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Tune
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First reply posted First like given First question asked

Hi

I have been looking at the datasheet for IMBF170R450M1

It has the following data:

Eoss@1000V=2.9uJ

When calculating the energy based on the output capacitance of 16pF@1000V I get 8uJ@1000V (0,5*16*10^-12*1000^2=8uJ) . So I would assume Eoss@1000V would be higher than 8uJ, since the output capacitance is voltage dependent and much higher at lower voltages.

Please explain (or correct the datasheet.)

 

----

The datasheet for IMBG120R030M1H makes more sense.

Eoss@800V=44uJ

When calculating the energy based on the output capacitance of 105pF@800V I get 33.4uJ@800V (0,5*105*10^-12*800^2=33.4uJ) . So the energy stated in the datasheet is larger due to the voltage dependency.

----

 

 

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AZIZ_HASSAN
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5 likes given 250 sign-ins 100 solutions authored

Hello,

Thanks for the explanation, the way you are calculating will not give the energy stored because, the capacitance is changing dynamically and to get the precise value, you need to integrate over the complete capacitance area from zero volt to 1000V then you will get close value. The way you are calculating will give you high value because of linearization of capacitance from zero to 1000V. But if you see the figure of capacitance w.r.t voltage, its not linear rather after some time, it gets almost sluggish constant so, under that curve duration, the area will be different which can be calculated precisely if you do integration by dividing the waveform in parts.

I hope you understood.

Thanks,

BR,

AZIZ

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AZIZ_HASSAN
Moderator
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5 likes given 250 sign-ins 100 solutions authored

Hi,

Thank you for posting at Infineon community.

Give me some time to clarify your doubt.

 

BR,

AZIZ

Hi,

We were analyzing your equation, but could not understand what are these value and is there any parameter missing?

(0,5*16*10^-12*1000^2=8uJ)

please clarify the values taken and the euqation.

Thanks,

BR,

AZIZ

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Hi,

Its energy stored in a capacitor. The formula is: 0,5*C*V^2

https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor

 

Coss@1000V=16pF is used. Found in table 5 in datasheet.

https://www.infineon.com/dgdl/Infineon-IMBF170R450M1-DataSheet-v02_03-EN.pdf?fileId=5546d462712ef9b7...

 

You can see the output capacitance (Coss) is changing with voltage. See figure 9.

So the energy stored at 1000V is actually higher than the 8uJ calculated. So this is normally stated in the datasheet, but 2.9uJ as you have in table 5, does not seem to fit with the capacitance data.

 

Regards,

Tune

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AZIZ_HASSAN
Moderator
Moderator
Moderator
5 likes given 250 sign-ins 100 solutions authored

Hello,

Thanks for the explanation, the way you are calculating will not give the energy stored because, the capacitance is changing dynamically and to get the precise value, you need to integrate over the complete capacitance area from zero volt to 1000V then you will get close value. The way you are calculating will give you high value because of linearization of capacitance from zero to 1000V. But if you see the figure of capacitance w.r.t voltage, its not linear rather after some time, it gets almost sluggish constant so, under that curve duration, the area will be different which can be calculated precisely if you do integration by dividing the waveform in parts.

I hope you understood.

Thanks,

BR,

AZIZ

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