Heatsink to junction thermal resistance (RthJH) clarification on the DDB2U40N12W1RF_B11 datasheet

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Arturo
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Hi there,

I'm wondering if anyone can help me clarify the the RthJH parameter on the DDB2U40N12W1RF_B11 datasheet. This parameter is on Table 4: Characteristic Values.

Below is a snippet of the mentioned table.

Arturo_0-1648001618439.png

Am I correct to say that for example, when the loss on the module is 30W, the temperature difference between heatsink to junction is calculated as follows:

Temperature increase = (30W)*(0.96K/W)*4 = 115.2K

Note that in the above calculation I've multiplied by 4 because on Table 4 it says per diode and since there are 4 diodes in the module I've multiplied by 4. I'm wondering if this calculation is correct.

Thanks in Advance.

 

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1 Solution
Meghana
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50 likes received 100 solutions authored 10 likes given

Hello Arturo,

Rthj-h represents the thermal resistance between the device junction and the heatsink. While calculating the junction temperature of the device, we need to consider the individual losses of the device instead of overall module, as the losses might vary from device to device based on the switching frequency and duty cycle. 

For example : In module DDB2U40N12W1RF_B11  if one diode is dissipating 30 watts, then junction temperature is calculated as below,

∆T = Pd * Rthj-h =  30* 0.964 = 28.92K, where  ∆T = Tj (Junction temp) - Th(Heat sink temp)

If 30 watts is the module loss and all diodes are conducting for equal time, then temperature rise on each diode would be

∆T = Pd* Rthj-h = (30/4)*0.964 = 7.23K

 

Regards

Meghana R

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Meghana
Moderator
Moderator
Moderator
50 likes received 100 solutions authored 10 likes given

Hello Arturo,

Rthj-h represents the thermal resistance between the device junction and the heatsink. While calculating the junction temperature of the device, we need to consider the individual losses of the device instead of overall module, as the losses might vary from device to device based on the switching frequency and duty cycle. 

For example : In module DDB2U40N12W1RF_B11  if one diode is dissipating 30 watts, then junction temperature is calculated as below,

∆T = Pd * Rthj-h =  30* 0.964 = 28.92K, where  ∆T = Tj (Junction temp) - Th(Heat sink temp)

If 30 watts is the module loss and all diodes are conducting for equal time, then temperature rise on each diode would be

∆T = Pd* Rthj-h = (30/4)*0.964 = 7.23K

 

Regards

Meghana R

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