 4
Distributor - NEXTY (Japan)
50 sign-ins 25 replies posted First like received
I have a question about the thermal advantage of the package.

How much difference is there in heat generation when comparing TOLL and ThinPAK8x8?
I would like to know if heat generation is not a problem if I replace TOLL with ThinPAK8x8.
Do you have any good references?

Also, is there any good way to actually measure or calculate Tj for the following 4 packages?

-TOLL
-DPAK
-ThinPAK5×6
-ThinPAK8x8
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1 Solution
Pablo_EG
Moderator
Moderator
Moderator
First question asked 250 sign-ins 250 replies posted

Hello Yuki,

Thank you for posting on Infineon Forums.

You can see the performance difference between TOLL and ThinPak8x8 in the following app note:
https://www.infineon.com/dgdl/Infineon-Package_MOSFET_Thermal_performance_of_surface_mount_semicondu...

footprint.png
Here different patterns are used to find the best heat transfer method.
We can see that if we apply the best pattern for both TOLL and ThinPak8x8, the TOLL device can achieve about 3.3 K/W Rth and the ThinPak8x8 achieves about 5.8 K/W Rth.
Therefore, there is a difference of 75% between them.
This Rth shows the thermal resistance between the device and the heatsink.
In your design, you should take into account the junction to case, case to the heatsink (App Note from above), and heatsink to air resistance.
Usually, TOLL and ThinPak8x8 Rthjc is the same for the same device.

In order to determine if ThinPak8x8 will not be a problem, you can use the following equation:
Tj = Tair + (Rthjc+Rthch+Rthha)*Ploss (Only applicable in steady state condition)

If Tj does not exceed the maximum limit of the device (with 80% derating), then it is safe to use.


Regarding your second question, please refer to the following thread:
https://www.infineonforums.com/threads/13258-IRF7759L2PbF-Tj-cool-down-time

You can either make a simulation or calculate it with:
Tj = Tcase + Rthjc*Ploss. (Only applicable in steady-state condition)

If you have any doubt, please let me know.

Best regards,
Pablo

View solution in original post

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1 Reply
Pablo_EG
Moderator
Moderator
Moderator
First question asked 250 sign-ins 250 replies posted

Hello Yuki,

Thank you for posting on Infineon Forums.

You can see the performance difference between TOLL and ThinPak8x8 in the following app note:
https://www.infineon.com/dgdl/Infineon-Package_MOSFET_Thermal_performance_of_surface_mount_semicondu...

footprint.png
Here different patterns are used to find the best heat transfer method.
We can see that if we apply the best pattern for both TOLL and ThinPak8x8, the TOLL device can achieve about 3.3 K/W Rth and the ThinPak8x8 achieves about 5.8 K/W Rth.
Therefore, there is a difference of 75% between them.
This Rth shows the thermal resistance between the device and the heatsink.
In your design, you should take into account the junction to case, case to the heatsink (App Note from above), and heatsink to air resistance.
Usually, TOLL and ThinPak8x8 Rthjc is the same for the same device.

In order to determine if ThinPak8x8 will not be a problem, you can use the following equation:
Tj = Tair + (Rthjc+Rthch+Rthha)*Ploss (Only applicable in steady state condition)

If Tj does not exceed the maximum limit of the device (with 80% derating), then it is safe to use.


Regarding your second question, please refer to the following thread:
https://www.infineonforums.com/threads/13258-IRF7759L2PbF-Tj-cool-down-time

You can either make a simulation or calculate it with:
Tj = Tcase + Rthjc*Ploss. (Only applicable in steady-state condition)

If you have any doubt, please let me know.

Best regards,
Pablo

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