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Fusik
Level 1
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5 replies posted 5 sign-ins First reply posted

Hi
In 2008, we solved the project of a BUCK and BOOST converter with a switching frequency of 10kHz and a module current of 100A to 200A. FD 800 R 17 KF6C B2 was used as the switching module. At the moment, we would need to make similar converters, and I found an available module FD800R17HP4-K_B2.
The total power dissipation of both modules is similar, but I was surprised by the RthCH parameter, where the newer module has 30.0K/kW and the old module had 8K/kW. The RthJC parameter for the transistor and the diode for both modules have similar values. My concern is whether the new module can cool. Why is the thermal resistance RthCH 4x higher in the new type of module.
well thank you

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Guru_Prasad
Moderator
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250 replies posted 100 solutions authored 25 likes received

Dear @Fusik 

The new device base plate technology is different 

The question is whether with the new FD800R17HP4-K_B2 module it will probably be a problem to cool the 2000W loss? The losses should be distributed in the ratio of 300W on the diode and 1700W on the transistor.

The 2000W losses is not problem. if you maintain junction temperature with in the limits . if it exceeding please work on heatsink Rth side.

At what temperature will it be necessary to maintain the temperature of the heatsink?

Ans)Here Heat sink temperature is not a main task. the main task is  when you change to new device The device junction  temperature is maintained with in he limits are not. 

Note: Please consider the your application temperature limits for junction temperature calculation  based on this try to calculate the junction temperature. in this process you may get heatsink temperature. this heat sink temperature also may create some problem to near by components.  to reduce the heat  please try with  below options   

1) change  heat sink or cooling system

2)optimize gate resistance.

3)proper Mechanical mounting.

  

Thank you

Guru

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Guru_Prasad
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250 replies posted 100 solutions authored 25 likes received

Dear @Fusik 

Greetings from Infineon.

Let me check the reason and feasible cooling mechanism. Please give us some time

Thank you

Guru 

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Guru_Prasad
Moderator
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250 replies posted 100 solutions authored 25 likes received

Dear @Fusik 

Greetings!

I looked at the old datasheet for the FD800R17KF6C_B2, and the RthCH indicated 25 K/KW.

You may notice a slight increase in RthCH because the device and base plate technologies have changed for the new (FD800R17HP4-K_B2) version.

As a result, the power handling capability has decreased to 5.2 kW from 6.25 kW.

Please see the picture below.

Guru_Prasad_0-1684991532930.png

Which document are you referring to, please?

Thank you

Guru

 

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Fusik
Level 1
Level 1
5 replies posted 5 sign-ins First reply posted

 

Greetings!
In the attachment, I am sending you my datasheet from which I started.
I am also sending a picture of the parameter "thermal resistance, case to heatsink" from my datasheet.
Thank you
Peter Fusik

Fusik_1-1684993689319.png

 

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Guru_Prasad
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250 replies posted 100 solutions authored 25 likes received

Dear @Fusik 

The parameter which you are referring is per module that is not IGBT RthJC. below i have attached exact RthJC for IGBT and the last decimal was missed.

Guru_Prasad_0-1684997335493.png

You are refereeing old datasheet which was made by EUPEC. kindly refer attached datasheet.

Thank you

Guru

 

 

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I realized the different manufacturer only now.
Thank you for the attached catalog sheet.
I recalculated the decimal place by looking at the different units of K/W and K/kW in the case of the datasheet from the manufacturer EUPEC and INFINEON.
The thermal resistance between the junction and the case did not surprise me because the value was approximately the same, but I was surprised by the thermal resistance between the case and the heatsink where the difference was more fundamental. It was a module that we used in the device from the manufacturer EUPEC and a new module that we want to use from the manufacturer INFINEON.
The question is whether with the new FD800R17HP4-K_B2 module it will probably be a problem to cool the 2000W loss? The losses should be distributed in the ratio of 300W on the diode and 1700W on the transistor. At what temperature will it be necessary to maintain the temperature of the heatsink?
Thank you
Peter Fusik

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Guru_Prasad
Moderator
Moderator
Moderator
250 replies posted 100 solutions authored 25 likes received

Dear @Fusik 

The new device base plate technology is different 

The question is whether with the new FD800R17HP4-K_B2 module it will probably be a problem to cool the 2000W loss? The losses should be distributed in the ratio of 300W on the diode and 1700W on the transistor.

The 2000W losses is not problem. if you maintain junction temperature with in the limits . if it exceeding please work on heatsink Rth side.

At what temperature will it be necessary to maintain the temperature of the heatsink?

Ans)Here Heat sink temperature is not a main task. the main task is  when you change to new device The device junction  temperature is maintained with in he limits are not. 

Note: Please consider the your application temperature limits for junction temperature calculation  based on this try to calculate the junction temperature. in this process you may get heatsink temperature. this heat sink temperature also may create some problem to near by components.  to reduce the heat  please try with  below options   

1) change  heat sink or cooling system

2)optimize gate resistance.

3)proper Mechanical mounting.

  

Thank you

Guru

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Fusik
Level 1
Level 1
5 replies posted 5 sign-ins First reply posted

Thank you for your reply, I will have to do a burn-in test and check the junction temperature.
For the flood module, I based the design on the following conditions:
- calculated temperature losses on the 2000W module
Considering the majority of losses on the IGBT module and simplifying the calculation, I calculated all losses of 2000W on the IGBT
- thermal resistance IGBT, junction to case 0.02W/K
- thermal resistance IGBT, case to heatsink 0.008W/K
- 2000W * (0.02W/K + 0.008W/K) = 56K temperature difference between junction and heatsink
- on the heatsink, the temperature protection of the device was set to 80°C, when the device will be blocked in case of overheating.
When I take the critical heatsink temperature of 80°C, I add to it the junction - heatsink temperature difference, so the junction temperature comes out to be 136°C, which is lower than the maximum junction temperature.
When I recalculate it analogously for the new module, the junction temperature is 170°C, which is higher than the maximum junction temperature of 150°C.
From this it follows that I will have to adjust the cooling system and reduce the heatsink temperature protection value to avoid thermal destruction of the junction.
Thank you
Peter Fusik

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Guru_Prasad
Moderator
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250 replies posted 100 solutions authored 25 likes received

Dear @Fusik 

The initial design can be done with mathematical calculations and simulations.

In your case the temperatures are increasing so try to optimize your system by calculating power loss (2000W)in the IPOSIM tool. The link is given below. 

https://iposim.infineon.com/application/ 

The cooling mechanism is also one of the main parameters for the cooldown of the device. this cooling mechanism is a secondary task 

Thank you

Guru

 

 

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Fusik
Level 1
Level 1
5 replies posted 5 sign-ins First reply posted

Thanks for the link, it's a great design tool.
Thank you
Peter Fusik

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