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The C2M0025120D datasheet, on page 3 (see attached picture) shows the characteristic curves of the drain-source on-state resistance Rds as a function of temperature T (on the right) and drain current Id (on the left).

I want to create a mathematical analytical loss model that specifies Rds as a function of T AND Id. For this reason, I want to use these two characteristic curves.

However, I noticed something that makes my approach impossible. Rds(T) (on the right) was measured at Id = 50 A. If we assume a gate-source voltage Vgs = 20 V, we can read the value: Rds(T = 150 °C) = 41.1 mOhm at Id = 50 A. I have assumed that this value matches Rds(Id) (on the left) at 150 °C, Id = 50 A and Vgs = 20 V. But if we take a closer look, we can see that at Id = 50 A and 150 °C a much higher resistance is reached: Rds(Id = 50 A) = 47.5 mOhm at 150 °C. So, the curves can't help me, because for some reason Rds(T = 150 °C) at 50 A is not equal to Rds(Id = 50 A) at 150 °C.

  • Have I understood this fact of the characteristic curves correctly?
  • Which characteristic curves could help me?
  • What are sources that I can use, dealing with the modelling of Rds as a function of T AND I. I have read some papers that describe the modelling of Rds(T), but found none that deal with Rds(T,I) or at least Rds(I).

enter image description here

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  • \$\begingroup\$ Very complete models for transistors already exist. They are used in SPICE simulators, and the equations behind them are well documented. Most transistor manufacturers will freely provide the appropriate coefficients for the various transistors that they make. \$\endgroup\$ – Elliot Alderson May 2 '19 at 12:28
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I think you spotted a mistake from the manufacturer the two points should be equal.

I took an arbitrary different datasheet of the manufacturer . See the image below.

At Id=1A with Tj=150C , the resistance is 2 ohm in figure 5.

in figure 6, at Tj=150C the resistance is also approximately 2 ohm.

enter image description here

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  • \$\begingroup\$ Thank you very much! Unfortunately, I have not thought of a mistake of the manufacturer. That definitely solves my first problem. \$\endgroup\$ – Noah May 2 '19 at 12:11

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