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Hello everyone, I hope that you can help me. I guess I have a problem to understand something. For my simulation in PLECS I want to define the "thermal Description" for my selected IMZ120R045M1 MOSFET.
For understanding and more I use the Application Note, V1.2, Jan. 2002 "How to Select the Right CoolMOS and its Power Handling Capability" and on the other hand an existing PLECS thermal description model (IMBG120R045M1H) for comparison.
PLECS accesses the LookUp Tables to calculate the power losses. In order to calculate the turn-on losses for the IMBG120R045M1H, custom tables were defined. Which are used for to calculate (for example) the turn on losses of the MOSFET.
To calculate the turn-on losses, the following lookup table "(lookup('e_i_on', 16)^3)*1e-6,0)" is used at the end of the formular. Since I'm just getting started with PLECS, I wanted to ask why this is being done? Isn't it enough to define E_on(v,i,T,Rgon,Rgoff) with the respective tables (without with the explained divisor)?
Thank you very much.
Solved! Go to Solution.
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Hi,
I have re-written the formula for better understanding.
Regards,
Rishabh
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Hi,
The turn on & Turn-off energy will vary with respect to the system parameters i.e. voltage(V), Current (I), Junction temp (T), Turn-off resistance (Rgoff) and Turn-on resistance (Rgon). You can feed your system parameters in variable tab. There are custom tables in the model f(x) v/s x, which is E_on(Rgon) v/s Rgon, E_on(I) v/s I, E_on(V) v/s V and E_on(T) v/s T. When you select formula as the computation method, the tool will take all the Eon as per your system parameters and multiply it and then will divide it by cube of Eon(@rated conditions, 25deg).
If you see the formula in numerator Eon is multiplied 4 times and in denominator cube of Eon is there to eventually get Eon at specified parameters.
Do revert for further clarification.
Regards,
Rishabh
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Hello Rishabh, first of all thank you very much for your fast response.
However, I'm not sure if I understood you correctly. As you have already described, the turn-on losses are calculated by using the function E_on(v,i,T,Rgon,Rgoff). That's how I understood it.
Now here's the point I don't understand:
The divisor 'e_i_on,16' on the other hand refers to the nominal value at 25 degrees, I_D = 16 A. Correct? (see also in the datasheet of the IMBG120R045M1H FET)
Why is this value raised to the power of 3 and multiplied with '1e-6.0'. Maybe to convert µJ to J?
Best regard
Michael
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Hi,
I have re-written the formula for better understanding.
Regards,
Rishabh
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Oh well. How obvious... 😁!
Sometimes you cannot see the wood for the trees. Very thank you!
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Hi,
If the issue is resolved, can I close the ticket?
Regards,
Rishabh
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Just one last question: Where I can find the voltage factor behaviour? If I look at the lookup-table of the FET, which I used to prefer or look at ones of some other MOSFETS, there are values entered that I can not find in the data sheet. Or are these calculated or assumed using the data sheet? On example of the voltage factor behaviour shown in the picture, which is in the attachment.
Am I correct in saying that this is an assumed curve that was calculated using I_DS and R_DS?
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Hi,
Actually tests are done on the switch with different Voltages keeping other parameter constant , data points are put in the custom table and linearized over the x axis (V) .
Regards,
Rishabh
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Okay! I get it. I just did something stupid and had a stupid thought error. Well...very thank you for your help/time. You can close this ticket.